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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 10,200 of 10,432    |
|    acer to Nasser M. Abbasi    |
|    Re: is this the correct antiderivative b    |
|    15 Jan 22 10:51:57    |
      From: maple@rogers.com              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)*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))              expand(simplify(convert(diff(y2,x),compose,        hypergeom,StandardFunctions))):              exp(-1/3*x^3)/x^2              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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