"Nasser M. Abbasi" schrieb:   
   >   
   > On 4/1/2017 3:04 AM, clicliclic@freenet.de wrote:   
   >   
   > > On the /index.htm page you are linking to /reports/rubi_4_11/index.htm,   
   > > which latter page did not seem to exist when I looked.   
   > >   
   >   
   > I had a small problem and had to upload everything again. It is all   
   > there now.   
   >   
      
   Yes, I see the data now. Two of the Timofeev integrals from Chapter 8   
   were 'saved' for Sympy by increasing the timeout from 3 to 5 minutes:   
      
     8.58 (#580) INT(1/COSH(x)^5, x)   
      
     8.109 (#644) INT(COS(x)*LN(SIN(x))/(1 + COS(x))^2, x)   
      
   While Sympy's antiderivative for the former integral is simply awful,   
   that for the latter looks quite reasonable:   
      
   -3*#i*LOG(TANH(x/2)-#i)*TANH(x/2)^8/(8*TANH(x/2)^8+32*TANH(x/2)^~   
   6+48*TANH(x/2)^4+32*TANH(x/2)^2+8)-12*#i*LOG(TANH(x/2)-#i)*TANH(~   
   x/2)^6/(8*TANH(x/2)^8+32*TANH(x/2)^6+48*TANH(x/2)^4+32*TANH(x/2)~   
   ^2+8)-18*#i*LOG(TANH(x/2)-#i)*TANH(x/2)^4/(8*TANH(x/2)^8+32*TANH~   
   (x/2)^6+48*TANH(x/2)^4+32*TANH(x/2)^2+8)-12*#i*LOG(TANH(x/2)-#i)~   
   *TANH(x/2)^2/(8*TANH(x/2)^8+32*TANH(x/2)^6+48*TANH(x/2)^4+32*TAN~   
   H(x/2)^2+8)-3*#i*LOG(TANH(x/2)-#i)/(8*TANH(x/2)^8+32*TANH(x/2)^6~   
   +48*TANH(x/2)^4+32*TANH(x/2)^2+8)+3*#i*LOG(TANH(x/2)+#i)*TANH(x/~   
   2)^8/(8*TANH(x/2)^8+32*TANH(x/2)^6+48*TANH(x/2)^4+32*TANH(x/2)^2~   
   +8)+12*#i*LOG(TANH(x/2)+#i)*TANH(x/2)^6/(8*TANH(x/2)^8+32*TANH(x~   
   /2)^6+48*TANH(x/2)^4+32*TANH(x/2)^2+8)+18*#i*LOG(TANH(x/2)+#i)*T~   
   ANH(x/2)^4/(8*TANH(x/2)^8+32*TANH(x/2)^6+48*TANH(x/2)^4+32*TANH(~   
   x/2)^2+8)+12*#i*LOG(TANH(x/2)+#i)*TANH(x/2)^2/(8*TANH(x/2)^8+32*~   
   TANH(x/2)^6+48*TANH(x/2)^4+32*TANH(x/2)^2+8)+3*#i*LOG(TANH(x/2)+~   
   #i)/(8*TANH(x/2)^8+32*TANH(x/2)^6+48*TANH(x/2)^4+32*TANH(x/2)^2+~   
   8)-10*TANH(x/2)^7/(8*TANH(x/2)^8+32*TANH(x/2)^6+48*TANH(x/2)^4+3~   
   2*TANH(x/2)^2+8)+6*TANH(x/2)^5/(8*TANH(x/2)^8+32*TANH(x/2)^6+48*~   
   TANH(x/2)^4+32*TANH(x/2)^2+8)-6*TANH(x/2)^3/(8*TANH(x/2)^8+32*TA~   
   NH(x/2)^6+48*TANH(x/2)^4+32*TANH(x/2)^2+8)+10*TANH(x/2)/(8*TANH(~   
   x/2)^8+32*TANH(x/2)^6+48*TANH(x/2)^4+32*TANH(x/2)^2+8)   
      
   -2*x/3+LOG(TAN(x/2)^2+1)*TAN(x/2)^3/6-LOG(TAN(x/2)^2+1)*TAN(x/2)~   
   /2-LOG(TAN(x/2))*TAN(x/2)^3/6+LOG(TAN(x/2))*TAN(x/2)/2-LOG(2)*TA~   
   N(x/2)^3/6-TAN(x/2)^3/18+LOG(2)*TAN(x/2)/2+5*TAN(x/2)/6   
      
   Both are correct, however.   
      
   Martin.   
      
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