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Topic: A bug in Reduce package 'algint'?
Replies: 9   Last Post: Feb 1, 2013 6:02 PM

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 Axel Vogt Posts: 1,068 Registered: 5/5/07
Re: A bug in Reduce package 'algint'?
Posted: Jan 22, 2013 2:08 PM

On 22.01.2013 18:42, clicliclic@freenet.de wrote:
>
> acer schrieb:

>>
>> Le lundi 21 janvier 2013 16:16:30 UTC-5, Axel Vogt a écrit :

>>> On 21.01.2013 19:55, clicliclic wrote:
>>>>
>>>> Wanting to refresh my knowledge of the capabilties of the Reduce algebra
>>>> system, I have recently browsed the website. The system comes with the
>>>> 'algint' package by J. Davenport which boosts the integrator
>>>> capabilities for algebraic functions. The package documentation
>>>>
>>>> <http://www.reduce-algebra.com/docs/algint.pdf>
>>>>
>>>> introduces the example integrand sqrt(sqrt(a^2 + x^2) + x)/x. A correct
>>>> antiderivative for this is
>>>>
>>>> 2*(sqrt(sqrt(a^2 + x^2) + x)
>>>> - sqrt(a)*atanh(sqrt(sqrt(a^2 + x^2) + x)/sqrt(a))
>>>> - sqrt(a)*atan(sqrt(sqrt(a^2 + x^2) + x)/sqrt(a)))
>>>>
>>>> The antiderivative printed in the documentation, however, is either very
>>>> wrong or garbled beyond recognition.
>>>>

>>>
>>> Maple 16 returns 2*sqrt(2*x)*hypergeom([-1/4, -1/4, 1/4],[1/2, 3/4],-a^2/x^2)
>>>
>>> May be right, but not that 'usefull' w.r.t. your result
>>>

>>
>> Just because Axel's brought up Maple (and not wishing to side-track Martin), in Maple 16.02,
>>
>> expr := sqrt(sqrt(a^2 + x^2) + x)/x:
>> p := u = sqrt(a^2 + x^2) + x:
>> new := student[changevar](p, Int(expr,x), u):
>> sol := eval(value(new),p):
>> lprint(sol);
>>
>> -((2*a^2*((a^2+x^2)^(1/2)+x)^2+a^4+((a^2+x^2)^(1/2)+x)^4)

> /((a^2+x^2)^(1/2)+x)^2)^(1/2)
> *((a^2+x^2)^(1/2)+x)*4^(1/2)*(-((a^2+x^2)^(1/2)+x)^(1/2)
> +a^(1/2)*arctan(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2))
> +a^(1/2)*arctanh(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2)))
> /(a^2+((a^2+x^2)^(1/2)+x)^2)

>>
>> lprint(other):
>>
>> 2*((a^2+x^2)^(1/2)+x)^(1/2)-2*a^(1/2)*arctan(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2))-2*a^(1/2)*arctanh(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2))
>>
>> What should we expect from the system, automatically?
>>

>
> My outside viewpoint: if the integrand is given in terms of elementary
> functions, an elementary antiderivative whenever one exists. Even if it
> can be expressed more compactly in terms of higher functions. But then
> an infinite 3F2 series is not really simple.
>
> The Wolfram Integrator produces an elementary answer, but includes an
> unnecessary extra factor. Perhaps this is arrived at by automatic
> simplification of 3F2(-1/4, -1/4, 1/4; 1/2, 3/4; -a^2/x^2) ?
>
> How does Maple arrive at 3F2 in the first place? Solution of an ODE?
> Series expansion of the integrand?
>
> Side-tracking won't harm my ramblings,
>
> Martin.
>

I have not checked, how Maple 'finds' the solution, but it does
not have implemented the 'catalog' of rules you have worked out
and at hand now. Unfortunately.