
Re: Rubi 4.5 released
Posted:
Jun 26, 2014 5:43 AM


"Nasser M. Abbasi" schrieb: > > Ok, an update before going more. I did the termbyterm 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 nonlogarithmic 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.

