analytic solution to definte integral

Question

I am looking for Analytic solution to a definite integral. Or an approriate transformation to apply. the conditions on $\alpha$ , $\beta$ being positive real numbers while $n$ is positive integer.the integral is given as $$ \int_{-\infty}^{\infty}x^ne^{-\beta x}\left(1 + \alpha e^{-\beta x}\right)^{-1/\alpha}\,dx $$

Answer 1

Consider $\int_{-\infty}^\infty e^{-\beta x}\left(1+\alpha e^{-bx}\right)^{-\frac{1}{\alpha}}~dx$ ,

$\int_{-\infty}^\infty e^{-\beta x}\left(1+\alpha e^{-bx}\right)^{-\frac{1}{\alpha}}~dx$

$=\int_\infty^0x^\frac{\beta}{b}~(1+\alpha x)^{-\frac{1}{\alpha}}~d\left(-\dfrac{\ln x}{b}\right)$

$=\dfrac{1}{b}\int_0^\infty x^{\frac{\beta}{b}-1}(1+\alpha x)^{-\frac{1}{\alpha}}~dx$

$=\dfrac{1}{b}\int_0^\infty\left(\dfrac{x}{\alpha}\right)^{\frac{\beta}{b}-1}(1+x)^{-\frac{1}{\alpha}}~d\left(\dfrac{x}{\alpha}\right)$

$=\dfrac{1}{\alpha^\frac{\beta}{b}~b}\int_0^\infty x^{\frac{\beta}{b}-1}(1+x)^{-\frac{1}{\alpha}}~dx$

$=\dfrac{1}{\alpha^\frac{\beta}{b}~b}B\left(\dfrac{\beta}{b},\dfrac{1}{\alpha}-\dfrac{\beta}{b}\right)$

$\therefore\int_{-\infty}^\infty x^ne^{-\beta x}\left(1+\alpha e^{-\beta x}\right)^{-\frac{1}{\alpha}}~dx=(-1)^n\dfrac{d^n}{d\beta^n}\left(\dfrac{1}{\alpha^\frac{\beta}{b}~b}B\left(\dfrac{\beta}{b},\dfrac{1}{\alpha}-\dfrac{\beta}{b}\right)\right)(b=\beta)$