You can't express this integral in terms of elementary functions.
Depending on what you need the integration for, you may choose a way of integration or another.
Integration via power series
Recall that #e^x# is analytic on #mathbb{R}#, so #forall x in \mathbb{R}# the following equality holds
#e^x=sum_{n=0}^{+infty}x^n/{n!}#
and this means that
#e^{x^3}=sum_{n=0}^{+infty}(x^3)^n/{n!}=sum_{n=0}^{+infty}{x^{3n}}/{n!}#
Now youcan integrate:
#int e^{x^3} dx=int (sum_{n=0}^{+infty}{x^{3n}}/{n!}) dx=c+sum_{n=0}^{+infty}{x^{3n+1}}/{(3n+1)n!}#
Integration via the Incomplete Gamma Function
First, substitute #t=-x^3#:
#int e^{x^3} dx = - 1/3 int e^{-t} t^{-2/3} dt#
The function #e^{x^3}# is continuous. This means that its primitive functions are #F:\mathbb{R} to \mathbb{R}# such that
#F(y) = c + int_0^y e^{x^3}dx=c- 1/3 int_0^{-y^3} e^{-t} t^{-2/3} dt#
and this is well defined because the function #f(t)=e^{-t}t^{-2/3}# is such that for #t to 0# it holds #f(t) ~~ t^{-2/3}#, so that the improper integral #int_0^s f(t) dt# is finite (I call #s=-y^3#).
So you have that
#int e^{x^3} dx=c- 1/3 int_0^s f(t)dt#
Remark that #t^{-2/3}< 1 hArr t>1#. This means that for #t to +infty# we get that #f(t)=e^{-t} * t^{-2/3} < e^{-t} * 1 = e^{-t}#, so that #|int_1^{+ infty} f(t)dt|<|int_1^{+infty} e^{-t}dt|=e#. So following improper integral of #f(t)# is finite:
#c'=int_0^{+infty}f(t)dt=int_0^{+infty} e^{-t}t^{1/3 -1}dt=Gamma(1/3)#.
We can write:
#int e^{x^3} dx=c-1/3 (int_0^{+infty} f(t)dt -int_s^{+infty} f(t)dt)#
that is
#int e^{x^3} dx=c-1/3 c' +1/3 int_s^{+infty} e^{-t}t^{1/3 -1}dt#.
In the end we get
#int e^{x^3} dx=C+1/3 Gamma(1/3,t) =C+1/3 Gamma(1/3,-x^3)#
