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   sci.math.symbolic      Symbolic algebra discussion      10,432 messages   

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   Message 9,421 of 10,432   
   antispam@math.uni.wroc.pl to Nasser M. Abbasi   
   Re: int( arcsin(x)/x,x), FriCas result   
   03 May 17 14:32:44   
   
   Nasser M. Abbasi  wrote:   
   > Hello;   
   >   
   > Why can't Fricas integrate asin(x)/x dx ?  Since there is standard   
   > substitution u=asin(x) which converts the integrand to   
   >   
   >        u*cos(u)/sin(u) du   
   >   
   > which Fricas can integrate with no problem?   
   >   
   > Illustration using Rubi:   
   >   
   > In[88]:= Int[ArcSin[x]/x,x]   
   > Out[88]= -(1/2) I ArcSin[x]^2+ArcSin[x] Log[1-E^(2 I ArcSin[x])]-1/2 I   
   PolyLog[2,E^(2 I ArcSin[x])]   
   >   
      
   > But Fricas does not like   
   >   
   >  ii:=integrate(asin(x)/x,x)   
   >   
   > but it can do the transformed version   
   >   
   > ii:=integrate(u*cos(u)/sin(u),u)   
   >   
   > (u*log((2*sin(u))/(sin(u)+((-1)*(-1)^(1/2)*cos(u)+(-1)*(-1)^(1/2))))+(u*log(   
   >   (2*sin(u))/(sin(u)+((-1)^(1/2)*cos(u)+(-1)^(1/2))))+(u*log((   
   *(-1)^(1/2)*cos(   
   >   u)+2*(-1)^(1/2))/(sin(u)+((-1)^(1/2)*cos(u)+(-1)^(1/2))))+(u   
   log(((-2)*(-1)^(   
   >   1/2)*cos(u)+(-2)*(-1)^(1/2))/(sin(u)+((-1)*(-1)^(1/2)*cos(u)   
   (-1)*(-1)^(1/2))   
   >   ))+((-1)^(1/2)*dilog((2*sin(u))/(sin(u)+((-1)*(-1)^(1/2)*cos   
   u)+(-1)*(-1)^(1/   
   >   2))))+((-1)*(-1)^(1/2)*dilog((2*sin(u))/(sin(u)+((-1)^(1/2)*   
   os(u)+(-1)^(1/2)   
   >   )))+((-1)*(-1)^(1/2)*dilog((2*(-1)^(1/2)*cos(u)+2*(-1)^(1/2)   
   /(sin(u)+((-1)^(   
   >   1/2)*cos(u)+(-1)^(1/2))))+(-1)^(1/2)*dilog(((-2)*(-1)^(1/2)*   
   os(u)+(-2)*(-1)^   
   >   (1/2))/(sin(u)+((-1)*(-1)^(1/2)*cos(u)+(-1)*(-1)^(1/2)))))))))))/2   
   >   
   > Just wondering why, that is all.   
      
   Simple thing: the second case is implemented the first not.  Note   
   that the integral requires polylog.  When main function is an   
   exponential, then there is relatively simple algorithm for   
   finding polylog terms.  When main function is a log finding   
   polylog terms is much more tricky and currently all FriCAS   
   can do is to try pattern matching (with _single_ pattern).   
   Concerning substitution 'u = asin(x)', such substition has   
   limited availability.  Also, it can make function harder   
   to integrate.  So FriCAS do not try it.   
      
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