From: nma@12000.org   
      
   On 1/26/2018 2:19 AM, clicliclic@freenet.de wrote:   
      
   > 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.   
   >   
      
   I am not good at GroebnerBasis, but using it and lots   
   of manual steps, it is possible to find many solutions.   
      
   For example here are two solution   
      
   a=1; w=2; b=1/2; u=3; v=3;   
   a=1; w=1; b=5; u=11/10; v=1;   
      
      
   ClearAll[a,b,u,v,w]   
   parms1={a->1,w->2,b->1/2,u->3,v->3};   
   parms2={a->1,w->1,b->5,u->11/10,v->1};   
      
   eq1=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*   
   *u*v*w+u*(v+w)+v*w==0 ;   
      
   eq2=-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 ;   
      
   eq3= 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 ;   
      
   eq4=-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 ;   
      
   eq5= 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+   
   *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*   
   ^2*w^2*(v+w)+v^3*w^3==0;   
      
   In[349]:= {eq1,eq2,eq3,eq4,eq5}/.parms1   
   Out[349]= {True,True,True,True,True}   
      
   In[350]:= {eq1,eq2,eq3,eq4,eq5}/.parms2   
   Out[350]= {True,True,True,True,True}   
      
   --Nasser   
      
   --- SoupGate-Win32 v1.05   
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