Be $\,a,b,c,d,s>0\,$ , $\,d\,$ an integer.
$(A)\hspace{1cm}$ Only an approximation for large positive $\,s\,$ .
Be $\,cd<ab\,$ . Be $\,s\,$ large enough so that $\,\displaystyle \left(1+\frac{x}{s}\right)^s\approx e^x$ and $\,\displaystyle \Gamma\left(x,y+\frac{z}{s}\right)\approx \Gamma(x,y)\,$ .
It's $\,\displaystyle \left(1-\frac{\Gamma(d,b+\frac{x}{n})}{\Gamma(d)}\right)^n\to 0\,$ for $\,n\to\infty\,$ .
Then we have
$\displaystyle \int\limits_0^\infty x^d (b+cx)^{ds} \left(1-\frac{\Gamma(d,b+cx)}{\Gamma(d)}\right)^{-s} e^{-asx}dx = $
$\displaystyle =\frac{b^{ds}}{(as)^{d+1}} \int\limits_0^\infty x^d \left(1+\frac{c}{abs}x\right)^{ds} \left(1-\frac{\Gamma(d,b+\frac{c}{as}x)}{\Gamma(d)}\right)^{-s} e^{-x}dx$
$\displaystyle \approx \frac{b^{ds}}{(as)^{d+1}\left(1-\frac{\Gamma(d,b)}{\Gamma(d)}\right)^s} \int\limits_0^\infty x^d e^{-(1-\frac{cd}{ab})x}dx = \frac{b^{ds}d!}{\left(as(1-\frac{cd}{ab})\right)^{d+1}\left(1-\frac{\Gamma(d,b)}{\Gamma(d)}\right)^s}\enspace$ .
$(B)\hspace{1cm}$ Series expansion.
It’s $\enspace\displaystyle\frac{\Gamma(d,x)}{\Gamma(d)}=\frac{1}{e^x}\sum\limits_{k=0}^{d-1}\frac{x^k}{k!}\enspace$ .
If we define $\enspace\displaystyle \sum\limits_{k=0}^{mn} a_{m,n,k}x^k := \left(\sum\limits_{k=0}^m \frac{x^k}{k!}\right)^n \enspace$ then we can calculate explicit
$\displaystyle \int\limits_0^\infty x^d (b+cx)^{ds} \left(1-\frac{\Gamma(d,b+cx)}{\Gamma(d)}\right)^{-s} e^{-asx}dx = $
$\displaystyle =\sum\limits_{n=0}^\infty {\binom {s+n-1}{n}} \frac{1}{e^{bn}} \sum\limits_{k=0}^{(d-1)n} a_{d-1,n,k} \sum\limits_{j=0}^\infty {\binom {ds+k}{j}} \frac{b^{ds+k-j} c^j (d+j)!}{(as+cn)^{d+j+1}}\enspace$ .
It makes sense to choose natural numbers for $\,s\,$ .
