Hot Integral - 8

I will prove a general integral -

$\displaystyle \int_{0}^{\infty}{\frac{{x}^{a}}{{\left(b + c{x}^{d}\right)}^{e}} dx} = \displaystyle \frac{\Gamma\left(\frac{a+1}{d}\right)\Gamma\left(e-\frac{a+1}{d}\right)}{\Gamma(e)d {b}^{e-\frac{a+1}{d}}{c}^{\frac{a+1}{d}}}$

I will do this using a more non-conventional way using Ramanujan Master Theorem .

Substituting $u = \frac{c}{b}{x}^{d}$ , $dx = \frac{b}{cd{x}^{d-1}}du$

$\displaystyle \frac{b}{cd{b}^{e}}\int_{0}^{\infty}{\frac{{x}^{a-d+1} du}{{(1+u)}^{e}}}$

$\displaystyle \frac{1}{cd{b}^{e-1}}\int_{0}^{\infty}{\frac{{\left(\frac{bu}{c}\right)}^{\frac{a-d+1}{d}} du}{{(1+u)}^{e}}}$

$\displaystyle \frac{1}{d {c}^{\frac{a+1}{d}} {b}^{e-\frac{a+1}{d}}} \displaystyle \int_{0}^{\infty}{\displaystyle \frac{{u}^{\frac{a+1}{d}-1}}{{(1+u)}^{e}}}$

Now the last integral is in the form of Mellin transform, so, we will use Ramanujan Master Theorem. We know that

$\displaystyle \frac{1}{{(1+x)}^{n}} = \sum_{k=0}^{\infty}{\frac{\Gamma(n+k)}{\Gamma(n)} \frac{{(-x)}^{k}}{k!}}$

We get that our last integral will be equal to -

$\displaystyle \Gamma\left(\frac{a+1}{d}\right)\Gamma\left(e - \frac{a+1}{d}\right)$

Hence our total integral becomes -

$\displaystyle\frac{\Gamma\left(\frac{a+1}{d}\right)\Gamma\left(e-\frac{a+1}{d}\right)}{\Gamma(e)d {b}^{e-\frac{a+1}{d}}{c}^{\frac{a+1}{d}}}$

QED

Actually, the most difficult part of the problem is not what I have done. It's the calculations which follow.