Albert Rich schrieb:   
   >   
   > On Thursday, January 25, 2018 at 10:18:57 PM UTC-10, clicl...@freenet.de   
   wrote:   
   > >   
   > > For the latter foursome (just in case that one or the other had not   
   > > been posted before) I have this:   
   > >   
   > > INT(1/((3 - x^2)*(1 - 3*x^2)^(1/3)), x)   
   > >  = SQRT(3)/48*(LN((1 - SQRT(3)*x)^3 + 8*(1 - 3*x^2))   
   > >  - 3*LN((1 - SQRT(3)*x) + 2*(1 - 3*x^2)^(1/3))   
   > >  + 2*SQRT(3)*ATAN(SQRT(3)/3*(1 - (1 - SQRT(3)*x)/(1 - 3*x^2)^(1/3))))   
   > >  - SQRT(3)/48*(LN((1 + SQRT(3)*x)^3 + 8*(1 - 3*x^2))   
   > >  - 3*LN((1 + SQRT(3)*x) + 2*(1 - 3*x^2)^(1/3))   
   > >  + 2*SQRT(3)*ATAN(1/SQRT(3)*(1 - (1 + SQRT(3)*x)/(1 - 3*x^2)^(1/3))))   
   > >   
   > > INT(1/((3 + x^2)*(1 + 3*x^2)^(1/3)), x)   
   > >  = 1/24*(LN((1 + x)^3 - (1 + 3*x^2))   
   > >  - 3*LN((1 + x) - (1 + 3*x^2)^(1/3))   
   > >  + 2*SQRT(3)*ATAN(1/SQRT(3)*(1 + 2*((1 + x)/(1 + 3*x^2)^(1/3)))))   
   > >  - 1/24*(LN((1 + 3*x^2) - (1 - x)^3)   
   > >  - 3*LN((1 + 3*x^2)^(1/3) - (1 - x))   
   > >  + 2*SQRT(3)*ATAN(1/SQRT(3)*(1 + 2*((1 - x)/(1 + 3*x^2)^(1/3)))))   
   > >   
   > > INT(1/((3 + x^2)*(1 - x^2)^(1/3)), x)   
   > >  = 2^(1/3)/24*(LN((1 - x)^3 + 2*(1 - x^2))   
   > >  - 3*LN((1 - x) + 2^(1/3)*(1 - x^2)^(1/3))   
   > >  + 2*SQRT(3)*ATAN(1/SQRT(3)*(1 - 2^(2/3)*((1 - x)/(1 - x^2)^(1/3)))))   
   > >  - 2^(1/3)/24*(LN((1 + x)^3 + 2*(1 - x^2))   
   > >  - 3*LN((1 + x) + 2^(1/3)*(1 - x^2)^(1/3))   
   > >  + 2*SQRT(3)*ATAN(1/SQRT(3)*(1 - 2^(2/3)*((1 + x)/(1 - x^2)^(1/3)))))   
   > >   
   > > INT(1/((3 - x^2)*(1 + x^2)^(1/3)), x)   
   > >  = SQRT(3)*2^(1/3)/72*(LN(6*SQRT(3)*(1 + x^2) - (SQRT(3) + x)^3)   
   > >  - 3*LN((6*SQRT(3))^(1/3)*(1 + x^2)^(1/3) - (SQRT(3) + x))   
   > >  + 2*SQRT(3)*ATAN(1/SQRT(3) + 2^(2/3)*(SQRT(3) + x)   
   > >    /(3*(1 + x^2)^(1/3))))   
   > >  - SQRT(3)*2^(1/3)/72*(LN(6*SQRT(3)*(1 + x^2) - (SQRT(3) - x)^3)   
   > >  - 3*LN((6*SQRT(3))^(1/3)*(1 + x^2)^(1/3) - (SQRT(3) - x))   
   > >  + 2*SQRT(3)*ATAN(1/SQRT(3) + 2^(2/3)*(SQRT(3) - x)   
   > >    /(3*(1 + x^2)^(1/3))))   
   > >   
   >   
   > For the above four integrals, Rubi 4.14.5 returns the following   
   antiderivatives (in Mathematica syntax):   
   >   
   > (1/4)*ArcTan[(1 - (1 - 3*x^2)^(1/3))/x] +   
   >   ArcTanh[x/Sqrt[3]]/(4*Sqrt[3]) -   
   >   ArcTanh[(1 - (1 - 3*x^2)^(1/3))^2/(3*Sqrt[3]*x)]/(4*Sqrt[3])}   
   >   
   > ArcTan[x/Sqrt[3]]/(4*Sqrt[3]) +   
   >   ArcTan[(1 - (1 + 3*x^2)^(1/3))^2/(3*Sqrt[3]*x)]/(4*Sqrt[3]) -   
   >   (1/4)*ArcTanh[(1 - (1 + 3*x^2)^(1/3))/x]}   
   >   
   > ArcTan[Sqrt[3]/x]/(2*2^(2/3)*Sqrt[3]) +   
   >   ArcTan[(Sqrt[3]*(1 - 2^(1/3)*(1 - x^2)^(1/3)))/x]/(2*2^(2/3)*Sqrt[3]) -   
   >   ArcTanh[x]/(6*2^(2/3)) +   
   >   ArcTanh[x/(1 + 2^(1/3)*(1 - x^2)^(1/3))]/(2*2^(2/3))   
   >   
   > -(ArcTan[x]/(6*2^(2/3))) +   
   >   ArcTan[x/(1 + 2^(1/3)*(1 + x^2)^(1/3))]/(2*2^(2/3)) -   
   >   ArcTanh[Sqrt[3]/x]/(2*2^(2/3)*Sqrt[3]) -   
   >   ArcTanh[(Sqrt[3]*(1 - 2^(1/3)*(1 + x^2)^(1/3)))/x]/(2*2^(2/3)*Sqrt[3])   
   >   
   > They were easily derived using substitution to transform integrands of   
   > the form 1/((a+b*x^2)^(1/3)*(c+d*x^2)) to the pseudo-elliptic form   
   > x/((a+b*x^3)*sqrt(c+d*x^3)), and then using the 3 or 4 arctan(h)   
   > antiderivatives you provided earlier.   
   >   
   > Is there anything wrong with these significantly simpler   
   > antiderivatives?   
   >   
      
   They look alright when plotted on Derive, except for the third one,   
   whose real part jumps at x^2 = 5. (The imaginary part also jumps at the   
   algebraic singularities x^2 = 1.)   
      
   Martin.   
      
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