is this the correct antiderivative by Maple?

[Nasser M. Abbasi]

As I was verifying solution to an ode. Even thought everything I did
was correct, Maple was saying my solution is not correct (using odetest).

I zoomed into the problem and it is comming from result of integrate step
I did using Maple.

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.

I tried the same thing in Mathematica,
and it gives result using incomplete Gamma function, not WhittakerM,
and it does differentiate back to the integrand:

-------------
ClearAll[x]
integrand = 1/(Exp[x^3/3]*x^2)
y2 = Integrate[integrand, x]
---------

-(((x^3)^(1/3) Gamma[-(1/3),x^3/3])/(3 3^(1/3) x))

And now

D[y2, x] // Simplify

gives

E^(-(x^3/3))/x^2

Which is the integrand. So its results is correct.

I do not know the relation between incomplete Gamma and WhittakerM
to try to translate Maple's result to Gamma and compare better.

question is: Is Maple's int() result correct?

Mathematica is V 13. Maple is 2021.2. All on windows 10

--Nasser

[acer]

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

[acer]

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

[Richard Fateman]

> >
Maxima 5.45..1 gives ...
(%i2) integrate(integrand,x);
(%o2) -gamma_incomplete(-1/3,x^3/3)/3^(4/3)
(%i3) diff(%,x);
(%o3) %e^(-x^3/3)/x^2


Interestingly, for the integral, Mathematica gives an expression that includes
the subexpression (x^3) ^ (1/3). Some people would expect that to be x.
Say x=-1, then we have (-1)^(1/3), which some people would expect to be -1.
Mathematica says %//N is 0.5 + 0.866025 I

which is indeed another cube root of -1.

I'm sure one can make an argument for this choice, but does it matter for the integral?

And shouldn't Mathematica simplify (x^3)^(1/3)/x to (-1)^(1/3) ? It doesn't.
Just asking :)

[Валерий Заподовников]

Integrate[1/(Exp[x^3/3]*x^2),x]//FullSimplify

returns -(ExpIntegralE[4/3,x^3/3]/(3 x)) 

Also (x^3)^(1/3) is not due to 13.0.0.

P.S. you should always use some form of simplify,
be it maple or mathematica to find this things out.

[Dr Huang]

these are too complicated. input your formula into mathHand.com, click the integrate button.

or click the link:
http://server.drhuang.com/input/?guess=integrate%281%2F%28exp%28x%5E3%2F3%29*x%5E2%29%29&inp=1%2F%28exp%28x%5E3%2F3%29*x%5E2%29

[Dr Huang]

mathHand.com auto plot both integrand and integral