# Symbolic Integrals¶

The integrals module in SymPy implements methods to calculate definite and indefinite integrals of expressions.

Principal method in this module is integrate()

• integrate(f, x) returns the indefinite integral $$\int f\,dx$$
• integrate(f, (x, a, b)) returns the definite integral $$\int_{a}^{b} f\,dx$$

## Examples¶

SymPy can integrate a vast array of functions. It can integrate polynomial functions:

>>> from sympy import *
>>> init_printing(use_unicode=False, wrap_line=False, no_global=True)
>>> x = Symbol('x')
>>> integrate(x**2 + x + 1, x)
3    2
x    x
-- + -- + x
3    2


Rational functions:

>>> integrate(x/(x**2+2*x+1), x)
1
log(x + 1) + -----
x + 1


Exponential-polynomial functions. These multiplicative combinations of polynomials and the functions exp, cos and sin can be integrated by hand using repeated integration by parts, which is an extremely tedious process. Happily, SymPy will deal with these integrals.

>>> integrate(x**2 * exp(x) * cos(x), x)
2  x           2  x                         x           x
x *e *sin(x)   x *e *cos(x)      x          e *sin(x)   e *cos(x)
------------ + ------------ - x*e *sin(x) + --------- - ---------
2              2                           2           2


even a few nonelementary integrals (in particular, some integrals involving the error function) can be evaluated:

>>> integrate(exp(-x**2)*erf(x), x)
____    2
\/ pi *erf (x)
--------------
4


## Integral Transforms¶

SymPy has special support for definite integrals, and integral transforms.

sympy.integrals.transforms.mellin_transform(f, x, s, **hints)[source]

Compute the Mellin transform $$F(s)$$ of $$f(x)$$,

$F(s) = \int_0^\infty x^{s-1} f(x) \mathrm{d}x.$
For all “sensible” functions, this converges absolutely in a strip
$$a < \operatorname{Re}(s) < b$$.

The Mellin transform is related via change of variables to the Fourier transform, and also to the (bilateral) Laplace transform.

This function returns (F, (a, b), cond) where F is the Mellin transform of f, (a, b) is the fundamental strip (as above), and cond are auxiliary convergence conditions.

If the integral cannot be computed in closed form, this function returns an unevaluated MellinTransform object.

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). If noconds=False, then only $$F$$ will be returned (i.e. not cond, and also not the strip (a, b)).

>>> from sympy.integrals.transforms import mellin_transform
>>> from sympy import exp
>>> from sympy.abc import x, s
>>> mellin_transform(exp(-x), x, s)
(gamma(s), (0, oo), True)

sympy.integrals.transforms.inverse_mellin_transform(F, s, x, strip, **hints)[source]

Compute the inverse Mellin transform of $$F(s)$$ over the fundamental strip given by strip=(a, b).

This can be defined as

$f(x) = \int_{c - i\infty}^{c + i\infty} x^{-s} F(s) \mathrm{d}s,$

for any $$c$$ in the fundamental strip. Under certain regularity conditions on $$F$$ and/or $$f$$, this recovers $$f$$ from its Mellin transform $$F$$ (and vice versa), for positive real $$x$$.

One of $$a$$ or $$b$$ may be passed as None; a suitable $$c$$ will be inferred.

If the integral cannot be computed in closed form, this function returns an unevaluated InverseMellinTransform object.

Note that this function will assume x to be positive and real, regardless of the sympy assumptions!

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit().

>>> from sympy.integrals.transforms import inverse_mellin_transform
>>> from sympy import oo, gamma
>>> from sympy.abc import x, s
>>> inverse_mellin_transform(gamma(s), s, x, (0, oo))
exp(-x)


The fundamental strip matters:

>>> f = 1/(s**2 - 1)
>>> inverse_mellin_transform(f, s, x, (-oo, -1))
(x/2 - 1/(2*x))*Heaviside(x - 1)
>>> inverse_mellin_transform(f, s, x, (-1, 1))
-x*Heaviside(-x + 1)/2 - Heaviside(x - 1)/(2*x)
>>> inverse_mellin_transform(f, s, x, (1, oo))
(-x/2 + 1/(2*x))*Heaviside(-x + 1)

sympy.integrals.transforms.laplace_transform(f, t, s, **hints)[source]

Compute the Laplace Transform $$F(s)$$ of $$f(t)$$,

$F(s) = \int_0^\infty e^{-st} f(t) \mathrm{d}t.$

For all “sensible” functions, this converges absolutely in a half plane $$a < \operatorname{Re}(s)$$.

This function returns (F, a, cond) where F is the Laplace transform of f, $$\operatorname{Re}(s) > a$$ is the half-plane of convergence, and cond are auxiliary convergence conditions.

If the integral cannot be computed in closed form, this function returns an unevaluated LaplaceTransform object.

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). If noconds=True, only $$F$$ will be returned (i.e. not cond, and also not the plane a).

>>> from sympy.integrals import laplace_transform
>>> from sympy.abc import t, s, a
>>> laplace_transform(t**a, t, s)
(s**(-a)*gamma(a + 1)/s, 0, -re(a) < 1)

sympy.integrals.transforms.inverse_laplace_transform(F, s, t, plane=None, **hints)[source]

Compute the inverse Laplace transform of $$F(s)$$, defined as

$f(t) = \int_{c-i\infty}^{c+i\infty} e^{st} F(s) \mathrm{d}s,$

for $$c$$ so large that $$F(s)$$ has no singularites in the half-plane $$\operatorname{Re}(s) > c-\epsilon$$.

The plane can be specified by argument plane, but will be inferred if passed as None.

Under certain regularity conditions, this recovers $$f(t)$$ from its Laplace Transform $$F(s)$$, for non-negative $$t$$, and vice versa.

If the integral cannot be computed in closed form, this function returns an unevaluated InverseLaplaceTransform object.

Note that this function will always assume $$t$$ to be real, regardless of the sympy assumption on $$t$$.

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit().

>>> from sympy.integrals.transforms import inverse_laplace_transform
>>> from sympy import exp, Symbol
>>> from sympy.abc import s, t
>>> a = Symbol('a', positive=True)
>>> inverse_laplace_transform(exp(-a*s)/s, s, t)
Heaviside(-a + t)

sympy.integrals.transforms.fourier_transform(f, x, k, **hints)[source]

Compute the unitary, ordinary-frequency Fourier transform of $$f$$, defined as

$F(k) = \int_{-\infty}^\infty f(x) e^{-2\pi i x k} \mathrm{d} x.$

If the transform cannot be computed in closed form, this function returns an unevaluated FourierTransform object.

For other Fourier transform conventions, see the function sympy.integrals.transforms._fourier_transform().

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). Note that for this transform, by default noconds=True.

>>> from sympy import fourier_transform, exp
>>> from sympy.abc import x, k
>>> fourier_transform(exp(-x**2), x, k)
sqrt(pi)*exp(-pi**2*k**2)
>>> fourier_transform(exp(-x**2), x, k, noconds=False)
(sqrt(pi)*exp(-pi**2*k**2), True)

sympy.integrals.transforms.inverse_fourier_transform(F, k, x, **hints)[source]

Compute the unitary, ordinary-frequency inverse Fourier transform of $$F$$, defined as

$f(x) = \int_{-\infty}^\infty F(k) e^{2\pi i x k} \mathrm{d} k.$

If the transform cannot be computed in closed form, this function returns an unevaluated InverseFourierTransform object.

For other Fourier transform conventions, see the function sympy.integrals.transforms._fourier_transform().

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). Note that for this transform, by default noconds=True.

>>> from sympy import inverse_fourier_transform, exp, sqrt, pi
>>> from sympy.abc import x, k
>>> inverse_fourier_transform(sqrt(pi)*exp(-(pi*k)**2), k, x)
exp(-x**2)
>>> inverse_fourier_transform(sqrt(pi)*exp(-(pi*k)**2), k, x, noconds=False)
(exp(-x**2), True)

sympy.integrals.transforms.sine_transform(f, x, k, **hints)[source]

Compute the unitary, ordinary-frequency sine transform of $$f$$, defined as

$F(k) = \sqrt{\frac{2}{\pi}} \int_{0}^\infty f(x) \sin(2\pi x k) \mathrm{d} x.$

If the transform cannot be computed in closed form, this function returns an unevaluated SineTransform object.

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). Note that for this transform, by default noconds=True.

>>> from sympy import sine_transform, exp
>>> from sympy.abc import x, k, a
>>> sine_transform(x*exp(-a*x**2), x, k)
sqrt(2)*k*exp(-k**2/(4*a))/(4*a**(3/2))
>>> sine_transform(x**(-a), x, k)
2**(-a + 1/2)*k**(a - 1)*gamma(-a/2 + 1)/gamma(a/2 + 1/2)

sympy.integrals.transforms.inverse_sine_transform(F, k, x, **hints)[source]

Compute the unitary, ordinary-frequency inverse sine transform of $$F$$, defined as

$f(x) = \sqrt{\frac{2}{\pi}} \int_{0}^\infty F(k) \sin(2\pi x k) \mathrm{d} k.$

If the transform cannot be computed in closed form, this function returns an unevaluated InverseSineTransform object.

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). Note that for this transform, by default noconds=True.

>>> from sympy import inverse_sine_transform, exp, sqrt, gamma, pi
>>> from sympy.abc import x, k, a
>>> inverse_sine_transform(2**((1-2*a)/2)*k**(a - 1)*
...     gamma(-a/2 + 1)/gamma((a+1)/2), k, x)
x**(-a)
>>> inverse_sine_transform(sqrt(2)*k*exp(-k**2/(4*a))/(4*sqrt(a)**3), k, x)
x*exp(-a*x**2)

sympy.integrals.transforms.cosine_transform(f, x, k, **hints)[source]

Compute the unitary, ordinary-frequency cosine transform of $$f$$, defined as

$F(k) = \sqrt{\frac{2}{\pi}} \int_{0}^\infty f(x) \cos(2\pi x k) \mathrm{d} x.$

If the transform cannot be computed in closed form, this function returns an unevaluated CosineTransform object.

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). Note that for this transform, by default noconds=True.

>>> from sympy import cosine_transform, exp, sqrt, cos
>>> from sympy.abc import x, k, a
>>> cosine_transform(exp(-a*x), x, k)
sqrt(2)*a/(sqrt(pi)*(a**2 + k**2))
>>> cosine_transform(exp(-a*sqrt(x))*cos(a*sqrt(x)), x, k)
a*exp(-a**2/(2*k))/(2*k**(3/2))

sympy.integrals.transforms.inverse_cosine_transform(F, k, x, **hints)[source]

Compute the unitary, ordinary-frequency inverse cosine transform of $$F$$, defined as

$f(x) = \sqrt{\frac{2}{\pi}} \int_{0}^\infty F(k) \cos(2\pi x k) \mathrm{d} k.$

If the transform cannot be computed in closed form, this function returns an unevaluated InverseCosineTransform object.

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). Note that for this transform, by default noconds=True.

>>> from sympy import inverse_cosine_transform, exp, sqrt, pi
>>> from sympy.abc import x, k, a
>>> inverse_cosine_transform(sqrt(2)*a/(sqrt(pi)*(a**2 + k**2)), k, x)
exp(-a*x)
>>> inverse_cosine_transform(1/sqrt(k), k, x)
1/sqrt(x)

sympy.integrals.transforms.hankel_transform(f, r, k, nu, **hints)[source]

Compute the Hankel transform of $$f$$, defined as

$F_\nu(k) = \int_{0}^\infty f(r) J_\nu(k r) r \mathrm{d} r.$

If the transform cannot be computed in closed form, this function returns an unevaluated HankelTransform object.

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). Note that for this transform, by default noconds=True.

>>> from sympy import hankel_transform, inverse_hankel_transform
>>> from sympy import gamma, exp, sinh, cosh
>>> from sympy.abc import r, k, m, nu, a

>>> ht = hankel_transform(1/r**m, r, k, nu)
>>> ht
2*2**(-m)*k**(m - 2)*gamma(-m/2 + nu/2 + 1)/gamma(m/2 + nu/2)

>>> inverse_hankel_transform(ht, k, r, nu)
r**(-m)

>>> ht = hankel_transform(exp(-a*r), r, k, 0)
>>> ht
a/(k**3*(a**2/k**2 + 1)**(3/2))

>>> inverse_hankel_transform(ht, k, r, 0)
exp(-a*r)

sympy.integrals.transforms.inverse_hankel_transform(F, k, r, nu, **hints)[source]

Compute the inverse Hankel transform of $$F$$ defined as

$f(r) = \int_{0}^\infty F_\nu(k) J_\nu(k r) k \mathrm{d} k.$

If the transform cannot be computed in closed form, this function returns an unevaluated InverseHankelTransform object.

For a description of possible hints, refer to the docstring of sympy.integrals.transforms.IntegralTransform.doit(). Note that for this transform, by default noconds=True.

>>> from sympy import hankel_transform, inverse_hankel_transform, gamma
>>> from sympy import gamma, exp, sinh, cosh
>>> from sympy.abc import r, k, m, nu, a

>>> ht = hankel_transform(1/r**m, r, k, nu)
>>> ht
2*2**(-m)*k**(m - 2)*gamma(-m/2 + nu/2 + 1)/gamma(m/2 + nu/2)

>>> inverse_hankel_transform(ht, k, r, nu)
r**(-m)

>>> ht = hankel_transform(exp(-a*r), r, k, 0)
>>> ht
a/(k**3*(a**2/k**2 + 1)**(3/2))

>>> inverse_hankel_transform(ht, k, r, 0)
exp(-a*r)


## Internals¶

There is a general method for calculating antiderivatives of elementary functions, called the Risch algorithm. The Risch algorithm is a decision procedure that can determine whether an elementary solution exists, and in that case calculate it. It can be extended to handle many nonelementary functions in addition to the elementary ones.

SymPy currently uses a simplified version of the Risch algorithm, called the Risch-Norman algorithm. This algorithm is much faster, but may fail to find an antiderivative, although it is still very powerful. SymPy also uses pattern matching and heuristics to speed up evaluation of some types of integrals, e.g. polynomials.

For non-elementary definite integrals, SymPy uses so-called Meijer G-functions. Details are described here:

## API reference¶

static integrals.integrate(f, var, ...)

Compute definite or indefinite integral of one or more variables using Risch-Norman algorithm and table lookup. This procedure is able to handle elementary algebraic and transcendental functions and also a huge class of special functions, including Airy, Bessel, Whittaker and Lambert.

var can be:

• a symbol – indefinite integration

• a tuple (symbol, a) – indefinite integration with result

given with $$a$$ replacing $$symbol$$

• a tuple (symbol, a, b) – definite integration

Several variables can be specified, in which case the result is multiple integration. (If var is omitted and the integrand is univariate, the indefinite integral in that variable will be performed.)

Indefinite integrals are returned without terms that are independent of the integration variables. (see examples)

Definite improper integrals often entail delicate convergence conditions. Pass conds=’piecewise’, ‘separate’ or ‘none’ to have these returned, respectively, as a Piecewise function, as a separate result (i.e. result will be a tuple), or not at all (default is ‘piecewise’).

Strategy

SymPy uses various approaches to definite integration. One method is to find an antiderivative for the integrand, and then use the fundamental theorem of calculus. Various functions are implemented to integrate polynomial, rational and trigonometric functions, and integrands containing DiracDelta terms.

SymPy also implements the part of the Risch algorithm, which is a decision procedure for integrating elementary functions, i.e., the algorithm can either find an elementary antiderivative, or prove that one does not exist. There is also a (very successful, albeit somewhat slow) general implementation of the heuristic Risch algorithm. This algorithm will eventually be phased out as more of the full Risch algorithm is implemented. See the docstring of Integral._eval_integral() for more details on computing the antiderivative using algebraic methods.

The option risch=True can be used to use only the (full) Risch algorithm. This is useful if you want to know if an elementary function has an elementary antiderivative. If the indefinite Integral returned by this function is an instance of NonElementaryIntegral, that means that the Risch algorithm has proven that integral to be non-elementary. Note that by default, additional methods (such as the Meijer G method outlined below) are tried on these integrals, as they may be expressible in terms of special functions, so if you only care about elementary answers, use risch=True. Also note that an unevaluated Integral returned by this function is not necessarily a NonElementaryIntegral, even with risch=True, as it may just be an indication that the particular part of the Risch algorithm needed to integrate that function is not yet implemented.

Another family of strategies comes from re-writing the integrand in terms of so-called Meijer G-functions. Indefinite integrals of a single G-function can always be computed, and the definite integral of a product of two G-functions can be computed from zero to infinity. Various strategies are implemented to rewrite integrands as G-functions, and use this information to compute integrals (see the meijerint module).

The option manual=True can be used to use only an algorithm that tries to mimic integration by hand. This algorithm does not handle as many integrands as the other algorithms implemented but may return results in a more familiar form. The manualintegrate module has functions that return the steps used (see the module docstring for more information).

In general, the algebraic methods work best for computing antiderivatives of (possibly complicated) combinations of elementary functions. The G-function methods work best for computing definite integrals from zero to infinity of moderately complicated combinations of special functions, or indefinite integrals of very simple combinations of special functions.

The strategy employed by the integration code is as follows:

• If computing a definite integral, and both limits are real, and at least one limit is +- oo, try the G-function method of definite integration first.
• Try to find an antiderivative, using all available methods, ordered by performance (that is try fastest method first, slowest last; in particular polynomial integration is tried first, meijer g-functions second to last, and heuristic risch last).
• If still not successful, try G-functions irrespective of the limits.

The option meijerg=True, False, None can be used to, respectively: always use G-function methods and no others, never use G-function methods, or use all available methods (in order as described above). It defaults to None.

Integral, Integral.doit

Examples

>>> from sympy import integrate, log, exp, oo
>>> from sympy.abc import a, x, y

>>> integrate(x*y, x)
x**2*y/2

>>> integrate(log(x), x)
x*log(x) - x

>>> integrate(log(x), (x, 1, a))
a*log(a) - a + 1

>>> integrate(x)
x**2/2


Terms that are independent of x are dropped by indefinite integration:

>>> from sympy import sqrt
>>> integrate(sqrt(1 + x), (x, 0, x))
2*(x + 1)**(3/2)/3 - 2/3
>>> integrate(sqrt(1 + x), x)
2*(x + 1)**(3/2)/3

>>> integrate(x*y)
Traceback (most recent call last):
...
ValueError: specify integration variables to integrate x*y


Note that integrate(x) syntax is meant only for convenience in interactive sessions and should be avoided in library code.

>>> integrate(x**a*exp(-x), (x, 0, oo)) # same as conds='piecewise'
Piecewise((gamma(a + 1), -re(a) < 1),
(Integral(x**a*exp(-x), (x, 0, oo)), True))

>>> integrate(x**a*exp(-x), (x, 0, oo), conds='none')
gamma(a + 1)

>>> integrate(x**a*exp(-x), (x, 0, oo), conds='separate')
(gamma(a + 1), -re(a) < 1)

static integrals.line_integrate(field, Curve, variables)

Compute the line integral.

integrate, Integral

Examples

>>> from sympy import Curve, line_integrate, E, ln
>>> from sympy.abc import x, y, t
>>> C = Curve([E**t + 1, E**t - 1], (t, 0, ln(2)))
>>> line_integrate(x + y, C, [x, y])
3*sqrt(2)

static integrals.deltaintegrate(f, x)

The idea for integration is the following:

• If we are dealing with a DiracDelta expression, i.e. DiracDelta(g(x)), we try to simplify it.

If we could simplify it, then we integrate the resulting expression. We already know we can integrate a simplified expression, because only simple DiracDelta expressions are involved.

If we couldn’t simplify it, there are two cases:

1. The expression is a simple expression: we return the integral, taking care if we are dealing with a Derivative or with a proper DiracDelta.
2. The expression is not simple (i.e. DiracDelta(cos(x))): we can do nothing at all.
• If the node is a multiplication node having a DiracDelta term:

First we expand it.

If the expansion did work, the we try to integrate the expansion.

If not, we try to extract a simple DiracDelta term, then we have two cases:

1. We have a simple DiracDelta term, so we return the integral.
2. We didn’t have a simple term, but we do have an expression with simplified DiracDelta terms, so we integrate this expression.

sympy.functions.special.delta_functions.DiracDelta, sympy.integrals.integrals.Integral

Examples

>>> from sympy.abc import x, y, z
>>> from sympy.integrals.deltafunctions import deltaintegrate
>>> from sympy import sin, cos, DiracDelta, Heaviside
>>> deltaintegrate(x*sin(x)*cos(x)*DiracDelta(x - 1), x)
sin(1)*cos(1)*Heaviside(x - 1)
>>> deltaintegrate(y**2*DiracDelta(x - z)*DiracDelta(y - z), y)
z**2*DiracDelta(x - z)*Heaviside(y - z)

static integrals.ratint(f, x, **flags)

Performs indefinite integration of rational functions.

Given a field $$K$$ and a rational function $$f = p/q$$, where $$p$$ and $$q$$ are polynomials in $$K[x]$$, returns a function $$g$$ such that $$f = g'$$.

>>> from sympy.integrals.rationaltools import ratint
>>> from sympy.abc import x

>>> ratint(36/(x**5 - 2*x**4 - 2*x**3 + 4*x**2 + x - 2), x)
(12*x + 6)/(x**2 - 1) + 4*log(x - 2) - 4*log(x + 1)


sympy.integrals.integrals.Integral.doit, ratint_logpart, ratint_ratpart

References

 [Bro05] M. Bronstein, Symbolic Integration I: Transcendental Functions, Second Edition, Springer-Verlag, 2005, pp. 35-70
static integrals.heurisch(f, x, rewrite=False, hints=None, mappings=None, retries=3, degree_offset=0, unnecessary_permutations=None)

Compute indefinite integral using heuristic Risch algorithm.

This is a heuristic approach to indefinite integration in finite terms using the extended heuristic (parallel) Risch algorithm, based on Manuel Bronstein’s “Poor Man’s Integrator”.

The algorithm supports various classes of functions including transcendental elementary or special functions like Airy, Bessel, Whittaker and Lambert.

Note that this algorithm is not a decision procedure. If it isn’t able to compute the antiderivative for a given function, then this is not a proof that such a functions does not exist. One should use recursive Risch algorithm in such case. It’s an open question if this algorithm can be made a full decision procedure.

This is an internal integrator procedure. You should use toplevel ‘integrate’ function in most cases, as this procedure needs some preprocessing steps and otherwise may fail.

sympy.integrals.integrals.Integral.doit, sympy.integrals.integrals.Integral, components

Examples

>>> from sympy import tan
>>> from sympy.integrals.heurisch import heurisch
>>> from sympy.abc import x, y

>>> heurisch(y*tan(x), x)
y*log(tan(x)**2 + 1)/2


See Manuel Bronstein’s “Poor Man’s Integrator”:

[2] K. Geddes, L. Stefanus, On the Risch-Norman Integration
Method and its Implementation in Maple, Proceedings of ISSAC‘89, ACM Press, 212-217.
[3] J. H. Davenport, On the Parallel Risch Algorithm (I),
Proceedings of EUROCAM‘82, LNCS 144, Springer, 144-157.
[4] J. H. Davenport, On the Parallel Risch Algorithm (III):
Use of Tangents, SIGSAM Bulletin 16 (1982), 3-6.
[5] J. H. Davenport, B. M. Trager, On the Parallel Risch
Algorithm (II), ACM Transactions on Mathematical Software 11 (1985), 356-362.

Specification

heurisch(f, x, rewrite=False, hints=None)

where

f : expression x : symbol

rewrite -> force rewrite ‘f’ in terms of ‘tan’ and ‘tanh’ hints -> a list of functions that may appear in anti-derivate

• hints = None –> no suggestions at all
• hints = [ ] –> try to figure out
• hints = [f1, ..., fn] –> we know better
static integrals.heurisch_wrapper(f, x, rewrite=False, hints=None, mappings=None, retries=3, degree_offset=0, unnecessary_permutations=None)

A wrapper around the heurisch integration algorithm.

This method takes the result from heurisch and checks for poles in the denominator. For each of these poles, the integral is reevaluated, and the final integration result is given in terms of a Piecewise.

heurisch

Examples

>>> from sympy.core import symbols
>>> from sympy.functions import cos
>>> from sympy.integrals.heurisch import heurisch, heurisch_wrapper
>>> n, x = symbols('n x')
>>> heurisch(cos(n*x), x)
sin(n*x)/n
>>> heurisch_wrapper(cos(n*x), x)
Piecewise((x, n == 0), (sin(n*x)/n, True))

static integrals.trigintegrate(f, x, conds='piecewise')

Integrate f = Mul(trig) over x

>>> from sympy import Symbol, sin, cos, tan, sec, csc, cot
>>> from sympy.integrals.trigonometry import trigintegrate
>>> from sympy.abc import x

>>> trigintegrate(sin(x)*cos(x), x)
sin(x)**2/2

>>> trigintegrate(sin(x)**2, x)
x/2 - sin(x)*cos(x)/2

>>> trigintegrate(tan(x)*sec(x), x)
1/cos(x)

>>> trigintegrate(sin(x)*tan(x), x)
-log(sin(x) - 1)/2 + log(sin(x) + 1)/2 - sin(x)


http://en.wikibooks.org/wiki/Calculus/Integration_techniques

sympy.integrals.integrals.Integral.doit, sympy.integrals.integrals.Integral

static integrals.manualintegrate(f, var)

Compute indefinite integral of a single variable using an algorithm that resembles what a student would do by hand.

Unlike integrate, var can only be a single symbol.

sympy.integrals.integrals.integrate, sympy.integrals.integrals.Integral.doit, sympy.integrals.integrals.Integral

Examples

>>> from sympy import sin, cos, tan, exp, log, integrate
>>> from sympy.integrals.manualintegrate import manualintegrate
>>> from sympy.abc import x
>>> manualintegrate(1 / x, x)
log(x)
>>> integrate(1/x)
log(x)
>>> manualintegrate(log(x), x)
x*log(x) - x
>>> integrate(log(x))
x*log(x) - x
>>> manualintegrate(exp(x) / (1 + exp(2 * x)), x)
atan(exp(x))
>>> integrate(exp(x) / (1 + exp(2 * x)))
RootSum(4*_z**2 + 1, Lambda(_i, _i*log(2*_i + exp(x))))
>>> manualintegrate(cos(x)**4 * sin(x), x)
-cos(x)**5/5
>>> integrate(cos(x)**4 * sin(x), x)
-cos(x)**5/5
>>> manualintegrate(cos(x)**4 * sin(x)**3, x)
cos(x)**7/7 - cos(x)**5/5
>>> integrate(cos(x)**4 * sin(x)**3, x)
cos(x)**7/7 - cos(x)**5/5
>>> manualintegrate(tan(x), x)
-log(cos(x))
>>> integrate(tan(x), x)
-log(sin(x)**2 - 1)/2

static manualintegrate.integral_steps(integrand, symbol, **options)

Returns the steps needed to compute an integral.

This function attempts to mirror what a student would do by hand as closely as possible.

SymPy Gamma uses this to provide a step-by-step explanation of an integral. The code it uses to format the results of this function can be found at https://github.com/sympy/sympy_gamma/blob/master/app/logic/intsteps.py.

Returns : rule : namedtuple The first step; most rules have substeps that must also be considered. These substeps can be evaluated using manualintegrate to obtain a result.

Examples

>>> from sympy import exp, sin, cos
>>> from sympy.integrals.manualintegrate import integral_steps
>>> from sympy.abc import x
>>> print(repr(integral_steps(exp(x) / (1 + exp(2 * x)), x)))
URule(u_var=_u, u_func=exp(x), constant=1,
substep=ArctanRule(context=1/(_u**2 + 1), symbol=_u),
context=exp(x)/(exp(2*x) + 1), symbol=x)
>>> print(repr(integral_steps(sin(x), x)))
TrigRule(func='sin', arg=x, context=sin(x), symbol=x)
>>> print(repr(integral_steps((x**2 + 3)**2 , x)))
RewriteRule(rewritten=x**4 + 6*x**2 + 9,
ConstantTimesRule(constant=6, other=x**2,
substep=PowerRule(base=x, exp=2, context=x**2, symbol=x),
context=6*x**2, symbol=x),
ConstantRule(constant=9, context=9, symbol=x)],
context=x**4 + 6*x**2 + 9, symbol=x), context=(x**2 + 3)**2, symbol=x)


The class $$Integral$$ represents an unevaluated integral and has some methods that help in the integration of an expression.

class sympy.integrals.Integral

Represents unevaluated integral.

is_commutative

Returns whether all the free symbols in the integral are commutative.

as_sum(n, method='midpoint')

Approximates the definite integral by a sum.

method ... one of: left, right, midpoint, trapezoid

These are all basically the rectangle method [1], the only difference is where the function value is taken in each interval to define the rectangle.

Integral.doit
Perform the integration using any hints

Examples

>>> from sympy import sin, sqrt
>>> from sympy.abc import x
>>> from sympy.integrals import Integral
>>> e = Integral(sin(x), (x, 3, 7))
>>> e
Integral(sin(x), (x, 3, 7))


For demonstration purposes, this interval will only be split into 2 regions, bounded by [3, 5] and [5, 7].

The left-hand rule uses function evaluations at the left of each interval:

>>> e.as_sum(2, 'left')
2*sin(5) + 2*sin(3)


The midpoint rule uses evaluations at the center of each interval:

>>> e.as_sum(2, 'midpoint')
2*sin(4) + 2*sin(6)


The right-hand rule uses function evaluations at the right of each interval:

>>> e.as_sum(2, 'right')
2*sin(5) + 2*sin(7)


The trapezoid rule uses function evaluations on both sides of the intervals. This is equivalent to taking the average of the left and right hand rule results:

>>> e.as_sum(2, 'trapezoid')
2*sin(5) + sin(3) + sin(7)
>>> (e.as_sum(2, 'left') + e.as_sum(2, 'right'))/2 == _
True


All but the trapexoid method may be used when dealing with a function with a discontinuity. Here, the discontinuity at x = 0 can be avoided by using the midpoint or right-hand method:

>>> e = Integral(1/sqrt(x), (x, 0, 1))
>>> e.as_sum(5).n(4)
1.730
>>> e.as_sum(10).n(4)
1.809
>>> e.doit().n(4)  # the actual value is 2
2.000


The left- or trapezoid method will encounter the discontinuity and return oo:

>>> e.as_sum(5, 'left')
oo
>>> e.as_sum(5, 'trapezoid')
oo

doit(**hints)

Perform the integration using any hints given.

sympy.integrals.trigonometry.trigintegrate, sympy.integrals.risch.heurisch, sympy.integrals.rationaltools.ratint

as_sum
Approximate the integral using a sum

Examples

>>> from sympy import Integral
>>> from sympy.abc import x, i
>>> Integral(x**i, (i, 1, 3)).doit()
Piecewise((2, log(x) == 0), (x**3/log(x) - x/log(x), True))

free_symbols

This method returns the symbols that will exist when the integral is evaluated. This is useful if one is trying to determine whether an integral depends on a certain symbol or not.

function, limits, variables

Examples

>>> from sympy import Integral
>>> from sympy.abc import x, y
>>> Integral(x, (x, y, 1)).free_symbols
set([y])

is_number

Return True if the Integral will result in a number, else False.

Integrals are a special case since they contain symbols that can be replaced with numbers. Whether the integral can be done or not is another issue. But answering whether the final result is a number is not difficult.

is_zero

Examples

>>> from sympy import Integral
>>> from sympy.abc import x, y
>>> Integral(x).is_number
False
>>> Integral(x, y).is_number
False
>>> Integral(x, (y, 1, x)).is_number
False
>>> Integral(x, (y, 1, 2)).is_number
False
>>> Integral(x, (y, 1, 1)).is_number
True
>>> Integral(x, (x, 1, 2)).is_number
True
>>> Integral(x*y, (x, 1, 2), (y, 1, 3)).is_number
True
>>> Integral(1, x, (x, 1, 2)).is_number
True

is_zero

Since Integral doesn’t autosimplify it it useful to see if it would simplify to zero or not in a trivial manner, i.e. when the function is 0 or two limits of a definite integral are the same.

This is a very naive and quick test, not intended to check for special patterns like Integral(sin(m*x)*cos(n*x), (x, 0, 2*pi)) == 0.

is_number

Examples

>>> from sympy import Integral
>>> from sympy.abc import x, y, z
>>> Integral(1, (x, 1, 1)).is_zero
True
>>> Integral(0, (x, y, z)).is_zero
True
>>> Integral(1, (x, 1, 2)).is_zero
False

transform(x, u, inverse=False)

Performs a change of variables from $$x$$ to $$u$$ using the relationship given by $$x$$ and $$u$$ which will define the transformations $$f$$ and $$F$$ (which are inverses of each other) as follows:

1. If $$x$$ is a Symbol (which is a variable of integration) then $$u$$ will be interpreted as some function, f(u), with inverse F(u). This, in effect, just makes the substitution of x with f(x).
2. If $$u$$ is a Symbol then $$x$$ will be interpreted as some function, F(x), with inverse f(u). This is commonly referred to as u-substitution.

The $$inverse$$ option will reverse $$x$$ and $$u$$. It is a deprecated option since $$x$$ and $$u$$ can just be passed in reverse order.

Once f and F have been identified, the transformation is made as follows:

$\int_a^b x \mathrm{d}x \rightarrow \int_{F(a)}^{F(b)} f(x) \frac{\mathrm{d}}{\mathrm{d}x}$

where $$F(x)$$ is the inverse of $$f(x)$$ and the limits and integrand have been corrected so as to retain the same value after integration.

variables
Lists the integration variables
as_dummy
Replace integration variables with dummy ones

Notes

The mappings, F(x) or f(u), must lead to a unique integral. Linear or rational linear expression, $$2*x$$, $$1/x$$ and $$sqrt(x)$$, will always work; quadratic expressions like $$x**2 - 1$$ are acceptable as long as the resulting integrand does not depend on the sign of the solutions (see examples).

The integral will be returned unchanged if $$x$$ is not a variable of integration.

$$x$$ must be (or contain) only one of of the integration variables. If $$u$$ has more than one free symbol then it should be sent as a tuple ($$u$$, $$uvar$$) where $$uvar$$ identifies which variable is replacing the integration variable. XXX can it contain another integration variable?

Examples

>>> from sympy.abc import a, b, c, d, x, u, y
>>> from sympy import Integral, S, cos, sqrt

>>> i = Integral(x*cos(x**2 - 1), (x, 0, 1))


transform can change the variable of integration

>>> i.transform(x, u)
Integral(u*cos(u**2 - 1), (u, 0, 1))


transform can perform u-substitution as long as a unique integrand is obtained:

>>> i.transform(x**2 - 1, u)
Integral(cos(u)/2, (u, -1, 0))


This attempt fails because x = +/-sqrt(u + 1) and the sign does not cancel out of the integrand:

>>> Integral(cos(x**2 - 1), (x, 0, 1)).transform(x**2 - 1, u)
Traceback (most recent call last):
...
ValueError:
The mapping between F(x) and f(u) did not give a unique integrand.


transform can do a substitution. Here, the previous result is transformed back into the original expression using “u-substitution”:

>>> ui = _
>>> _.transform(sqrt(u + 1), x) == i
True


We can accomplish the same with a regular substitution:

>>> ui.transform(u, x**2 - 1) == i
True


If the $$x$$ does not contain a symbol of integration then the integral will be returned unchanged. Integral $$i$$ does not have an integration variable $$a$$ so no change is made:

>>> i.transform(a, x) == i
True


When $$u$$ has more than one free symbol the symbol that is replacing $$x$$ must be identified by passing $$u$$ as a tuple:

>>> Integral(x, (x, 0, 1)).transform(x, (u + a, u))
Integral(a + u, (u, -a, -a + 1))
>>> Integral(x, (x, 0, 1)).transform(x, (u + a, a))
Integral(a + u, (a, -u, -u + 1))


## TODO and Bugs¶

There are still lots of functions that SymPy does not know how to integrate. For bugs related to this module, see http://code.google.com/p/sympy/issues/list?q=label:Integration

# Numeric Integrals¶

SymPy has functions to calculate points and weights for Gaussian quadrature of any order and any precision:

Computes the Gauss-Legendre quadrature [R254] points and weights.

The Gauss-Legendre quadrature approximates the integral:

$\int_{-1}^1 f(x)\,dx \approx \sum_{i=1}^n w_i f(x_i)$

The nodes $$x_i$$ of an order $$n$$ quadrature rule are the roots of $$P_n$$ and the weights $$w_i$$ are given by:

$w_i = \frac{2}{\left(1-x_i^2\right) \left(P'_n(x_i)\right)^2}$
Parameters : n : the order of quadrature n_digits : number of significant digits of the points and weights to return (x, w) : the x and w are lists of points and weights as Floats. The points $$x_i$$ and weights $$w_i$$ are returned as (x, w) tuple of lists.

gauss_laguerre, gauss_gen_laguerre, gauss_hermite, gauss_chebyshev_t, gauss_chebyshev_u, gauss_jacobi

References

Examples

>>> from sympy.integrals.quadrature import gauss_legendre
>>> x, w = gauss_legendre(3, 5)
>>> x
[-0.7746, 0, 0.7746]
>>> w
[0.55556, 0.88889, 0.55556]
>>> x, w = gauss_legendre(4, 5)
>>> x
[-0.86114, -0.33998, 0.33998, 0.86114]
>>> w
[0.34786, 0.65215, 0.65215, 0.34786]


Computes the Gauss-Laguerre quadrature [R256] points and weights.

The Gauss-Laguerre quadrature approximates the integral:

$\int_0^{\infty} e^{-x} f(x)\,dx \approx \sum_{i=1}^n w_i f(x_i)$

The nodes $$x_i$$ of an order $$n$$ quadrature rule are the roots of $$L_n$$ and the weights $$w_i$$ are given by:

$w_i = \frac{x_i}{(n+1)^2 \left(L_{n+1}(x_i)\right)^2}$
Parameters : n : the order of quadrature n_digits : number of significant digits of the points and weights to return (x, w) : the x and w are lists of points and weights as Floats. The points $$x_i$$ and weights $$w_i$$ are returned as (x, w) tuple of lists.

gauss_legendre, gauss_gen_laguerre, gauss_hermite, gauss_chebyshev_t, gauss_chebyshev_u, gauss_jacobi

References

Examples

>>> from sympy.integrals.quadrature import gauss_laguerre
>>> x, w = gauss_laguerre(3, 5)
>>> x
[0.41577, 2.2943, 6.2899]
>>> w
[0.71109, 0.27852, 0.010389]
>>> x, w = gauss_laguerre(6, 5)
>>> x
[0.22285, 1.1889, 2.9927, 5.7751, 9.8375, 15.983]
>>> w
[0.45896, 0.417, 0.11337, 0.010399, 0.00026102, 8.9855e-7]


Computes the Gauss-Hermite quadrature [R258] points and weights.

The Gauss-Hermite quadrature approximates the integral:

$\int_{-\infty}^{\infty} e^{-x^2} f(x)\,dx \approx \sum_{i=1}^n w_i f(x_i)$

The nodes $$x_i$$ of an order $$n$$ quadrature rule are the roots of $$H_n$$ and the weights $$w_i$$ are given by:

$w_i = \frac{2^{n-1} n! \sqrt{\pi}}{n^2 \left(H_{n-1}(x_i)\right)^2}$
Parameters : n : the order of quadrature n_digits : number of significant digits of the points and weights to return (x, w) : the x and w are lists of points and weights as Floats. The points $$x_i$$ and weights $$w_i$$ are returned as (x, w) tuple of lists.

gauss_legendre, gauss_laguerre, gauss_gen_laguerre, gauss_chebyshev_t, gauss_chebyshev_u, gauss_jacobi

References

Examples

>>> from sympy.integrals.quadrature import gauss_hermite
>>> x, w = gauss_hermite(3, 5)
>>> x
[-1.2247, 0, 1.2247]
>>> w
[0.29541, 1.1816, 0.29541]

>>> x, w = gauss_hermite(6, 5)
>>> x
[-2.3506, -1.3358, -0.43608, 0.43608, 1.3358, 2.3506]
>>> w
[0.00453, 0.15707, 0.72463, 0.72463, 0.15707, 0.00453]


Computes the generalized Gauss-Laguerre quadrature [R261] points and weights.

The generalized Gauss-Laguerre quadrature approximates the integral:

$\int_{0}^\infty x^{\alpha} e^{-x} f(x)\,dx \approx \sum_{i=1}^n w_i f(x_i)$

The nodes $$x_i$$ of an order $$n$$ quadrature rule are the roots of $$L^{\alpha}_n$$ and the weights $$w_i$$ are given by:

$w_i = \frac{\Gamma(\alpha+n)}{n \Gamma(n) L^{\alpha}_{n-1}(x_i) L^{\alpha+1}_{n-1}(x_i)}$
Parameters : n : the order of quadrature alpha : the exponent of the singularity, $$\alpha > -1$$ n_digits : number of significant digits of the points and weights to return (x, w) : the x and w are lists of points and weights as Floats. The points $$x_i$$ and weights $$w_i$$ are returned as (x, w) tuple of lists.

gauss_legendre, gauss_laguerre, gauss_hermite, gauss_chebyshev_t, gauss_chebyshev_u, gauss_jacobi

References

Examples

>>> from sympy import S
>>> x, w = gauss_gen_laguerre(3, -S.Half, 5)
>>> x
[0.19016, 1.7845, 5.5253]
>>> w
[1.4493, 0.31413, 0.00906]

>>> x, w = gauss_gen_laguerre(4, 3*S.Half, 5)
>>> x
[0.97851, 2.9904, 6.3193, 11.712]
>>> w
[0.53087, 0.67721, 0.11895, 0.0023152]


Computes the Gauss-Chebyshev quadrature [R263] points and weights of the first kind.

The Gauss-Chebyshev quadrature of the first kind approximates the integral:

$\int_{-1}^{1} \frac{1}{\sqrt{1-x^2}} f(x)\,dx \approx \sum_{i=1}^n w_i f(x_i)$

The nodes $$x_i$$ of an order $$n$$ quadrature rule are the roots of $$T_n$$ and the weights $$w_i$$ are given by:

$w_i = \frac{\pi}{n}$
Parameters : n : the order of quadrature n_digits : number of significant digits of the points and weights to return (x, w) : the x and w are lists of points and weights as Floats. The points $$x_i$$ and weights $$w_i$$ are returned as (x, w) tuple of lists.

gauss_legendre, gauss_laguerre, gauss_hermite, gauss_gen_laguerre, gauss_chebyshev_u, gauss_jacobi

References

Examples

>>> from sympy import S
>>> x, w = gauss_chebyshev_t(3, 5)
>>> x
[0.86602, 0, -0.86602]
>>> w
[1.0472, 1.0472, 1.0472]

>>> x, w = gauss_chebyshev_t(6, 5)
>>> x
[0.96593, 0.70711, 0.25882, -0.25882, -0.70711, -0.96593]
>>> w
[0.5236, 0.5236, 0.5236, 0.5236, 0.5236, 0.5236]


Computes the Gauss-Chebyshev quadrature [R265] points and weights of the second kind.

The Gauss-Chebyshev quadrature of the second kind approximates the integral:

$\int_{-1}^{1} \sqrt{1-x^2} f(x)\,dx \approx \sum_{i=1}^n w_i f(x_i)$

The nodes $$x_i$$ of an order $$n$$ quadrature rule are the roots of $$U_n$$ and the weights $$w_i$$ are given by:

$w_i = \frac{\pi}{n+1} \sin^2 \left(\frac{i}{n+1}\pi\right)$
Parameters : n : the order of quadrature n_digits : number of significant digits of the points and weights to return (x, w) : the x and w are lists of points and weights as Floats. The points $$x_i$$ and weights $$w_i$$ are returned as (x, w) tuple of lists.

gauss_legendre, gauss_laguerre, gauss_hermite, gauss_gen_laguerre, gauss_chebyshev_t, gauss_jacobi

References

Examples

>>> from sympy import S
>>> x, w = gauss_chebyshev_u(3, 5)
>>> x
[0.70711, 0, -0.70711]
>>> w
[0.3927, 0.7854, 0.3927]

>>> x, w = gauss_chebyshev_u(6, 5)
>>> x
[0.90097, 0.62349, 0.22252, -0.22252, -0.62349, -0.90097]
>>> w
[0.084489, 0.27433, 0.42658, 0.42658, 0.27433, 0.084489]


Computes the Gauss-Jacobi quadrature [R267] points and weights.

The Gauss-Jacobi quadrature of the first kind approximates the integral:

$\int_{-1}^1 (1-x)^\alpha (1+x)^\beta f(x)\,dx \approx \sum_{i=1}^n w_i f(x_i)$

The nodes $$x_i$$ of an order $$n$$ quadrature rule are the roots of $$P^{(\alpha,\beta)}_n$$ and the weights $$w_i$$ are given by:

$w_i = -\frac{2n+\alpha+\beta+2}{n+\alpha+\beta+1}\frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)} {\Gamma(n+\alpha+\beta+1)(n+1)!} \frac{2^{\alpha+\beta}}{P'_n(x_i) P^{(\alpha,\beta)}_{n+1}(x_i)}$
Parameters : n : the order of quadrature alpha : the first parameter of the Jacobi Polynomial, $$\alpha > -1$$ beta : the second parameter of the Jacobi Polynomial, $$\beta > -1$$ n_digits : number of significant digits of the points and weights to return (x, w) : the x and w are lists of points and weights as Floats. The points $$x_i$$ and weights $$w_i$$ are returned as (x, w) tuple of lists.

gauss_legendre, gauss_laguerre, gauss_hermite, gauss_gen_laguerre, gauss_chebyshev_t, gauss_chebyshev_u

References

Examples

>>> from sympy import S
>>> x, w = gauss_jacobi(3, S.Half, -S.Half, 5)
>>> x
[-0.90097, -0.22252, 0.62349]
>>> w
[1.7063, 1.0973, 0.33795]

>>> x, w = gauss_jacobi(6, 1, 1, 5)
>>> x
[-0.87174, -0.5917, -0.2093, 0.2093, 0.5917, 0.87174]
>>> w
[0.050584, 0.22169, 0.39439, 0.39439, 0.22169, 0.050584]