From: Albert_Rich@msn.com   
      
   On Monday, March 9, 2020 at 7:20:15 AM UTC-10, clicl...@freenet.de wrote:   
      
   > Perhaps the new version needs to be broken in like a new pair of shoes?   
   > Here are some tough Risch integration exercises involving cube roots:   
   >    
   >   integrate((x + 3)/((x - 1)^2*(x^2 - 1)^(1/3)), x)   
   >    
   >   integrate((x + 1)/((x + 3)*(2*x + 1)*(x^2 + 1)^(1/3)), x)   
   >    
   >   integrate((x + 1)/((x - 1)*(2*x + 1)*(3*x^2 - 1)^(1/3)), x)   
   >    
   > Martin.   
      
                               Spoiler Alert!   
      
   An antiderivative of (x + 1)/((x + 3)*(2*x + 1)*(x^2 + 1)^(1/3)) in   
   Mathematica syntax is   
      
   -Sqrt[3]*ArcTan[(5^(2/3) + (2*2^(1/3)*(2 - x))/(1 + x^2)^(1/3))/   
   Sqrt[3]*5^(2/3))]/(5*10^(1/3)) +    
      Log[5^(2/3) - (2^(1/3)*(2 - x))/(1 + x^2)^(1/3)]/(5*10^(1/3)) -    
      Log[5*5^(1/3) + (2^(2/3)*(2 - x)^2)/(1 + x^2)^(2/3) + (2^(1/3)*5^(2/3)*(2 -   
   x))/(1 + x^2)^(1/3)]/(10*10^(1/3))   
      
   An antiderivative of (x + 1)/((x - 1)*(2*x + 1)*(3*x^2 - 1)^(1/3)) is   
      
   -ArcTan[(1 + (2*2^(1/3)*x)/(-1 + 3*x^2)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]) +    
    Log[2^(1/3)*3^(2/3) - (6^(2/3)*x)/(-1 + 3*x^2)^(1/3)]/(3*2^(1/3)) -    
    Log[(2^(2/3)*x^2 + 2^(1/3)*x*(-1 + 3*x^2)^(1/3) + (-1 + 3*x^2)^(2/3))/(-1 +   
   3*x^2)^(2/3)]/(6*2^(1/3))   
      
   Unfortunately version 4.16.1 of Rubi currently available at http   
   ://rulebasedintegration.org/ does not produce elementary antideriatives for   
   these expressions.  However, the release of Rubi currently under development   
   does have rules for types of    
   integrals.   
      
   Albert   
      
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