Albert Rich schrieb:   
   >   
   > On Thursday, November 14, 2013 10:11:17 AM UTC-10, clicl...@freenet.de wrote:   
   >   
   > >> For the integral of (1+x)^n/x, Rubi currently returns   
   > >>   
   > >> -(1+x)^(1+n) * 2F1(1,1+n,2+n,1+x) / (1+n)   
   > >>   
   > >> In light of your comments above, would it be better to return   
   > >>   
   > >> (1+x)^n * 2F1(-n,-n,1-n,-1/x) / (n*(1+1/x)^n)  ?   
   > >   
   > > I think so. For positive x and non-integer n you are no longer sitting   
   > > right on the edge of a cliff then - the thought alone makes me dizzy. An   
   > > equivalent (by Euler's transformation) but simpler antiderivative is:   
   > >   
   > >   (1+x)^(1+n) * 2F1(1,1,1-n,-1/x) / (n*x)   
   > >   
   > > Note that the singularity at x=0 is already present in the integrand.   
   > > And Pfaff's transformation of these two puts one on the brink of the   
   > > chasm when x is negative and small:   
   > >   
   > >   (1+x)^n * 2F1(-n,1,1-n,1/(1+x)) / n   
   >   
   > Ok, for integrands of the form (c+d x)^n/(a+b x) when n is symbolic,   
   > the next version of Rubi will return   
   >   
   >     (c+d*x)^n/(b*n*(b*(c+d*x)/(d*(a+b*x)))^n)*   
   >       2F1(-n,-n,1-n,-(b*c-a*d)/(d*(a+b*x)))   
   >   
   > The simpler equivalent rule derived using Euler's transformation is   
   > not used since it is harder to simplify its derivative back to the   
   > original integrand.   
   >   
      
   But this is just an arbitrary property of the differentiator, right?   
   Another differentiator may give the result you would now obtain by   
   applying Euler's transformation first (and undoing it on non-elementary   
   hypergeometrics that remain in the derivative).   
      
   So your reason is no good reason; you are just bending to the dictate of   
   WRI. I suggest that Rubi redefines 2F1 differentiation instead. The   
   optimality of Rubi's antiderivatives determines the rules to which WRI   
   must bend!   
      
   I have spoken.   
      
   Martin.   
      
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