Albert Rich schrieb:   
   >   
   > On Wednesday, April 26, 2017 at 9:09:43 AM UTC-10, clicl...@freenet.de wrote:   
   > > "Nasser M. Abbasi" schrieb:   
   > > >   
   > > > On 4/26/2017 5:15 AM, Albert Rich wrote:   
   > > > >   
   > > > > An antiderivative of arctanh(t)*log(t)/(t-1) wrt t is:   
   > > > >   
   > > > > -1/2*Log[2]*Log[t]*Log[(1 + t)/2] + 1/4*Log[2]*Log[(1 + t)/2]^2 +   
   > > > >  1/2*Log[1 - t]*Log[t]*Log[1 + t] -   
   > > > >  1/2*Log[2]*Log[(1 + t)/2]*Log[(1 + t)/(2*t)] +   
   > > > >  1/4*Log[1 - t]*Log[(1 + t)/(2*t)]^2 +   
   > > > >  1/4*Log[-(1/t)]*Log[(1 + t)/(2*t)]^2 -   
   > > > >  1/4*Log[(-1 + t)/(2*t)]*Log[(1 + t)/(2*t)]^2 +   
   > > > >  1/2*Log[(1 + t)/(2*t)]*PolyLog[2, 1 + 1/t] -   
   > > > >  ArcTanh[t]*PolyLog[2, 1 - t] +   
   > > > >  1/2*Log[1 + t]*PolyLog[2, 1 - t] + 1/2*Log[1 + t]*PolyLog[2, t] -   
   > > > >  1/2*Log[(1 + t)/(2*t)]*PolyLog[2, t] +   
   > > > >  1/2*Log[t]*PolyLog[2, (1 + t)/2] +   
   > > > >  1/2*Log[(1 + t)/(2*t)]*PolyLog[2, (1 + t)/2] -   
   > > > >  1/2*Log[(1 + t)/(2*t)]*PolyLog[2, (1 + t)/(2*t)] -   
   > > > >  1/2*PolyLog[3, 1 + 1/t] -   
   > > > >  1/2*PolyLog[3, 1 - t] - (1/2)*PolyLog[3, t] -   
   > > > >  1/2*PolyLog[3, (1 + t)/2] +   
   > > > >  1/2*PolyLog[3, (1 + t)/(2*t)]   
   > > > >   
   > > > > Taking the difference of evaluating its limits as t approaches 1   
   > > > > and 0 gives 1/8 pi^2 log(2).   
   > > >   
   > > > Amazing. Well done. I assume you found a rule for this and it   
   > > > will be in next Rubi version.   
   > >   
   > > Perhaps a better nose when and how to integrate by parts is all   
   > > that's needed?   
   >   
   > Yes, for integrands involving inverse elementary functions,   
   > integration by parts is obviously the way to go. The following partial   
   > derivation of Nasser's integral uses parts twice:   
   >   
   > Int[ArcTanh[x]*Log[x]/(x-1),x]   
   >   
   > == -ArcTanh[x]*PolyLog[2,1-x] +   
   >    Int[PolyLog[2,1-x]/(1-x^2),x]   
   >   
   > == -ArcTanh[x]*PolyLog[2,1-x] +   
   >    1/2*Int[PolyLog[2,1-x]/(1-x),x] +   
   >    1/2*Int[PolyLog[2,1-x]/(1+x),x]   
   >   
   > == -ArcTanh[x]*PolyLog[2,1-x] -   
   >    1/2*PolyLog[3,1-x] +   
   >    1/2*Int[PolyLog[2,1-x]/(1+x),x]   
   >   
   > == -ArcTanh[x]*PolyLog[2,1-x] -   
   >    1/2*PolyLog[3,1-x] +   
   >    1/2*Log[1+x]*PolyLog[2,1-x] +   
   >    1/2*Int[Log[x]*Log[1+x]/(x-1),x]   
   >   
   > Then I got stuck.  Fortunately however, Mathematica's built-in   
   > integrator could integrate log(x)*log(1+x)/(x-1), although for some   
   > reason it could not integrate the original problem.   
   >   
   > So if somebody has a better nose than I how to step through the   
   > integration of log(a+b*x)*log(c+d*x)/(e+f*x), Rubi and I would very   
   > much like to know...   
   >   
      
   The current web-version of FriCAS returns even   
      
     integrate(log(x)*log(1+x)/(x-1), x)   
      
   unevaluated. So one would have to delve into polylogarithm theory or   
   sniff out experimentally what cases of log(a+b*x)*log(c+d*x)/(e+f*x) are   
   integrable on Mathematica (where one of the three linear terms may be   
   mapped onto x for simplicity). We know already that the antiderivative   
   may involve 18 terms, and that not just the polylogarithm arguments x   
   and 1-x and 1+1/x appear, but also (1+x)/2 and (1+x)/(2*x), the latter   
   pair in 12 terms. One might learn something from collecting terms with   
   related arguments and examining their derivatives.   
      
   Martin.   
      
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