clicliclic@freenet.de schrieb:   
   >   
   > antispam@math.uni.wroc.pl schrieb:   
   > >   
   > > My calculations indicate that points of order 3 correspond to   
   > > u^2 = 1 +- 2*sqrt(3)/3.  FriCAS says that integal is nonelementary,   
   > > so something is wrong.  Theory behind example is simple, so   
   > > I believe it.  I am not sure about points, but this is probably   
   > > correct.  So it remains to check FriCAS :(   
   > >   
   >   
   > A simple theory should allow to read off simple antiderivatives for   
   > the order-3 integrands.   
   >   
   > I suppose you have tried the integral on Axiom as well - the problem   
   > would have to be quite old then.   
   >   
   > 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) = ?   
   >   
   > So your result for order 3 is confirmed. I derived these x^2 - u^2 from   
   > a recursive definition of the division polynomials and then checked the   
   > points [u,v] by iterating the group operation.   
   >   
   > I should mention that my statements about Goursat integrability   
   > referred to Moebius mappings realizing the dihedral group D_2   
   > (isomorphic to the Klein four-group K_4) only; the special nature of   
   > the radicand x^3 - x may open other possibilities (thus, x <-> 1/x   
   > swaps two roots and preserves the remaining two). Compare Goursat pp.   
   > 113-118 for x^3 - 1.   
   >   
      
   Actually, Goursat considered the special radicand x^3 - x as well (on   
   pp. 118-119) and concludes that no special pseudo-elliptic cases arise.   
   I tend to believe him.   
      
   The substitution x <-> 1/x takes the integral into itself, with u   
   replaced by 1/u. Thus my pair of order-6 integrals is equivalent to the   
   order-3 one.   
      
   Meanwhile I can confirm that the order-3 integrals are elementary. The   
   antiderivatives take less than 1000 Bytes to write down.   
      
   Martin.   
      
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