antispam@math.uni.wroc.pl schrieb:   
   >   
   > clicliclic@freenet.de wrote:   
   > >   
   > > antispam@math.uni.wroc.pl schrieb:   
   > > >   
   > > > clicliclic@freenet.de wrote:   
   > > > >   
   > > > > 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 Thanks to the (still unpublished?)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.   
   > > >   
   > > > To be clear: above P is fixed.  And b is a function of a (there is   
   > > > at most one b for which the integral is elementary).  I think that   
   > > > b can be given as a simple expression, but it would take extra   
   > > > effort to find it.  If you are really bothered by b you can pass   
   > > > to Jacobi form, that is take P = (1 - x^2)*(1 - m*x^2).  Then   
   > > > b = 0.   
   > > >   
   > >   
   > > There is no need to fix P = (x-u)*(x-v)*(x-w) in INT((1/(x-a) - b)/   
   > > SQRT(P), x). But the search for pseudo-elliptic instances can be   
   > > narrowed down to u = 0, v = 1 without loss of generality: any linear   
   > > transformation x' = p*x + q just takes the integrand into some   
   > > (1/(x'-a') - b')/SQRT(P') of the same form with P' = (x'-u')*(x'-v')*   
   > > (x'-w').   
   >   
   > My point is that you can find solutions for _any_ P of allowed   
   > form.  Of course which 'a' lead to integrable case depend on P.   
   > But for each fixed P you get coutable set of 'a'.  This is   
   > answer to Albert question: can finite number of rules cover   
   > integrals of form:   
   >   
   > integrate(1/((x - a)*sqrt(P)), x)   
   >   
   > When P is of form (1 - x^2)*(1 - m*x^2) the answer is: no   
   > finite system of reasonable rules will cover cases integrable   
   > using elementary functions.  Namely, assuming polynomial type   
   > conditions on 'a' each rule can catch only finite number of   
   > 'a': there are infintely many nonitegrable cases so polynomial   
   > can be zero only for finite number of 'a'.  Also, there   
   > are infintely many different forms of integral.  More preciely   
   > you can write integral as   
   >   
   > log((p + q*sqrt(P))/(p + q*sqrt(P)))   
   >   
   > where p and q are polynomials, but there is no bound on   
   > degree.  In fact degrees must grow to infinity.   
   >   
   > For P of degree 3 with no multiple factors no finite system   
   > of resonable rules will cover all case integrable using   
   > elementary functions for family:   
   >   
   > integrate((1/(x - a) - b)*(1/sqrt(P)), x)   
   >   
   > If you insist to handle only 'b = 0' than in general   
   > the integral   
   >   
   > integrate(1/((x - a)*sqrt(P)), x)   
   >   
   > need elliptic integral F.  However, if you insist on "optimal"   
   > integral than presumably you will be not satisfied with   
   > generic answer which is sum of elliptic integral F and   
   > elliptic integral Pi and you will want logarithm in cases   
   > when elliptic integral Pi is elementary.  Again, no finite   
   > system of rules is enough to cover cases when integral F   
   > + logarith is elementary.   
   >   
   > And of course special role of 'b' somewhat shakes Rubi   
   > belief that it can split integrals at will: splitting   
   >   
   > integrate((1/(x - a) - b)*(1/sqrt(P)), x)   
   >   
   > will give   
   >   
   > integrate((1/((x - a)*sqrt(P)), x)   
   >   
   > and   
   >   
   > integrate(b/sqrt(P), x)   
   >   
   > The second one always leads to elliptic integral F.   
   > If the first one gives elliptic integral F + elementary   
   > part than F parts can cancel.  But if you take easy   
   > way of generating elliptic integral Pi, then it will   
   > remain in final result, even for cases when sum is   
   > elementary.   
   >   
      
   There is no doubt that a Rubi-type integrator cannot detect all of the   
   pseudo-elliptic cases of arbitrary elliptic integrands and return their   
   elementary antiderivatives. It is nonetheless not silly in my view to   
   equip Rubi with rules that handle a finite number of the simpler   
   pseudo-elliptic cases (say those covered by Goursat's approach) for   
   certain simple integrand patterns - even if infinitely many such cases   
   (of increasing complexity) are known to exist for them.   
      
   Meanwhile Goursat's conditions for your integrand 1/((x - a)*sqrt(P))   
   with P = (x-u)*(x-v)*(x-w) have been solved assuming u=0, v=1, without   
   loss of generality. Three pseudo-elliptic cases result, and according   
   to Nasser, FriCAS can handle two of them all the time, and the third   
   one 50% of the time.   
      
   For P = (1 - x^2)*(1 - m*x^2), Goursat's approach detects the trivial   
   case of a=0 only. I expect Rubi to be able to handle this already.   
      
   Martin.   
      
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