Re: Mathematica integration Vs Sympy

*To*: mathgroup at smc.vnet.net*Subject*: [mg130564] Re: Mathematica integration Vs Sympy*From*: Brentt <brenttnewman at gmail.com>*Date*: Sun, 21 Apr 2013 05:16:26 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <20130420094229.B63996A64@smc.vnet.net>

The result: In[0]: Integrate[1/(x*(1 - a*(1 - x))), x] Out[0]: (Log[1 + a (-1 + x)] - Log[x])/(-1 + a) Seems to be true for all complex a and x . Why do you think it assumes a>1? On Sat, Apr 20, 2013 at 2:42 AM, Sergio R <sergiorquestion at gmail.com> wrote: > Hello all, > > Just for fun a put an integral I was doing via mathematica > WolframAlpha > [ > http://www.wolframalpha.com/input/?i=Integrate[1%2F%28x*%281-a*%281-x%29%29%29%2Cx] > ] > into the online sympy [ http://live.sympy.org/ ] console > the following: > > a = Symbol('a'); g = 1/(x*(1-a*(1-x))) ; u=simplify(integrate(g,x)) > > Then, to display the result, at the sympy ">>>" prompt, type u > and hit return. > > To my surprise, sympy seems to give the right result without any > assumption, while mathematica's result seems to assume a>1, which is > not specified. Also for this case (a>1) sympy gives an extra constant > which is not present in the mathematica result. > > Is there a way to make mathematica to output a general result like > sympy > in this case? > > Sergio > >

**Follow-Ups**:**Re: Re: Mathematica integration Vs Sympy***From:*Alex Krasnov <akrasnov@eecs.berkeley.edu>

**Re: Mathematica integration Vs Sympy***From:*Murray Eisenberg <murray@math.umass.edu>

**References**:**Mathematica integration Vs Sympy***From:*Sergio R <sergiorquestion@gmail.com>