60270560   
   Andreas Dieckmann schrieb:   
   >   
   > The shortest form I could find is   
   > Integrate[(Log[a + b*x]*Log[c + d*x])/(e + f*x), x] =   
   > (1/f)*(Log[a + b*x]*Log[c + d*x]*Log[(b*(e + f*x))/(b*e - a*f)] -   
   >   (1/2)*Log[(b*(c*f - d*e))/(d*(a*f - b*e))]*   
   >     Log[((a*f - b*e)*(a + b*x))/f]*Log[(f*(a + b*x))/(a*f - b*e)] +   
   >   Log[c + d*x]*PolyLog[2, (f*(a + b*x))/(a*f - b*e)] +   
   >   Log[a + b*x]*PolyLog[2, (f*(c + d*x))/(c*f - d*e)] -   
   >   Log[((a*f - b*e)*(c + d*x))/((c*f - d*e)*(a + b*x))]*   
   >     (Log[(b*(c*f - d*e))/(d*(a*f - b*e))]*Log[a + b*x] -   
   >     PolyLog[2, (f*(c + d*x))/(c*f - d*e)] +   
   >     PolyLog[2, (f*(a + b*x))/(a*f - b*e)] -   
   >     PolyLog[2, (b*(c + d*x))/(d*(a + b*x))] +   
   >     PolyLog[2, ((a*f - b*e)*(c + d*x))/((c*f - d*e)*(a + b*x))]) -   
   >   PolyLog[3, (f*(a + b*x))/(a*f - b*e)] -   
   >   PolyLog[3, (f*(c + d*x))/(c*f - d*e)] -   
   >   PolyLog[3, (b*(c + d*x))/(d*(a + b*x))] +   
   >   PolyLog[3, ((a*f - b*e)*(c + d*x))/((c*f - d*e)*(a + b*x))])   
   >   
      
   Putting a = 0, b = 1, c = 1, d = 1, e = -1, f = 1, I obtain:   
      
   INT(LOG(x)*LOG(x + 1)/(x - 1), x)   
    = LOG(x)*LN(x + 1)*LOG(1 - x) - 1/2*LOG(2)*LOG(x)^2   
    + LOG(x + 1)*polylog(2, x) + LOG(x)*polylog(2, (x + 1)/2)   
    - LOG((x + 1)/(2*x))*(LOG(2)*LOG(x)   
      - polylog(2, (x + 1)/2) + polylog(2, x)   
      - polylog(2, 1/x + 1) + polylog(2, (x + 1)/(2*x)))   
    - polylog(3, x) - polylog(3, (x + 1)/2)   
    - polylog(3, 1/x + 1) + polylog(3, (x + 1)/(2*x))   
      
   With my definition of polylog(n, x), the imaginary part of this 13-term   
   antiderivative is not (piecewise) constant for x > 1, as it should   
   obviously be for all positive x.   
      
   The 17-term antiderivative posted by Albert Rich on Wed, 26 Apr 2017   
   03:15:05 -0700 (PDT), on the other hand, holds on the entire real line   
   with my definition of polylog(n, x).   
      
   Martin.   
      
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