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Topic: Mathematica integration Vs Sympy
Replies: 7   Last Post: Apr 23, 2013 9:38 PM

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Posts: 60
Registered: 9/24/11
Re: Mathematica integration Vs Sympy
Posted: Apr 21, 2013 5:18 AM
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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@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

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