From: Albert_Rich@msn.com   
      
   On Thursday, April 27, 2017 at 9:10:35 AM UTC-10, clicl...@freenet.de wrote:   
   > > 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.   
      
   The derivation of the antiderivative of log(a+b*x)*log(c+d*x)/(a+b*x) using   
   parts is relatively straight-forward:   
      
   Int[Log[a+b*x]*Log[c+d*x]/(a+b*x), x]   
   =   
   Log[a+b*x]^2*Log[c+d*x]/(2*b) -   
   d/(2*b)*Int[Log[a+b*x]^2/(c+d*x), x]   
   =   
   Log[a+b*x]^2*Log[c+d*x]/(2*b) -   
   Log[a+b*x]^2*Log[b*(c+d*x)/(b*c-a*d)]/(2*b) +   
   Int[Log[a + b*x]*Log[b*(c+d*x)/(b*c-a*d)]/(a+b*x), x]   
   =   
   Log[a+b*x]^2*Log[c+d*x]/(2*b) -   
   Log[a+b*x]^2*Log[b*(c+d*x)/(b*c-a*d)]/(2*b) -   
   Log[a+b*x]*PolyLog[2,-d*(a+b*x)/(b*c-a*d)]/b +   
   Int[PolyLog[2,-d*(a+b*x)/(b*c-a*d)]/(a+b*x), x]   
   =   
   Log[a+b*x]^2*Log[c+d*x]/(2*b) -   
   Log[a+b*x]^2*Log[b*(c+d*x)/(b*c-a*d)]/(2*b) -   
   Log[a+b*x]*PolyLog[2,-d*(a+b*x)/(b*c-a*d)]/b +   
   PolyLog[3,-d*(a+b*x)/(b*c-a*d)]/b   
      
   But I don't know how to generalize that for log(a+b*x)*log(c+d*x)/(e+f*x).   
      
   Albert   
      
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