Simplifies the given expression.
Simplification is not a well defined term and the exact strategies this function tries can change in the future versions of SymPy. If your algorithm relies on “simplification” (whatever it is), try to determine what you need exactly  is it powsimp()?, radsimp()?, together()?, logcombine()?, or something else? And use this particular function directly, because those are well defined and thus your algorithm will be robust.
Nonetheless, especially for interactive use, or when you don’t know anything about the structure of the expression, simplify() tries to apply intelligent heuristics to make the input expression “simpler”. For example:
>>> from sympy import simplify, cos, sin
>>> from sympy.abc import x, y
>>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2)
>>> a
(x**2 + x)/(x*sin(y)**2 + x*cos(y)**2)
>>> simplify(a)
x + 1
Note that we could have obtained the same result by using specific simplification functions:
>>> from sympy import trigsimp, cancel
>>> b = trigsimp(a)
>>> b
(x**2 + x)/x
>>> c = cancel(b)
>>> c
x + 1
In some cases, applying simplify() may actually result in some more complicated expression. The default ratio=1.7 prevents more extreme cases: if (result length)/(input length) > ratio, then input is returned unmodified. The measure parameter lets you specify the function used to determine how complex an expression is. The function should take a single argument as an expression and return a number such that if expression a is more complex than expression b, then measure(a) > measure(b). The default measure function is count_ops(), which returns the total number of operations in the expression.
For example, if ratio=1, simplify output can’t be longer than input.
>>> from sympy import sqrt, simplify, count_ops, oo
>>> root = 1/(sqrt(2)+3)
Since simplify(root) would result in a slightly longer expression, root is returned unchanged instead:
>>> simplify(root, ratio=1) == root
True
If ratio=oo, simplify will be applied anyway:
>>> count_ops(simplify(root, ratio=oo)) > count_ops(root)
True
Note that the shortest expression is not necessary the simplest, so setting ratio to 1 may not be a good idea. Heuristically, the default value ratio=1.7 seems like a reasonable choice.
You can easily define your own measure function based on what you feel should represent the “size” or “complexity” of the input expression. Note that some choices, such as lambda expr: len(str(expr)) may appear to be good metrics, but have other problems (in this case, the measure function may slow down simplify too much for very large expressions). If you don’t know what a good metric would be, the default, count_ops, is a good one.
For example:
>>> from sympy import symbols, log
>>> a, b = symbols('a b', positive=True)
>>> g = log(a) + log(b) + log(a)*log(1/b)
>>> h = simplify(g)
>>> h
log(a*b**(log(1/a) + 1))
>>> count_ops(g)
8
>>> count_ops(h)
6
So you can see that h is simpler than g using the count_ops metric. However, we may not like how simplify (in this case, using logcombine) has created the b**(log(1/a) + 1) term. A simple way to reduce this would be to give more weight to powers as operations in count_ops. We can do this by using the visual=True option:
>>> print(count_ops(g, visual=True))
2*ADD + DIV + 4*LOG + MUL
>>> print(count_ops(h, visual=True))
ADD + DIV + 2*LOG + MUL + POW
>>> from sympy import Symbol, S
>>> def my_measure(expr):
... POW = Symbol('POW')
... # Discourage powers by giving POW a weight of 10
... count = count_ops(expr, visual=True).subs(POW, 10)
... # Every other operation gets a weight of 1 (the default)
... count = count.replace(Symbol, type(S.One))
... return count
>>> my_measure(g)
8
>>> my_measure(h)
15
>>> 15./8 > 1.7 # 1.7 is the default ratio
True
>>> simplify(g, measure=my_measure)
log(a)*log(b) + log(a) + log(b)
Note that because simplify() internally tries many different simplification strategies and then compares them using the measure function, we get a completely different result that is still different from the input expression by doing this.
Collect additive terms of an expression.
This function collects additive terms of an expression with respect to a list of expression up to powers with rational exponents. By the term symbol here are meant arbitrary expressions, which can contain powers, products, sums etc. In other words symbol is a pattern which will be searched for in the expression’s terms.
The input expression is not expanded by collect(), so user is expected to provide an expression is an appropriate form. This makes collect() more predictable as there is no magic happening behind the scenes. However, it is important to note, that powers of products are converted to products of powers using the expand_power_base() function.
There are two possible types of output. First, if evaluate flag is set, this function will return an expression with collected terms or else it will return a dictionary with expressions up to rational powers as keys and collected coefficients as values.
Examples
>>> from sympy import S, collect, expand, factor, Wild
>>> from sympy.abc import a, b, c, x, y, z
This function can collect symbolic coefficients in polynomials or rational expressions. It will manage to find all integer or rational powers of collection variable:
>>> collect(a*x**2 + b*x**2 + a*x  b*x + c, x)
c + x**2*(a + b) + x*(a  b)
The same result can be achieved in dictionary form:
>>> d = collect(a*x**2 + b*x**2 + a*x  b*x + c, x, evaluate=False)
>>> d[x**2]
a + b
>>> d[x]
a  b
>>> d[S.One]
c
You can also work with multivariate polynomials. However, remember that this function is greedy so it will care only about a single symbol at time, in specification order:
>>> collect(x**2 + y*x**2 + x*y + y + a*y, [x, y])
x**2*(y + 1) + x*y + y*(a + 1)
Also more complicated expressions can be used as patterns:
>>> from sympy import sin, log
>>> collect(a*sin(2*x) + b*sin(2*x), sin(2*x))
(a + b)*sin(2*x)
>>> collect(a*x*log(x) + b*(x*log(x)), x*log(x))
x*(a + b)*log(x)
You can use wildcards in the pattern:
>>> w = Wild('w1')
>>> collect(a*x**y  b*x**y, w**y)
x**y*(a  b)
It is also possible to work with symbolic powers, although it has more complicated behavior, because in this case power’s base and symbolic part of the exponent are treated as a single symbol:
>>> collect(a*x**c + b*x**c, x)
a*x**c + b*x**c
>>> collect(a*x**c + b*x**c, x**c)
x**c*(a + b)
However if you incorporate rationals to the exponents, then you will get well known behavior:
>>> collect(a*x**(2*c) + b*x**(2*c), x**c)
x**(2*c)*(a + b)
Note also that all previously stated facts about collect() function apply to the exponential function, so you can get:
>>> from sympy import exp
>>> collect(a*exp(2*x) + b*exp(2*x), exp(x))
(a + b)*exp(2*x)
If you are interested only in collecting specific powers of some symbols then set exact flag in arguments:
>>> collect(a*x**7 + b*x**7, x, exact=True)
a*x**7 + b*x**7
>>> collect(a*x**7 + b*x**7, x**7, exact=True)
x**7*(a + b)
You can also apply this function to differential equations, where derivatives of arbitrary order can be collected. Note that if you collect with respect to a function or a derivative of a function, all derivatives of that function will also be collected. Use exact=True to prevent this from happening:
>>> from sympy import Derivative as D, collect, Function
>>> f = Function('f') (x)
>>> collect(a*D(f,x) + b*D(f,x), D(f,x))
(a + b)*Derivative(f(x), x)
>>> collect(a*D(D(f,x),x) + b*D(D(f,x),x), f)
(a + b)*Derivative(f(x), x, x)
>>> collect(a*D(D(f,x),x) + b*D(D(f,x),x), D(f,x), exact=True)
a*Derivative(f(x), x, x) + b*Derivative(f(x), x, x)
>>> collect(a*D(f,x) + b*D(f,x) + a*f + b*f, f)
(a + b)*f(x) + (a + b)*Derivative(f(x), x)
Or you can even match both derivative order and exponent at the same time:
>>> collect(a*D(D(f,x),x)**2 + b*D(D(f,x),x)**2, D(f,x))
(a + b)*Derivative(f(x), x, x)**2
Finally, you can apply a function to each of the collected coefficients. For example you can factorize symbolic coefficients of polynomial:
>>> f = expand((x + a + 1)**3)
>>> collect(f, x, factor)
x**3 + 3*x**2*(a + 1) + 3*x*(a + 1)**2 + (a + 1)**3
Note
Arguments are expected to be in expanded form, so you might have to call expand() prior to calling this function.
Separates variables in an expression, if possible. By default, it separates with respect to all symbols in an expression and collects constant coefficients that are independent of symbols.
If dict=True then the separated terms will be returned in a dictionary keyed to their corresponding symbols. By default, all symbols in the expression will appear as keys; if symbols are provided, then all those symbols will be used as keys, and any terms in the expression containing other symbols or nonsymbols will be returned keyed to the string ‘coeff’. (Passing None for symbols will return the expression in a dictionary keyed to ‘coeff’.)
If force=True, then bases of powers will be separated regardless of assumptions on the symbols involved.
Notes
The order of the factors is determined by Mul, so that the separated expressions may not necessarily be grouped together.
Although factoring is necessary to separate variables in some expressions, it is not necessary in all cases, so one should not count on the returned factors being factored.
Examples
>>> from sympy.abc import x, y, z, alpha
>>> from sympy import separatevars, sin
>>> separatevars((x*y)**y)
(x*y)**y
>>> separatevars((x*y)**y, force=True)
x**y*y**y
>>> e = 2*x**2*z*sin(y)+2*z*x**2
>>> separatevars(e)
2*x**2*z*(sin(y) + 1)
>>> separatevars(e, symbols=(x, y), dict=True)
{'coeff': 2*z, x: x**2, y: sin(y) + 1}
>>> separatevars(e, [x, y, alpha], dict=True)
{'coeff': 2*z, alpha: 1, x: x**2, y: sin(y) + 1}
If the expression is not really separable, or is only partially separable, separatevars will do the best it can to separate it by using factoring.
>>> separatevars(x + x*y  3*x**2)
x*(3*x  y  1)
If the expression is not separable then expr is returned unchanged or (if dict=True) then None is returned.
>>> eq = 2*x + y*sin(x)
>>> separatevars(eq) == eq
True
>>> separatevars(2*x + y*sin(x), symbols=(x, y), dict=True) == None
True
Rationalize the denominator by removing square roots.
Note: the expression returned from radsimp must be used with caution since if the denominator contains symbols, it will be possible to make substitutions that violate the assumptions of the simplification process: that for a denominator matching a + b*sqrt(c), a != +/b*sqrt(c). (If there are no symbols, this assumptions is made valid by collecting terms of sqrt(c) so the match variable a does not contain sqrt(c).) If you do not want the simplification to occur for symbolic denominators, set symbolic to False.
If there are more than max_terms radical terms do not simplify.
Examples
>>> from sympy import radsimp, sqrt, Symbol, denom, pprint, I
>>> from sympy.abc import a, b, c
>>> radsimp(1/(I + 1))
(1  I)/2
>>> radsimp(1/(2 + sqrt(2)))
(sqrt(2) + 2)/2
>>> x,y = list(map(Symbol, 'xy'))
>>> e = ((2 + 2*sqrt(2))*x + (2 + sqrt(8))*y)/(2 + sqrt(2))
>>> radsimp(e)
sqrt(2)*(x + y)
Terms are collected automatically:
>>> r2 = sqrt(2)
>>> r5 = sqrt(5)
>>> pprint(radsimp(1/(y*r2 + x*r2 + a*r5 + b*r5)))
___ ___
\/ 5 *(a  b) + \/ 2 *(x + y)

2 2 2 2
 5*a  10*a*b  5*b + 2*x + 4*x*y + 2*y
If radicals in the denominator cannot be removed, the original expression will be returned. If the denominator was 1 then any square roots will also be collected:
>>> radsimp(sqrt(2)*x + sqrt(2))
sqrt(2)*(x + 1)
Results with symbols will not always be valid for all substitutions:
>>> eq = 1/(a + b*sqrt(c))
>>> eq.subs(a, b*sqrt(c))
1/(2*b*sqrt(c))
>>> radsimp(eq).subs(a, b*sqrt(c))
nan
If symbolic=False, symbolic denominators will not be transformed (but numeric denominators will still be processed):
>>> radsimp(eq, symbolic=False)
1/(a + b*sqrt(c))
Returns a pair with expression’s numerator and denominator. If the given expression is not a fraction then this function will return the tuple (expr, 1).
This function will not make any attempt to simplify nested fractions or to do any term rewriting at all.
If only one of the numerator/denominator pair is needed then use numer(expr) or denom(expr) functions respectively.
>>> from sympy import fraction, Rational, Symbol
>>> from sympy.abc import x, y
>>> fraction(x/y)
(x, y)
>>> fraction(x)
(x, 1)
>>> fraction(1/y**2)
(1, y**2)
>>> fraction(x*y/2)
(x*y, 2)
>>> fraction(Rational(1, 2))
(1, 2)
This function will also work fine with assumptions:
>>> k = Symbol('k', negative=True)
>>> fraction(x * y**k)
(x, y**(k))
If we know nothing about sign of some exponent and ‘exact’ flag is unset, then structure this exponent’s structure will be analyzed and pretty fraction will be returned:
>>> from sympy import exp
>>> fraction(2*x**(y))
(2, x**y)
>>> fraction(exp(x))
(1, exp(x))
>>> fraction(exp(x), exact=True)
(exp(x), 1)
reduces expression by using known trig identities
Notes
deep:  Apply trigsimp inside all objects with arguments
recursive:  Use common subexpression elimination (cse()) and apply trigsimp recursively (this is quite expensive if the expression is large)
Examples
>>> from sympy import trigsimp, sin, cos, log, cosh, sinh
>>> from sympy.abc import x, y
>>> e = 2*sin(x)**2 + 2*cos(x)**2
>>> trigsimp(e)
2
>>> trigsimp(log(e))
log(2*sin(x)**2 + 2*cos(x)**2)
>>> trigsimp(log(e), deep=True)
log(2)
Simplify besseltype functions.
This routine tries to simplify besseltype functions. Currently it only works on the Bessel J and I functions, however. It works by looking at all such functions in turn, and eliminating factors of “I” and “1” (actually their polar equivalents) in front of the argument. After that, functions of halfinteger order are rewritten using trigonometric functions.
>>> from sympy import besselj, besseli, besselsimp, polar_lift, I, S
>>> from sympy.abc import z, nu
>>> besselsimp(besselj(nu, z*polar_lift(1)))
exp(I*pi*nu)*besselj(nu, z)
>>> besselsimp(besseli(nu, z*polar_lift(I)))
exp(I*pi*nu/2)*besselj(nu, z)
>>> besselsimp(besseli(S(1)/2, z))
sqrt(2)*cosh(z)/(sqrt(pi)*sqrt(z))
reduces expression by combining powers with similar bases and exponents.
Notes
If deep is True then powsimp() will also simplify arguments of functions. By default deep is set to False.
If force is True then bases will be combined without checking for assumptions, e.g. sqrt(x)*sqrt(y) > sqrt(x*y) which is not true if x and y are both negative.
You can make powsimp() only combine bases or only combine exponents by changing combine=’base’ or combine=’exp’. By default, combine=’all’, which does both. combine=’base’ will only combine:
a a a 2x x
x * y => (x*y) as well as things like 2 => 4
and combine=’exp’ will only combine
a b (a + b)
x * x => x
combine=’exp’ will strictly only combine exponents in the way that used to be automatic. Also use deep=True if you need the old behavior.
When combine=’all’, ‘exp’ is evaluated first. Consider the first example below for when there could be an ambiguity relating to this. This is done so things like the second example can be completely combined. If you want ‘base’ combined first, do something like powsimp(powsimp(expr, combine=’base’), combine=’exp’).
Examples
>>> from sympy import powsimp, exp, log, symbols
>>> from sympy.abc import x, y, z, n
>>> powsimp(x**y*x**z*y**z, combine='all')
x**(y + z)*y**z
>>> powsimp(x**y*x**z*y**z, combine='exp')
x**(y + z)*y**z
>>> powsimp(x**y*x**z*y**z, combine='base', force=True)
x**y*(x*y)**z
>>> powsimp(x**z*x**y*n**z*n**y, combine='all', force=True)
(n*x)**(y + z)
>>> powsimp(x**z*x**y*n**z*n**y, combine='exp')
n**(y + z)*x**(y + z)
>>> powsimp(x**z*x**y*n**z*n**y, combine='base', force=True)
(n*x)**y*(n*x)**z
>>> x, y = symbols('x y', positive=True)
>>> powsimp(log(exp(x)*exp(y)))
log(exp(x)*exp(y))
>>> powsimp(log(exp(x)*exp(y)), deep=True)
x + y
Radicals with Mul bases will be combined if combine=’exp’
>>> from sympy import sqrt, Mul
>>> x, y = symbols('x y')
Two radicals are automatically joined through Mul: >>> a=sqrt(x*sqrt(y)) >>> a*a**3 == a**4 True
But if an integer power of that radical has been autoexpanded then Mul does not join the resulting factors: >>> a**4 # auto expands to a Mul, no longer a Pow x**2*y >>> _*a # so Mul doesn’t combine them x**2*y*sqrt(x*sqrt(y)) >>> powsimp(_) # but powsimp will (x*sqrt(y))**(5/2) >>> powsimp(x*y*a) # but won’t when doing so would violate assumptions x*y*sqrt(x*sqrt(y))
Simplify combinatorial expressions.
This function takes as input an expression containing factorials, binomials, Pochhammer symbol and other “combinatorial” functions, and tries to minimize the number of those functions and reduce the size of their arguments. The result is be given in terms of binomials and factorials.
The algorithm works by rewriting all combinatorial functions as expressions involving rising factorials (Pochhammer symbols) and applies recurrence relations and other transformations applicable to rising factorials, to reduce their arguments, possibly letting the resulting rising factorial to cancel. Rising factorials with the second argument being an integer are expanded into polynomial forms and finally all other rising factorial are rewritten in terms more familiar functions. If the initial expression contained any combinatorial functions, the result is expressed using binomial coefficients and gamma functions. If the initial expression consisted of gamma functions alone, the result is expressed in terms of gamma functions.
If the result is expressed using gamma functions, the following three additional steps are performed:
All transformation rules can be found (or was derived from) here:
Examples
>>> from sympy.simplify import combsimp
>>> from sympy import factorial, binomial
>>> from sympy.abc import n, k
>>> combsimp(factorial(n)/factorial(n  3))
n*(n  2)*(n  1)
>>> combsimp(binomial(n+1, k+1)/binomial(n, k))
(n + 1)/(k + 1)
Given combinatorial term f(k) simplify its consecutive term ratio i.e. f(k+1)/f(k). The input term can be composed of functions and integer sequences which have equivalent representation in terms of gamma special function.
The algorithm performs three basic steps:
If f(k) is hypergeometric then as result we arrive with a quotient of polynomials of minimal degree. Otherwise None is returned.
For more information on the implemented algorithm refer to:
Returns True if ‘f’ and ‘g’ are hypersimilar.
Similarity in hypergeometric sense means that a quotient of f(k) and g(k) is a rational function in k. This procedure is useful in solving recurrence relations.
For more information see hypersimp().
Find a simple representation for a number or, if there are free symbols or if rational=True, then replace Floats with their Rational equivalents. If no change is made and rational is not False then Floats will at least be converted to Rationals.
For numerical expressions, a simple formula that numerically matches the given numerical expression is sought (and the input should be possible to evalf to a precision of at least 30 digits).
Optionally, a list of (rationally independent) constants to include in the formula may be given.
A lower tolerance may be set to find less exact matches. If no tolerance is given then the least precise value will set the tolerance (e.g. Floats default to 15 digits of precision, so would be tolerance=10**15).
With full=True, a more extensive search is performed (this is useful to find simpler numbers when the tolerance is set low).
See also
Examples
>>> from sympy import nsimplify, sqrt, GoldenRatio, exp, I, exp, pi
>>> nsimplify(4/(1+sqrt(5)), [GoldenRatio])
2 + 2*GoldenRatio
>>> nsimplify((1/(exp(3*pi*I/5)+1)))
1/2  I*sqrt(sqrt(5)/10 + 1/4)
>>> nsimplify(I**I, [pi])
exp(pi/2)
>>> nsimplify(pi, tolerance=0.01)
22/7
Return expr with terms having common square roots collected together. If evaluate is False a count indicating the number of sqrtcontaining terms will be returned and the returned expression will be an unevaluated Add with args ordered by default_sort_key.
Note: since I = sqrt(1), it is collected, too.
Examples
>>> from sympy import sqrt
>>> from sympy.simplify.simplify import collect_sqrt
>>> from sympy.abc import a, b
>>> r2, r3, r5 = [sqrt(i) for i in [2, 3, 5]]
>>> collect_sqrt(a*r2 + b*r2)
sqrt(2)*(a + b)
>>> collect_sqrt(a*r2 + b*r2 + a*r3 + b*r3)
sqrt(2)*(a + b) + sqrt(3)*(a + b)
>>> collect_sqrt(a*r2 + b*r2 + a*r3 + b*r5)
sqrt(3)*a + sqrt(5)*b + sqrt(2)*(a + b)
If evaluate is False then the arguments will be sorted and returned as a list and a count of the number of sqrtcontaining terms will be returned:
>>> collect_sqrt(a*r2 + b*r2 + a*r3 + b*r5, evaluate=False)
((sqrt(2)*(a + b), sqrt(3)*a, sqrt(5)*b), 3)
>>> collect_sqrt(a*sqrt(2) + b, evaluate=False)
((b, sqrt(2)*a), 1)
>>> collect_sqrt(a + b, evaluate=False)
((a + b,), 0)
A nongreedy collection of terms with similar number coefficients in an Add expr. If vars is given then only those constants will be targeted.
Examples
>>> from sympy import sqrt
>>> from sympy.abc import a, s
>>> from sympy.simplify.simplify import collect_const
>>> collect_const(sqrt(3) + sqrt(3)*(1 + sqrt(2)))
sqrt(3)*(sqrt(2) + 2)
>>> collect_const(sqrt(3)*s + sqrt(7)*s + sqrt(3) + sqrt(7))
(sqrt(3) + sqrt(7))*(s + 1)
>>> s = sqrt(2) + 2
>>> collect_const(sqrt(3)*s + sqrt(3) + sqrt(7)*s + sqrt(7))
(sqrt(2) + 3)*(sqrt(3) + sqrt(7))
>>> collect_const(sqrt(3)*s + sqrt(3) + sqrt(7)*s + sqrt(7), sqrt(3))
sqrt(7) + sqrt(3)*(sqrt(2) + 3) + sqrt(7)*(sqrt(2) + 2)
If no constants are provided then a leading Rational might be returned:
>>> collect_const(2*sqrt(3) + 4*a*sqrt(5))
2*(2*sqrt(5)*a + sqrt(3))
>>> collect_const(2*sqrt(3) + 4*a*sqrt(5), sqrt(3))
4*sqrt(5)*a + 2*sqrt(3)
Return eq (with generic symbols made positive) and a restore dictionary.
Any symbol that has positive=None will be replaced with a positive dummy symbol having the same name. This replacement will allow more symbolic processing of expressions, especially those involving powers and logarithms.
A dictionary that can be sent to subs to restore eq to its original symbols is also returned.
>>> from sympy import posify, Symbol, log
>>> from sympy.abc import x
>>> posify(x + Symbol('p', positive=True) + Symbol('n', negative=True))
(_x + n + p, {_x: x})
>> log(1/x).expand() # should be log(1/x) but it comes back as log(x) log(1/x)
>>> log(posify(1/x)[0]).expand() # take [0] and ignore replacements
log(_x)
>>> eq, rep = posify(1/x)
>>> log(eq).expand().subs(rep)
log(x)
>>> posify([x, 1 + x])
([_x, _x + 1], {_x: x})
Collect exponents on powers as assumptions allow.
Given a product of powers raised to a power, (bb1**be1 * bb2**be2...)**e, simplification can be done as follows:
Setting force to True will make symbols that are not explicitly negative behave as though they are positive, resulting in more denesting.
Setting polar to True will do simplifications on the riemann surface of the logarithm, also resulting in more denestings.
When there are sums of logs in exp() then a product of powers may be obtained e.g. exp(3*(log(a) + 2*log(b)))  > a**3*b**6.
Examples
>>> from sympy.abc import a, b, x, y, z
>>> from sympy import Symbol, exp, log, sqrt, symbols, powdenest
>>> powdenest((x**(2*a/3))**(3*x))
(x**(2*a/3))**(3*x)
>>> powdenest(exp(3*x*log(2)))
2**(3*x)
Assumptions may prevent expansion:
>>> powdenest(sqrt(x**2))
sqrt(x**2)
>>> p = symbols('p', positive=True)
>>> powdenest(sqrt(p**2))
p
No other expansion is done.
>>> i, j = symbols('i,j', integer=True)
>>> powdenest((x**x)**(i + j)) # X> (x**x)**i*(x**x)**j
x**(x*(i + j))
But exp() will be denested by moving all nonlog terms outside of the function; this may result in the collapsing of the exp to a power with a different base:
>>> powdenest(exp(3*y*log(x)))
x**(3*y)
>>> powdenest(exp(y*(log(a) + log(b))))
(a*b)**y
>>> powdenest(exp(3*(log(a) + log(b))))
a**3*b**3
If assumptions allow, symbols can also be moved to the outermost exponent:
>>> i = Symbol('i', integer=True)
>>> p = Symbol('p', positive=True)
>>> powdenest(((x**(2*i))**(3*y))**x)
((x**(2*i))**(3*y))**x
>>> powdenest(((x**(2*i))**(3*y))**x, force=True)
x**(6*i*x*y)
>>> powdenest(((p**(2*a))**(3*y))**x)
p**(6*a*x*y)
>>> powdenest(((x**(2*a/3))**(3*y/i))**x)
((x**(2*a/3))**(3*y/i))**x
>>> powdenest((x**(2*i)*y**(4*i))**z, force=True)
(x*y**2)**(2*i*z)
>>> n = Symbol('n', negative=True)
>>> powdenest((x**i)**y, force=True)
x**(i*y)
>>> powdenest((n**i)**x, force=True)
(n**i)**x
Takes logarithms and combines them using the following rules:
These identities are only valid if x and y are positive and if a is real, so the function will not combine the terms unless the arguments have the proper assumptions on them. Use logcombine(func, force=True) to automatically assume that the arguments of logs are positive and that coefficients are real. Note that this will not change any assumptions already in place, so if the coefficient is imaginary or the argument negative, combine will still not combine the equations. Change the assumptions on the variables to make them combine.
Examples
>>> from sympy import Symbol, symbols, log, logcombine
>>> from sympy.abc import a, x, y, z
>>> logcombine(a*log(x)+log(y)log(z))
a*log(x) + log(y)  log(z)
>>> logcombine(a*log(x)+log(y)log(z), force=True)
log(x**a*y/z)
>>> x,y,z = symbols('x,y,z', positive=True)
>>> a = Symbol('a', real=True)
>>> logcombine(a*log(x)+log(y)log(z))
log(x**a*y/z)
Denests sqrts in an expression that contain other square roots if possible, otherwise returns the expr unchanged. This is based on the algorithms of [1].
See also
sympy.solvers.solvers.unrad
References
[1] http://www.almaden.ibm.com/cs/people/fagin/symb85.pdf
[2] D. J. Jeffrey and A. D. Rich, ‘Symplifying Square Roots of Square Roots by Denesting’ (available at http://www.cybertester.com/data/denest.pdf)
Examples
>>> from sympy.simplify.sqrtdenest import sqrtdenest
>>> from sympy import sqrt
>>> sqrtdenest(sqrt(5 + 2 * sqrt(6)))
sqrt(2) + sqrt(3)
Returns all possible subsets of the set (0, 1, ..., n1) except the empty set, listed in reversed lexicographical order according to binary representation, so that the case of the fourth root is treated last.
Examples
>>> from sympy.simplify.sqrtdenest import subsets
>>> subsets(2)
[[1, 0], [0, 1], [1, 1]]
Perform common subexpression elimination on an expression.
Parameters :  exprs : list of sympy expressions, or a single sympy expression
symbols : infinite iterator yielding unique Symbols
optimizations : list of (callable, callable) pairs, optional
postprocess : a function which accepts the two return values of cse and


Returns :  replacements : list of (Symbol, expression) pairs
reduced_exprs : list of sympy expressions

Expand hypergeometric functions. If allow_hyper is True, allow partial simplification (that is a result different from input, but still containing hypergeometric functions).
Examples
>>> from sympy.simplify.hyperexpand import hyperexpand
>>> from sympy.functions import hyper
>>> from sympy.abc import z
>>> hyperexpand(hyper([], [], z))
exp(z)
Nonhyperegeometric parts of the expression and hypergeometric expressions that are not recognised are left unchanged:
>>> hyperexpand(1 + hyper([1, 1, 1], [], z))
hyper((1, 1, 1), (), z) + 1
Use func to transform expr at the given level.
Examples
>>> from sympy import use, expand
>>> from sympy.abc import x, y
>>> f = (x + y)**2*x + 1
>>> use(f, expand, level=2)
x*(x**2 + 2*x*y + y**2) + 1
>>> expand(f)
x**3 + 2*x**2*y + x*y**2 + 1
Manipulate expressions using paths.
EPath grammar in EBNF notation:
literal ::= /[AZaz_][AZaz_09]*/
number ::= /?\d+/
type ::= literal
attribute ::= literal "?"
all ::= "*"
slice ::= "[" number? (":" number? (":" number?)?)? "]"
range ::= all  slice
query ::= (type  attribute) ("" (type  attribute))*
selector ::= range  query range?
path ::= "/" selector ("/" selector)*
See the docstring of the epath() function.
Modify parts of an expression selected by a path.
Examples
>>> from sympy.simplify.epathtools import EPath
>>> from sympy import sin, cos, E
>>> from sympy.abc import x, y, z, t
>>> path = EPath("/*/[0]/Symbol")
>>> expr = [((x, 1), 2), ((3, y), z)]
>>> path.apply(expr, lambda expr: expr**2)
[((x**2, 1), 2), ((3, y**2), z)]
>>> path = EPath("/*/*/Symbol")
>>> expr = t + sin(x + 1) + cos(x + y + E)
>>> path.apply(expr, lambda expr: 2*expr)
t + sin(2*x + 1) + cos(2*x + 2*y + E)
Retrieve parts of an expression selected by a path.
Examples
>>> from sympy.simplify.epathtools import EPath
>>> from sympy import sin, cos, E
>>> from sympy.abc import x, y, z, t
>>> path = EPath("/*/[0]/Symbol")
>>> expr = [((x, 1), 2), ((3, y), z)]
>>> path.select(expr)
[x, y]
>>> path = EPath("/*/*/Symbol")
>>> expr = t + sin(x + 1) + cos(x + y + E)
>>> path.select(expr)
[x, x, y]
Manipulate parts of an expression selected by a path.
This function allows to manipulate large nested expressions in single line of code, utilizing techniques to those applied in XML processing standards (e.g. XPath).
If func is None, epath() retrieves elements selected by the path. Otherwise it applies func to each matching element.
Note that it is more efficient to create an EPath object and use the select and apply methods of that object, since this will compile the path string only once. This function should only be used as a convenient shortcut for interactive use.
This is the supported syntax:
Equivalent of for arg in args:.
Supports standard Python’s slice syntax.
Emulates isinstance().
Emulates hasattr().
Parameters :  path : str  EPath
expr : Basic  iterable
func : callable (optional)
args : tuple (optional)
kwargs : dict (optional)


Examples
>>> from sympy.simplify.epathtools import epath
>>> from sympy import sin, cos, E
>>> from sympy.abc import x, y, z, t
>>> path = "/*/[0]/Symbol"
>>> expr = [((x, 1), 2), ((3, y), z)]
>>> epath(path, expr)
[x, y]
>>> epath(path, expr, lambda expr: expr**2)
[((x**2, 1), 2), ((3, y**2), z)]
>>> path = "/*/*/Symbol"
>>> expr = t + sin(x + 1) + cos(x + y + E)
>>> epath(path, expr)
[x, x, y]
>>> epath(path, expr, lambda expr: 2*expr)
t + sin(2*x + 1) + cos(2*x + 2*y + E)