This one: $\int_{0}^{\bar{x}}e^{x^{a}}x^{b}(1x)^{c}dx,a,b,c\ge0$. When $a=1,c=0,\bar{x}=\infty$ it is the gamma function.

$\begingroup$ The integral does not even converge for $a>0$. $\endgroup$ – Bullet51 Jun 13 at 1:41

$\begingroup$ no name, no closed form evaluation. $\endgroup$ – Carlo Beenakker Jun 13 at 5:47
Maple evaluates $\int e^{x^a}x^b\;dx$ in terms of the Whittaker M function, which may also be written as ${}_1F_1$ hypergeometrics. $$ \int \!{{\rm e}^{{x}^{a}}}{x}^{b}\,{\rm d}x= \\ {\frac {{{\rm e}^{{ x}^{a}/2}}}{ \left( b+1 \right) \left( a+b+1 \right) \left( 2\,a+b+1 \right) } \left( \left( \left( a+b+1 \right) {x}^{b/2+1/23/2\,a}+{ x}^{b/2+1/2a/2}a \right) a{{\rm M}_{{\frac {ab1}{2a}},\,{ \frac {2\,a+b+1}{2a}}}\left({x}^{a}\right)}+{x}^{b/2+1/23/2\,a}{ {\rm M}_{{\frac {a+b+1}{2a}},\,{\frac {2\,a+b+1}{2a}}}\left({x }^{a}\right)} \left( a+b+1 \right) ^{2} \right) } $$