"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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