From: Albert_Rich@msn.com   
      
   On Wednesday, November 13, 2013 1:56:55 AM UTC-10, clicl...@freenet.de wrote:   
      
   > The hypergeometric series here (and in Example 3) terminate for a   
   > non-negative integer exponent n, while the series for your evaluation of   
   > Examples 6a,b terminate (after applying Euler's transformation) for a   
   > negative integer exponent n. The latter type of representation is   
   > somewhat more compact; a disadvantage is that the antiderivative 6a for   
   > positive a,m,x ends up on the branch cut of the hypergeometric function.   
   > I therefore propose to normalize to the former type:   
      
   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)  ?   
      
   Albert   
      
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