Albert Rich schrieb:   
   >   
   > What Rubi needs to know is how to derive it.   
   >   
      
   Here are some relations for the specific original problem, to be broken   
   down further:   
      
   INT(LOG(x)*LOG(1 + x)/(x - 1), x)   
    = LOG(1 - x)*LOG(x)*LOG(1 + x)   
    + LOG(1 + x)*polylog(2, x)   
    - polylog(3, x)   
    - polylog(3, 1 + 1/x)   
    - INT(LOG(x)*LOG(1 - x)/(x + 1), x)   
    + INT(polylog(2, x)/(x*(x + 1)), x)   
    - INT(polylog(2, 1 + 1/x)/(x*(x + 1)), x)   
      
   - INT(LOG(x)*LOG(1 - x)/(x + 1), x)   
    = - LOG(2)*LOG(x)*LOG((1 + x)/2)   
    + 1/2*LOG(2)*LOG((1 + x)/2)^2   
    + LOG(x)*polylog(2, (1 + x)/2)   
    - polylog(3, (1 + x)/2)   
    + INT(LOG(2)*LOG((x + 1)/2)/(x*(x + 1)), x)   
    - INT(polylog(2, (x + 1)/2)/(x*(x + 1)), x)   
      
   INT(polylog(2, x)/(x*(x + 1)), x)   
    - INT(polylog(2, 1 + 1/x)/(x*(x + 1)), x)   
    = - LOG(2)*LOG((1 + x)/2)*LOG((1 + x)/(2*x))   
    + 1/2*LOG(1 - x)*LOG((1 + x)/(2*x))^2   
    + 1/2*LOG(- 1/x)*LOG((1 + x)/(2*x))^2   
    - 1/2*LOG((-1 + x)/(2*x))*LOG((1 + x)/(2*x))^2   
    + LOG((1 + x)/(2*x))*polylog(2, 1 + 1/x)   
    - LOG((1 + x)/(2*x))*polylog(2, x)   
    + LOG((1 + x)/(2*x))*polylog(2, (1 + x)/2)   
    - LOG((1 + x)/(2*x))*polylog(2, (1 + x)/(2*x))   
    + polylog(3, (1 + x)/(2*x))   
    - INT(LOG(2)*LOG((x + 1)/2)/(x*(x + 1)), x)   
    + INT(polylog(2, (x + 1)/2)/(x*(x + 1)), x)   
      
   The final two integrals drop out. Still, his may be hard to cast into a   
   sequence of goal-directed rewrite rules. Perhaps Rubi should adopt the   
   generic Mathematica antiderivative wholesale?   
      
   Martin.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)