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Topic: Rubi 4.5 released
Replies: 21   Last Post: Jun 29, 2014 4:29 AM

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clicliclic@freenet.de

Posts: 988
Registered: 4/26/08
Re: Rubi 4.5 released
Posted: Jun 26, 2014 5:43 AM
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"Nasser M. Abbasi" schrieb:
>
> Ok, an update before going more. I did the term-by-term method.
>
> Using the outer integral you gave, converted it to M code, expanded
> it, and obtained 94 additive terms. Now added a Loop to integrate term
> by term. The problem, Rubi 4.5 is taking too long just on the first
> term, so I stopped it after waiting 10 minutes, since when I use
> Mathematica Integrate on the same first term, it does it in about
> 5 seconds. So before I spend more time on this, I will show the first
> term, and lets see first why Rubi is not doing this. May be
> I did something wrong. There are 94 terms, so if each takes that
> long, this whole process will really take too long.
>
> The first term is: (this is after replacing c by Cos[p], in the
> outer integral expression)
>
> -((9*a^14*z^2*Cos[p]^2)/(2*(a^2 + 4*z^2)^(3/2)*
> (4*z^2 + a^2*(1 - Cos[p]^2))^2*(z^2 + a^2*Cos[p]*
> (Cos[p] - Sqrt[-1 + Cos[p]^2]))^2*
> (z^2 + a^2*Cos[p]*(Cos[p] + Sqrt[-1 + Cos[p]^2]))^2))
>
> Mathematica Integrate gives (in about 5 seconds)
>
> Assuming[Element[{r, p, a, z}, Reals] && z > 0, Integrate[term, p]]
>
> -((1/(32*(a^2 + 3*z^2)^6*(a^2 + 4*z^2)^(3/2)))*
> (9*a^14*z^2*(-((8*(a^10 + 10*a^8*z^2 + 33*a^6*z^4 +
> 44*a^4*z^6 + 16*a^2*z^8 - 8*z^10)*ArcTan[(1 + a^2/z^2)*Cot[p]])/
> (a^2*z^2*(a^2 + z^2)^3)) + ((a^6 + 22*a^4*z^2 + 105*a^2*z^4 +
> 128*z^6)*ArcTan[(Sqrt[a^2 + 4*z^2]*Tan[p])/(2*z)])/
> (a^2*z^3*Sqrt[a^2 + 4*z^2]) + (4*(a^4 + 5*a^2*z^2 + 6*z^4)^2*Sin[p])/
> ((a^2 + z^2)^3*((a^2 + z^2)*Cos[p] - I*z^2*Sin[p])) +
> (4*(a^4 + 5*a^2*z^2 + 6*z^4)^2*Sin[p])/
> ((a^2 + z^2)^3*((a^2 + z^2)*Cos[p] + I*z^2*Sin[p])) +
> (2*(a^2 + 3*z^2)^2*Sin[2*p])/(z^2*(a^2 + 8*z^2 - a^2*Cos[2*p])))))
>
> While Rubi4.5:
>
> Assuming[Element[{r, p, a, z}, Reals] && z > 0, Integrate[term, p]]
>
> No response, busy. MathKernel CPU 100%, I can wait more......
> but why so long? Will keep it running to see....
>


This is primarily for Albert to make sense of. Perhaps Rubi can do this
integral if all occurences of Sqrt[-1 + Cos[p]^2] in the integrand are
replaced by I*Sin[p]. And Mathematica might become faster then ...

Actually, I didn't mean to split the integrand into sooooo many parts;
what I had in mind was to separate the 9+9 logarithms into 9 terms of
1+1 logarithm (pairing with the corresponding logarithm from the other
group serves to keep the integral real), with a tenth term formed by the
non-logarithmic remainder. Thus, my second integral covers the (easy to
integrate) full remainder, and my third integral the first (relatively
hard to integrate) logarithm from each group of nine.

Martin.



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