From: Albert_Rich@msn.com   
      
   On Friday, March 24, 2017 at 11:52:26 PM UTC-10, oldk...@gmail.com wrote:   
   > On Saturday, March 25, 2017 at 5:25:59 PM UTC+8, Albert Rich wrote:   
   > > On Friday, March 24, 2017 at 6:03:19 PM UTC-10, oldk...@gmail.com wrote:   
   > > > Off topic a little, I find it interesting that Rubi fails Timofeev 493,   
   > > >   
   > > >      int(x^2/(x*cos(x)-sin(x))^2,x)  --> (cos(x) + x*sin(x))/(x*cos(x) -   
   sin(x))   
   > > >   
   > > > A pure transcendental function, and easy to solve by heuristic   
   > > > methods.   
   > >   
   > > I'm a little slow.  What is the heuristic method used to integrate it?   
   > >   
   > > Albert   
   >   
   > What I meant is:   
   >   
   > This integral can be solved if transcendental case of Risch algorithm   
   > is implemented.   
   >   
   > Otherwise, a CAS system might try to solve it using heuristic method,   
   > aka integration by parts, guess, etc.  For this integral, as Timofeev said,   
   > it equals to int( x/sin(x) d(1/(x*cos(x)-sin(x)))), then integration by part,   
   > result is x/sin(x)/(x*cos(x)-sin(x)) + int(1/sin(x)^2,x) .   
      
   The best I could come up with using integration by parts as you suggested was   
   the following formula:   
      
   int(x^m/(cos(x)^(m-2)*(cos(x)+x*sin(x))^2),x) -->   
      
       -x^(m-1)/(cos(x)^(m-1)*(cos(x)+x*sin(x))) +   
       (m-1)*int(x^(m-2)/cos(x)^m,x)   
      
   If m>1 is an integer, x^(m-2)/cos(x)^m is integrable in terms of the   
   polylogarithm function.  The Timofeev Chapter 7, problem 11 on page 344 is the   
   special case when m=2.   
      
   If anyone can derive a more general formula, please let me and Rubi know.   
      
   Albert   
      
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