On Sunday, May 3, 2020 at 10:09:47 PM UTC+10, clicl...@freenet.de wrote:   
   > samuel.thomas.blake@gmail.com schrieb:   
   > >    
   > > On Sunday, May 3, 2020 at 8:12:30 PM UTC+10, clicl...@freenet.de wrote:   
   > > >   
   > > > Here is a pair of Kauers candidates involving 4th roots:   
   > > >   
   > > >   integrate((3 - x^2)/((1 - x^2)*(1 - 6*x^2 + x^4)^(1/4)), x)   
   > > >   
   > > >   integrate((1 + x^2)^2/((1 - x^2)*(1 - 6*x^2 + x^4)^(3/4)), x)   
   > > >   
   > > > which also causes FriCAS 1.3.6 to squawk:   
   > > >   
   > > > >> Error detected within library code:   
   > > >    integrate: implementation incomplete   
   > > >    (residue poly has multiple non-linear factors)   
   > > >   
   > >    
   > > These are two very nice integrals. Kauer's package (at least the   
   > > version I use which uses Mathematica's GroebnerBasis routine in place   
   > > of calling Singular) cannot integrate these.   
   > >    
   > > What substitution do you propose to compute these integrals?   
   > >    
   >    
   > These two are from a posthumous publication by Euler (Nova Acta   
   > Academiae Scientiarum Imperialis Petropolitanae IX, 1795 for 1791, pp.   
   > 118-126 and 127-131); they must be split for substitution to work.   
   > Simple but complex solutions are:   
   >    
   > INT((3 - x^2)/((1 - x^2)*(1 - 6*x^2 + x^4)^(1/4)), x) =    
   > ATANH((x + #i)/(1 - 6*x^2 + x^4)^(1/4))   
   >  + ATAN((x + #i)/(1 - 6*x^2 + x^4)^(1/4))   
   >  + ATANH((x - #i)/(1 - 6*x^2 + x^4)^(1/4))   
   >  + ATAN((x - #i)/(1 - 6*x^2 + x^4)^(1/4))   
   >    
   > and:   
   >    
   > INT((1 + x^2)^2/((1 - x^2)*(1 - 6*x^2 + x^4)^(3/4)), x) =   
   >  - ATANH((x + #i)/(1 - 6*x^2 + x^4)^(1/4))   
   >  + ATAN((x + #i)/(1 - 6*x^2 + x^4)^(1/4))   
   >  - ATANH((x - #i)/(1 - 6*x^2 + x^4)^(1/4))   
   >  + ATAN((x - #i)/(1 - 6*x^2 + x^4)^(1/4))   
   >    
   > Clearly two components suffice.   
   >    
   > The Kauers-solvable cube-root integrals posted earlier were (in    
   > essence) by Euler and by Legendre:   
   >    
   > INT((a + b*x)/((3 - x^2)*(1 - 3*x^2)^(1/3)), x) =    
   > SQRT(3)*(a - SQRT(3)*b)/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)*(a + SQRT(3)*b)/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))))   
   >    
   > INT((a + b*x)/((3 + x^2)*(1 - x^2)^(1/3)), x) =    
   > 2^(1/3)*(a + 3*b)/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)*(a - 3*b)/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)))))   
   >    
   > These solutions are real; again two components suffice. The case a=0 is   
   > trivial; Euler and Legendre considered the case b=0. Compare Legendre's   
   > famous Traité des fonctions elliptiques I, 1825, Ch. XXVI, pp. 165-175   
   > (untranslated into English, I believe).    
   >    
   > Martin.   
      
   Hi Martin,    
      
   Can you please double check those solutions?    
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)