Albert Rich schrieb:   
   >   
   > On Tuesday, January 31, 2017 at 6:40:08 AM UTC-10, clicl...@freenet.de wrote:   
   > >   
   > > This can still be simplified considerably:   
   > >   
   > >   1/2*ATANH(x*(1 - x^2)/SQRT(1 - x^4))   
   > >   + 1/2*ATAN(x*(1 + x^2)/SQRT(1 - x^4))   
   > >   
   >   
   > Generalizing the above for integrands of the form   
   > sqrt(a+b*x^4)/(c+d*x^4) when a*d+b*c=0 yields the following   
   > antiderivative:   
   >   
   > a/(2*c*(-a*b)^(1/4))*   
   >     arctan((-a*b)^(1/4)*x*(a+sqrt(-a*b)*x^2)/(a*sqrt(a+b*x^4))) +   
   > a/(2*c*(-a*b)^(1/4))*   
   >     arctanh((-a*b)^(1/4)*x*(a-sqrt(-a*b)*x^2)/(a*sqrt(a+b*x^4)))   
   >   
   > The next release of Rubi will use this rule when a*b<0. Any ideas for   
   > when a*b>0?   
   >   
      
   In view of its ability to handle Timofeev's Chapter 4 Examples 113-114   
   and 117-118, Rubi knows enough already:   
      
     INT(SQRT(1 + p*x^2 + x^4)/(1 - x^4), x)   
     = SQRT(2 + p)/4*ATANH(SQRT(2 + p)*x/SQRT(1 + p*x^2 + x^4))   
     + SQRT(2 - p)/4*ATAN(SQRT(2 - p)*x/SQRT(1 + p*x^2 + x^4))   
      
   Watching with amazement how your decision tree sprouts more and more   
   pseudo-elliptic leaves,   
      
   Martin.   
      
   PS: Your current elliptic antiderivative contains a sum of canonical   
   integrals of the third kind with complex-conjugate first arguments. But   
   according to Legendre's "Traité", Tome I, Chapitre XXIV, such sums can   
   be always re-expressed in real terms. Free djvu versions of all three   
   volumes are offered under .   
      
   --- SoupGate-Win32 v1.05   
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