Waldek Hebisch schrieb:   
   >   
   > clicliclic@freenet.de wrote:   
   > >   
   > > clicliclic@freenet.de schrieb:   
   > > >   
   > > > integrate((2 - (k + 1)*x)/((1 - (k + 1)*x)   
   > > >   *(x*(1 - x)*(1 - k*x))^(1/3)), x)   
   > > >   
   > > > >> Error detected within library code:   
   > > >    impossible   
   > > >   
   > > > It is good to be cognizant of one's limitations ;).   
   > > >   
   > >   
   > > ... and another quick FriCAS 1.2.3 on-line experiment:   
   > >   
   > > integrate((a + b*x + c*x^2)/((1 - x + x^2)*(1 - x^3)^(1/3)), x)   
   > >   
   > > >> Error detected within library code:   
   > >    integrate: implementation incomplete   
   > >    (residue poly has multiple non-linear factors)   
   > >   
   > > Is there no workable treatment of "multiple non-linear factors" other   
   > > than Kauer's Groebner-basis heuristics?   
   >   
   > Beware that Kauer's heuristics is incomplete.  AFAICS the   
   > your first example is one when this heurisctic fail.  Namely,   
   > rationalizing substitution gives integral of the form   
   >   
   > \sum w*log(x - w*(x*(1 - x)*(1 - k*x))^(1/3))   
   >   
   > where w runs over cube roots of 1/k.  Derivative of the   
   > log has pole at 0 and at point over 1/(k+1).  Trager   
   > method depends on rearanging integral in such a form that   
   > logands are regular at 0.  So Trager method produces   
   > 3 divisors, each one corresponding to point with   
   > x = 1/(k+1) and y beeing one ot the roots of (x*(1 - x)*(1 - k*x)).   
   > The same happens in Kaures method.  Next Kaures tries   
   > to find function having zero as specified by the divisor   
   > or its power and arbitrary behaviour at infinity.  For this   
   > to succeed the divisor have to be torsion divisor.  But   
   > this is unlikely to be the case.  At least my partial   
   > implementation of Kauers method gives up on this example.   
   >   
   > The example is amenable to Trager method.  If I replace k   
   > by k^3, then FriCAS can compute the integral.  Such replacement   
   > is equivalent to extending field of constants by cube root   
   > of k and FriCAS could do this automatically.  It is not   
   > done ATM because this can lead to excessive execution time   
   > (in this case it is 287.49 sec on fast machine).  I did   
   > not analyze the second example, but it seem to be similar.   
   >   
   > Let me add that that trying Kauers method I substituted   
   > 5 for k.  Without such substitution Groebner basis   
   > computations were quite slow.   
   >   
      
   Wow, Kauers is done in! [How about publishing a corresponding note   
   explaining why torsion divisors are needed here and Kauers' Groebner   
   bases cannot furnish them? (I could supply the theory of Goursat   
   pseudo-elliptics for cube roots of quadratics and cubics.)]   
      
   Yes, my antiderivative of the first example contains (-k)^(1/3):   
      
   INT((2 - (k + 1)*x)/((1 - (k + 1)*x)*(x*(1 - x)*(1 - k*x))^(1/3)), x)   
    = - 1/(2*(-k)^(1/3))*LN(x*(1 - x)*(1 - k*x) - k*x^3)   
    + 3/(2*(-k)^(1/3))*LN((x*(1 - x)*(1 - k*x))^(1/3) + (-k)^(1/3)*x)   
    - SQRT(3)/(-k)^(1/3)   
    *ATAN(1/SQRT(3)*(1 - 2*(-k)^(1/3)*x/(x*(1 - x)*(1 - k*x))^(1/3)))   
      
   Expecting the FriCAS subsubversion for an improved Trager procedure to   
   be x < 9,   
      
   Martin.   
      
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