Albert Rich schrieb:   
   >   
   > On Monday, February 19, 2018 at 7:26:06 AM UTC-10, Nasser M. Abbasi wrote:   
   > > Rubi 4.14.6  can integrate   
   > >   
   > >         (x Exp[x])/(1 - Exp[x])     (1)   
   > >   
   > > but not   
   > >   
   > >         (x Exp[x])/(1 - Exp[-x])    (2)   
   > >   
   > > I was just wondering why that is. Could not figure what changes   
   > > in rules needed. Since the result are very similar (in terms   
   > > of PolyLog, which Rubi supprts). For example, for (1) it   
   > > gives   
   > >   
   > > In[59]:= rubi = Int[(x Exp[x])/(1 - Exp[x]), x]   
   > > Out[59]= -x Log[1 - E^x] - PolyLog[2, E^x]   
   > >   
   > > But for (2) it does not evaluate it. Mathematica gives for (2)   
   > >   
   > > In[60]:= mma = Integrate[(x Exp[x])/(1 - Exp[-x]), x]   
   > > Out[60]= E^x (-1 + x) + x Log[1 - E^x] + PolyLog[2, E^x]   
   > >   
   > > thanks   
   > > --Nasser   
   >   
   > Thank you for reporting this embarrassing deficiency in Rubi. The just   
   > released version 4.14.7 includes a new rule which algebraically   
   > expands your example integrand into x*E^x+x/(1-E^(-x)). This sum is   
   > then integrated in 7 steps using existing rules.   
   >   
   > Among other enhancements, Rubi 4.14.7 can find elementary   
   > antiderivatives for integrands of the form x/((c+d*x^3)*sqrt(a+b*x^3))   
   > when b*c-4*a*d=0, or b*c+8*a*d=0, or b*c-2*(5+3*sqrt(3))*a*d=0.  These   
   > pseudo-elliptic integrals are discussed in the sci.math.symbolic   
   > thread "More teething help".   
   >   
   > I am still trying to find optimal antiderivatives for when   
   > b*c-2*(5-3*sqrt(3))*a*d=0. The best so far has 3 arctan(h) terms, but   
   > involves the imaginary unit. Maybe some system or person can do   
   > better...   
   >   
      
   Oops, are you perhaps still looking for some of these:   
      
   INT(x/((x^3 + 6*SQRT(3) + 10)*SQRT(x^3 + 1)), x)   
    = SQRT(2)*3^(3/4)*(2 - SQRT(3))/36*   
   (2*ATANH(12^(1/4)*(2*x - SQRT(3) - 1)/(2*SQRT(x^3 + 1)))   
    + ATANH(12^(1/4)*(SQRT(3) - 1)*(x + 1)/(2*SQRT(x^3 + 1))))   
    + 12^(1/4)*(2 - SQRT(3))/36*   
   (2*ATAN(12^(1/4)*(SQRT(3) - 1)*SQRT(x^3 + 1)/6)   
    - 3*ATAN(12^(1/4)*(SQRT(3) + 1)*(x + 1)/(2*SQRT(x^3 + 1))))   
      
   INT(x/((x^3 - 6*SQRT(3) + 10)*SQRT(x^3 + 1)), x)   
    = 12^(1/4)*(2 + SQRT(3))/36*   
   (2*ATANH(12^(1/4)*(SQRT(3) + 1)*SQRT(x^3 + 1)/6)   
    - 3*ATANH(12^(1/4)*(SQRT(3) - 1)*(x + 1)/(2*SQRT(x^3 + 1))))   
    + SQRT(2)*3^(3/4)*(2 + SQRT(3))/36*   
   (2*ATAN(12^(1/4)*(2*x + SQRT(3) - 1)/(2*SQRT(x^3 + 1)))   
    - ATAN(12^(1/4)*(SQRT(3) + 1)*(x + 1)/(2*SQRT(x^3 + 1))))   
      
   INT(x/((x^3 - 6*SQRT(3) - 10)*SQRT(x^3 - 1)), x)   
    = 12^(1/4)*(2 - SQRT(3))/36*   
   (2*ATANH(12^(1/4)*(SQRT(3) - 1)*SQRT(x^3 - 1)/6)   
    - 3*ATANH(12^(1/4)*(SQRT(3) + 1)*(x - 1)/(2*SQRT(x^3 - 1))))   
    + SQRT(2)*3^(3/4)*(2 - SQRT(3))/36*   
   (2*ATAN(12^(1/4)*(2*x + SQRT(3) + 1)/(2*SQRT(x^3 - 1)))   
    + ATAN(12^(1/4)*(SQRT(3) - 1)*(x - 1)/(2*SQRT(x^3 - 1))))   
      
   INT(x/((x^3 + 6*SQRT(3) - 10)*SQRT(x^3 - 1)), x)   
    = SQRT(2)*3^(3/4)*(2 + SQRT(3))/36*   
   (2*ATANH(12^(1/4)*(2*x - SQRT(3) + 1)/(2*SQRT(x^3 - 1)))   
    - ATANH(12^(1/4)*(SQRT(3) + 1)*(x - 1)/(2*SQRT(x^3 - 1))))   
    + 12^(1/4)*(2 + SQRT(3))/36*   
   (2*ATAN(12^(1/4)*(SQRT(3) + 1)*SQRT(x^3 - 1)/6)   
    - 3*ATAN(12^(1/4)*(SQRT(3) - 1)*(x - 1)/(2*SQRT(x^3 - 1))))   
      
   ? To flip the sign of SQRT(3), one must systematically replace 3^(1/4)   
   by #i*3^(1/4), SQRT(3) by -SQRT(3), and 3^(3/4) by -#i*3^(3/4) in both   
   the integrand and the evaluation. Risch's disciples will know why.   
      
   Or are you trying to exorcise jumps with imaginary units?   
      
   Martin.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)