antispam@math.uni.wroc.pl schrieb:   
   >   
   > clicliclic@freenet.de wrote:   
   > >   
   > > Thanks to the (still unpublished?) Masser-Zanier counterexample   
   > > Waldek informed us about, we know that there are infinitely many   
   > > (though increasingly complicated algebraic) u^2 for which   
   > >   
   > >   INT(x/((x^2 - u^2)*SQRT(x^3 - x)), x)   
   > >   
   > > has a solution (also of increasing complexity) in terms of   
   > > elementary functions. Consequently, via the usual Moebius   
   > > transformation of the integration variable, there are also   
   > > infinitely many v = (u^2 - 1)/(u^2 + 1) for which   
   > >   
   > >   INT((y^2 - 1)/((y^2 + 2*v*y + 1)*SQRT(y^4 - 1)), y)   
   > >   
   > > has such an elementary solution. I too expect these patterns to be   
   > > exceptions rather than the rule, the rule being that the number of   
   > > pseudo-elliptic cases is finite and small. But does this pair   
   > > already exhaust the store of patterns involving simple square-root   
   > > radicands and giving rise to infinitely many pseudo-elliptic   
   > > integrals?   
   >   
   > Well, any torsion point leads to a pseudo-elliptic integral.  Over   
   > algebraically closed field there are infintely many torsion points   
   > on any curve.  So example with _single_ parameter may be exceptional,   
   > put once you allow algebraic extentions there is see of strange   
   > examples -- just two parameters family:   
   >   
   > integrate((1/(x - a)  - b)/sqrt(P), x)   
   >   
   > has no elementary integral valid for continuous family of a and b   
   > and infintely many integrable cases when P is a polynomial of   
   > degree 3 without multiple factors.   
   >   
      
   Yes, my comment was about the dependence on a single parameter, with any   
   additional parameters kept fixed. Taking P = (x - u)*(x - v)*(x - w),   
   Goursat pseudo-elliptic instances of your integral correspond to the   
   solutions of:   
      
   4*a^3*b+a^2*(3-4*b*(u+v+w))+2*a*(2*b*(u*(v+w)+v*w)-u-v-w)-4*b*u*~   
   v*w+u*(v+w)+v*w=0 AND -4*(a^4*b+a^3-2*a^2*b*(u*(v+w)+v*w)+a*(8*b~   
   *u*v*w-u*(v+w)-v*w)+b*(u^2*(v^2-2*v*w+w^2)-2*u*v*w*(v+w)+v^2*w^2~   
   )+2*u*v*w)=0 AND 2*(2*a^4*b*(u+v+w)-2*a^3*(2*b*(u*(v+w)+v*w)-u-v~   
   -w)-3*a^2*(u*(v+w)+v*w)-4*a*b*(u^2*(v^2-2*v*w+w^2)-2*u*v*w*(v+w)~   
   +v^2*w^2)+2*b*(u^3*(v^2-2*v*w+w^2)+u^2*(v+w)*(v^2-2*v*w+w^2)-u*v~   
   *w*(2*v^2+v*w+2*w^2)+v^2*w^2*(v+w))-u^2*(v^2-2*v*w+w^2)+2*u*v*w*~   
   (v+w)-v^2*w^2)=0 AND -4*(a^4*b*(u*(v+w)+v*w)-a^3*(8*b*u*v*w-u*(v~   
   +w)-v*w)-2*a^2*(b*(u^2*(v^2-2*v*w+w^2)-2*u*v*w*(v+w)+v^2*w^2)+3*~   
   u*v*w)-a*(u^2*(v^2-2*v*w+w^2)-2*u*v*w*(v+w)+v^2*w^2)+b*(u^3*(v+w~   
   )*(v^2-2*v*w+w^2)-u^2*v*w*(v^2-2*v*w+w^2)-u*v^2*w^2*(v+w)+v^3*w^~   
   3))=0 AND 4*a^4*b*u*v*w+4*a^3*(b*(u^2*(v^2-2*v*w+w^2)-2*u*v*w*(v~   
   +w)+v^2*w^2)+u*v*w)+a^2*(3*(u^2*(v^2-2*v*w+w^2)-2*u*v*w*(v+w)+v^~   
   2*w^2)-4*b*(u^3*(v^2-2*v*w+w^2)+u^2*(v^3-v^2*w-v*w^2+w^3)-u*v*w*~   
   (2*v^2+v*w+2*w^2)+v^2*w^2*(v+w)))+2*a*(2*b*(u^3*(v^3-v^2*w-v*w^2~   
   +w^3)-u^2*v*w*(v^2-2*v*w+w^2)-u*v^2*w^2*(v+w)+v^3*w^3)-u^3*(v^2-~   
   2*v*w+w^2)-u^2*(v^3-v^2*w-v*w^2+w^3)+u*v*w*(2*v^2+v*w+2*w^2)-v^2~   
   *w^2*(v+w))+u^3*(v^3-v^2*w-v*w^2+w^3)-u^2*v*w*(v^2-2*v*w+w^2)-u*~   
   v^2*w^2*(v+w)+v^3*w^3=0   
      
   which earlier experience suggests to correspond to torsion points of   
   order four. Can anybody determine (some or all) solutions to the five   
   polynomial equations in five unknowns? FriCAS could subsequently solve   
   the corresponding integrals, I hope.   
      
   But while the latest FriCAS is performing well on square-root integrands   
   even for higher-order points, the system is not yet prepared to handle   
   cube-root pseudo-elliptics reliably, such as my infamous   
      
     INT((1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)), x)   
      
   Martin.   
      
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