How to evaluate this definite integral - $ \int_{0}^{\infty} x \dfrac{p(a+cx)(b+ck) + ab-c}{(1+ak+bx+ckx)^{p+2}} \mathrm{d}x$? I'm trying to evaluate this integral, with no luck so far:
$$\displaystyle \int_{0}^{\infty} x \dfrac{p(a+cx)(b+ck) + ab-c}{(1+ak+bx+ckx)^{p+2}} \mathrm{d}x$$
Here, $a, b, c, k, p > 0$ are constants.
I'm at a loss on how to proceed. Can anyone give me any hints or point to references that can help evaluate this integral as a closed form expression?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}x\,{p\pars{a + cx}\pars{b + ck} + ab - c \over \pars{1 + ak + bx + ckx}^{\, p + 2}}\,\dd x}:\
{\Large ?}.\quad a,b,c,k,p > 0\ \mbox{are}}$ constants.

After some rescaling you are entitled to evaluate the following integral:
$\ds{\left.I_{\alpha} \equiv \int_{0}^{\infty}{x^{\alpha - 1} \over \pars{1 + x }^{\, p + 2}}\,\dd x\,\right\vert_{\ 0\ <\ \Re\pars{\alpha}\ <\ p\ +\ 2}}$.
In particular, your initial evaluation requires $\ds{I_{2}\ \mbox{and}\ I_{3}}$. $\ds{I_{\alpha}}$ will be convenient evaluated by means of the Ramanujan's Master Theorem.

Note that 
\begin{align}
{1 \over \pars{1 + x}^{\, p + 2}} & =
\sum_{k = 0}^{\infty}{-p - 2 \choose k}x^{k} =
\sum_{k = 0}^{\infty}
\bracks{{k + p + 1 \choose k}\pars{-1}^{k}}x^{k}
\\[5mm] & =
\sum_{k = 0}^{\infty}
\color{red}{\Gamma\pars{k + p + 2} \over \Gamma\pars{p + 2}}\,{\pars{-x}^{k} \over k!}
\end{align}
Then,
\begin{align}
I_{\alpha} & \equiv
\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{x^{\alpha - 1} \over
\pars{1 + x }^{\, p + 2}}\,\dd x
\,\right\vert_{\ 0\ <\ \Re\pars{\alpha}\ <\ p\ +\ 2}} =
\bbx{\Gamma\pars{\alpha}\,
{\Gamma\pars{-\alpha + p + 2} \over \Gamma\pars{p + 2}}} \\ &
\end{align}
$$
\left\{\begin{array}{lclcl}
\ds{I_{2}} & \ds{=} &
\ds{{\Gamma\pars{p} \over \Gamma\pars{p + 2}}} & \ds{=} &
\ds{1 \over \pars{p + 1}p}
\\[2mm]
\ds{I_{3}} & \ds{=} &
\ds{2\,{\Gamma\pars{p - 1} \over \Gamma\pars{p + 2}}} & \ds{=} &
\ds{2 \over \pars{p + 1}p\pars{p - 1}}
\end{array}\right.
$$
