I need help with this integration: $\int_{0}^{\infty}x^{a}e^{x^{b}}e^{-(e^{x^{b}}-1)^c}dx$ I am trying to fined a closed form for this integration
$$\int_{0}^{\infty}x^{a}e^{x^{b}}e^{-(e^{x^{b}}-1)^c}dx,$$
where $a,b,c>0$
I think the generalized integro-exponential ($E_{s}^{r}(z)=\frac{1}{r!} \int_{1}^{\infty}(\log t)^{r}t^{-s}e^{-zt}dt$) function can be used here.
 A: For c=1, If we consider THE following integral
$I=\int_{0}^{\infty}x^{a}e^{x^{b}}e^{-\lambda(e^{x^{b}}-1)}dy$
then by letting $y=x^{b}$ we get
I=$\frac{e^{\lambda}}{b^{a+1}}\int_{1}^{\infty} (log~y)^{\frac{a}{b}}e^{-\lambda y}dy$
if we consider $\frac{a}{b}$ is an integer then 
$I=\frac{e^{\lambda}}{b^{a+1}} (\frac{a}{b})! E_{0}^{\frac{a}{b}}(\lambda)$
where $E_{s}^{r}(z)=\frac{1}{r!} \int_{1}^{\infty}(log t)^{r}t^{-s}e^{-zt}dt$ is the generalized integro-exponential function, (Milgram 1985).
Note that, $E_{0}^{r}(z)=\frac{E_{1}^{r-1}(z)}{z}$ and $E_{1}^{0}(z)=E_{1}(z)=\Gamma(0,z)$, where $\Gamma(s,z)=\int_{z}^{\infty}t^{s-1} e^{-t}dt$ is the complementary incomplete gamma function.
Also, it can be written as a Meijer G-function as 
$E_{s}^{r}(z)=G_{r+1,r+2}^{r+2,0}\left(z \mid \begin{array}{c}
                                          ;s,\dots,s \\
                                          0,s-1,\dots,s-1;
                                        \end{array} \right)$
For real values of $\frac{a}{b}>-1$ we may consider an extension of generalized integro-exponential function  (Pogany et al. 2017)
