From: maple@rogers.com   
      
   On Saturday, January 15, 2022 at 1:51:58 PM UTC-5, acer wrote:   
   > On Saturday, January 15, 2022 at 12:19:28 PM UTC-5, Nasser M. Abbasi wrote:   
   > ...   
   > > Here is the short version:   
   > >   
   > > restart;   
   > > integrand:=1/(exp(x^3/3)*x^2);   
   > > y2:= int( integrand, x);   
   > >   
   > > Maple gives   
   > >   
   > > 1/9*3^(2/3)*(-9/10*x^2*3^(2/3)/(x^3)^(1/3)*exp(-1/6*x^3)*   
   > > WhittakerM(1/3,5/6,1/3*x^3)-9/2/x^4*3^(2/3)*(x^3+2)/(x^3)^(1/3)*   
   > > exp(-1/6*x^3)*WhittakerM(4/3,5/6,1/3*x^3))   
   > >   
   > > but this does not differentiate back to the integrand. I tried   
   > > simplify and assumptions. No luck.   
   > >   
   > > I think the Maple antiderivative is wrong but I am still not sure.   
   > Using Maple 2021.1,   
   >   
   > restart;   
   > integrand:=1/(exp(x^3/3)*x^2):   
   > y2:= int( integrand, x):   
   >   
   > 1/9*3^(2/3)*(-9/10*x^2*3^(2/3)/(x^3)^(1/3)*exp(-1/6*x^3)*Whitt   
   kerM(1/3,5/6,1/3   
   > *x^3)-9/2/x^4*3^(2/3)*(x^3+2)/(x^3)^(1/3)*exp(-1/6*x^3)*Whitta   
   erM(4/3,5/6,1/3*   
   > x^3))   
   >   
   > expand(simplify(convert(diff(y2,x),compose,   
   > hypergeom,StandardFunctions))):   
   >   
   > exp(-1/3*x^3)/x^2   
      
   Better yet, that conversion can be applied to the generated antiderivative.   
      
   restart;   
   integrand:=1/(exp(x^3/3)*x^2):   
   y2:=int(integrand, x):   
      
   y3:=collect(simplify(expand(convert(y2,compose,   
                                       hypergeom,StandardFunctions))),   
               exp,simplify);   
      
      -exp(-1/3*x^3)/x-1/3*3^(2/3)*x^2*(GAMMA(2/3)-GAMMA(2/3,1/3*x^3))/(x^3)^(2/3)   
      
   simplify(diff(y3,x));   
      
      exp(-1/3*x^3)/x^2   
      
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