"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?   
      
   Martin.   
      
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