From: Albert_Rich@msn.com   
      
   On Thursday, January 25, 2018 at 10:18:57 PM UTC-10, clicl...@freenet.de wrote:   
   > >   
   > > As promised it returns optimal elementary antiderivatives for   
   > > pseudo-elliptic integrands of the form x/((a+b*x^3)*sqrt(c+d*x^3))   
   > > when 4*b*c-a*d=0 or when 8*b*c+a*d=0; and of the form   
   > > 1/((a+b*x^2)^(1/3)*(c+d*x^2)) when b*c+3*a*d=0 or when b*c-9*a*d=0.   
   > > Of course, it also returns such antiderivatives when reduction rules   
   > > reduce integrands to any of these forms.   
   >   
   > 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))))   
   >   
   > Martin.   
      
   For the above four integrals, Rubi 4.14.5 returns the following    
   ntiderivatives (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?   
      
   Albert   
      
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