clicliclic@freenet.de wrote:   
   >   
   > There can be no doubt that something is wrong with FriCAS since Gauss   
   > has provided me with a hand-written finite elementary evaluation of one   
   > order-3 integral. And I find his evaluation to hold up.   
      
   > > > The points of order 3, 6 and 8 determined by me correspond to the   
   > > > following three pairs of integrals:   
   > > >   
   > > > INT(x/((3*x^2 + 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?   
   > > >   
   > > > INT(x/((3*x^2 - 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?   
   > > >   
   > > > INT(x/((x^2 + 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?   
   > > >   
   > > > INT(x/((x^2 - 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?   
   > > >   
   > > > INT(x/((x^2 + 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),   
   > > > x) = ?   
   > > >   
   > > > INT(x/((x^2 - 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),   
   > > > x) = ?   
      
   FriCAS had a silly bug due to which it treated correctly computed   
   integral as nonelementary.  For finding the bug it did not help   
   that point that you gave in integral for order 3 and 6 were   
   wrong (you wrote SQRT(2) while SQRT(3) is correct).   
      
   > Thanks for the details. As noted above, the special denominators x^2 -   
   > u^2 appears as factors of the so-called division polynomials for the   
   > elliptic curve v^2 = u^3 - u:   
      
      
   Using division polynomials I have found points of order 5,   
   one corresponds to u^2 = (1 + 2*sqrt(-1))/5.  FriCAS have   
   now computed integrals for orders 3, 5, 6, 8.  The results   
   are rather lengthy, so instead of posting them here I have   
   put them at:   
      
   http://www.math.uni.wroc.pl/~hebisch/fricas/p3   
      
   (3 above means order 3, replace 3 by 5, 6, 8 for higher order).   
      
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                                 Waldek Hebisch   
      
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