Albert Rich schrieb:   
   >   
   > On Friday, February 16, 2018 at 10:26:51 PM UTC-10, clicl...@freenet.de   
   wrote:   
   > >   
   > > clicliclic@freenet.de schrieb:   
   > >   
   > > Meanwhile I have found the time to work out the (rather an) answer:   
   > >   
   > > INT(x/((x^3 + 10 + 6*SQRT(3))*SQRT(x^3 + 1)), x) =   
   > > SQRT(2)*3^(3/4)*(2 - SQRT(3))/36   
   > > *(ATANH(12^(1/4)*(2*x - SQRT(3) - 1)/(2*SQRT(x^3 + 1)))   
   > >  + ATANH(SQRT(2)*3^(3/4)*(SQRT(3) - 1)*(x + 1)^2/(6*SQRT(x^3 + 1))))   
   > >  + 12^(1/4)*(2 - SQRT(3))/36   
   > > *(2*ATAN(12^(1/4)*(SQRT(3) - 1)*SQRT(x^3 + 1)/6)   
   > >  - 2*ATAN(12^(1/4)*(SQRT(3) + 1)*(x + 1)/(2*SQRT(x^3 + 1)))   
   > >  - ATAN(12^(1/4)*(2*x + SQRT(3) - 1)/(2*SQRT(x^3 + 1)))   
   > >  + ATAN(SQRT(2)*3^(3/4)*(SQRT(3) + 1)*(x + 1)^2/(6*SQRT(x^3 + 1))))   
   > >   
   > > Be warned that this antiderivative has a few jumps still. The length   
   > > without spaces is less than 400 Bytes. Compare this with the FriCAS   
   > > result! I suppose the latter jumps too?   
   > >   
   >   
   > Your last 3 arctan terms for the antiderivative of   
   > x/((10+6*sqrt(3)+x^3)*sqrt(1+x^3)) can be merged into one, yielding   
   > the 4 term antiderivative   
   >   
   > ArcTan[((-1 + Sqrt[3])*Sqrt[1 + x^3])/(Sqrt[2]*3^(3/4))]/   
   >   (3*Sqrt[72 + 42*Sqrt[3]]) +   
   > ArcTan[(Sqrt[2 - Sqrt[3]]*Sqrt[1 + x^3])/(3^(1/4)*(1 + x))]/   
   >   (2*Sqrt[72 + 42*Sqrt[3]]) -   
   > ArcTanh[(3^(1/4)*(1 + Sqrt[3] - 2*x))/(Sqrt[2]*Sqrt[1 + x^3])]/   
   >   (6*Sqrt[24 + 14*Sqrt[3]]) +   
   > ArcTanh[((-1 + Sqrt[3])*(1 + x)^2)/(Sqrt[2]*3^(1/4)*Sqrt[1 + x^3])]/   
   >   (6*Sqrt[24 + 14*Sqrt[3]])   
   >   
   > Although simpler, this antiderivative is also misbehaved on the real   
   > line, so not yet optimal...   
   >   
   > Also I think a simpler split of the integrand is   
   >   
   > x^2/   
   >   (2*(1+Sqrt[3])*Sqrt[1+x^3]*(10+6*Sqrt[3]+x^3)) +   
   > ((2*(1+Sqrt[3])-x)*x)/   
   >   (2*(1+Sqrt[3])*Sqrt[1+x^3]*(10+6*Sqrt[3]+x^3))   
   >   
   > since the first term of the split trivially integrates using the   
   > substitution t=x^3 to the first term of the antiderivative above.   
   >   
      
   Excellent observation! I had overlooked that (differentially):   
      
   ATAN(SQRT(2)*3^(3/4)*(SQRT(3) + 1)*(x + 1)^2/(6*SQRT(x^3 + 1)))   
    = 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)))   
      
   by use of which three of the ATANs collapse into one. Similarly one has   
   (differentially):   
      
   ATANH(SQRT(2)*3^(3/4)*(SQRT(3) - 1)*(x + 1)^2/(6*SQRT(x^3 + 1)))   
    = 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)))   
      
   by use of which the two ATANHs can be rewritten, though neither their   
   number be reduced nor the jump at x = SQRT(3) - 1 be removed.   
      
   The length of the simplified antiderivative:   
      
   INT(x/((x^3 + 10 + 6*SQRT(3))*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))))   
      
   without spaces is 252 Bytes. Beat that!   
      
   The integrand splitting was as prescribed by Goursat (immediately below   
   his eq. 5), where I used the one and only real K_4 = D_2 Möbius mapping   
   among the radicand zeros. This prescription is guaranteed to work,   
   relying solely on symmetry properties of the integrand. One can often   
   do better once the antiderivative is known and broken onto components.   
      
   Martin.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)