How to solve the given integral avoiding infinite series sum? Question: How to solve the following integral?
$$I = \int_0^\infty \dfrac{x^{N_a + N_b - 1}}{(p \Omega_1 + \Omega_2 x)^{N_a + 1}} \ln (1 + Qx) \, _2F_1\left( N_b + 1, N_b; N_b +1; \dfrac{-\Omega_3}{\Omega_4}x\right)dx, \tag{1}$$
where $N_a, N_b \in \mathbb Z_+$ and $\Omega_1, \Omega_2, \Omega_3, \Omega_4, p, Q \in \mathbb R_+$. I am looking for a closed-form solution for the above integral. A solution in terms of any special function will also be good enough. 
Any leads appreciated. 
My attempt: Representing $\log(1 + Qx)$ in terms of Meijer's $G$ function, we have 
\begin{align}
 I = {} & \, \Omega_2^{-(N_a + 1)}\int_0^{\infty}x^{N_d + N_b - 1}\left( x + \dfrac{p\Omega_1}{\Omega_2}\right)^{-(N_a + 1)}G_{2, 2}^{1, 2}\left( Qx \left\vert \begin{smallmatrix} 1, & 1\\ 1, & 0\end{smallmatrix}\right.\right) \\ 
& \hspace{6cm}\times\, _2F_1\left( N_b + 1, N_b; N_b +1; \dfrac{-\Omega_3}{\Omega_4} x\right) \, dx \tag{2}
\end{align}
A solution to (2) exists in [1, eqn. 2.2], resulting into infinite series summation. 
Can anyone suggest any alternate solution that doesn't contain infinite series sum? 
 A: Too long for a comment. The hypergeometric function inside the integral is actually elementary because of parameters: $_2F_1(\ldots)=\left(1+\frac{\Omega_3}{\Omega_4}x\right)^{-N_b}$. The integral can therefore be reduced to the form
\begin{align}
\int_{0}^{\infty}\left(\frac{x}{\alpha+x}\right)^{N_a+1}\left(\frac{x}{\beta+x}\right)^{N_b}\frac{\ln(1+Qx)}{x^2}\,dx=\tag{1}\\
=\int_{0}^{\infty}(1+\alpha y)^{-N_a-1}(1+\beta y)^{-N_b}\ln(1+Qy^{-1})\,dy.\tag{2}
\end{align}


*

*Now, if $N_a$, $N_b$ were arbitrary (admissible) complex numbers and if we could replace $\ln(1+Qx)$ by $(1+Qx)^s$ with some $s$, then the integral could be expressed in terms of Lauricella function $F_D^{(2)}$. When we have logarithm instead, we could write the result as a derivative of this function w.r.t. to parameter $s$ evaluated at $s=0$.

*For arbitrary but fixed integer $N_a$, $N_b$ the integrals (1)-(2) look computable to me (e.g. doing partial fraction expansion first). I would not expect something more complicated than dilogarithms. Moreover for $\alpha=\beta$ the result is clearly an elementary function (integration by parts).
A: Here is an alternate way to solve. 
\begin{align}
 & \int_{0}^{\infty} \dfrac{x^{N_a + N_b - 1}}{(p \Omega_1 + \Omega_2 x)^{N_a + 1}} \ln(1 + Qx) \, _2F_1\left( N_b + 1, N_b,; N_b +1 , \dfrac{-\Omega_3}{\Omega_4}x\right)dx \\
= & (p\Omega_1)^{-(N_a + 1)} \int_{0}^{\infty} x^{N_a + N_b - 1} \left( 1 + \dfrac{\Omega_2}{p\Omega_1} x\right)^{-(N_a + 1)} \ln(1 + Qx) \, _2F_1\left( N_b + 1, N_b,; N_b +1 , \dfrac{-\Omega_3}{\Omega_4}x\right)dx \\
= & \dfrac{1}{(p\Omega_1)^{(N_a + 1)} \Gamma(N_a + 1)\Gamma(N_b)} \int_{0}^{\infty} x^{N_a + N_b - 1} G_{1, 1}^{1, 1} \left( \left.\dfrac{\Omega_2}{p\Omega_1} x\right\vert \begin{smallmatrix} -N_a \\ 0\end{smallmatrix}\right) G_{2, 2}^{1, 2} \left( Qx \left\vert \begin{smallmatrix} 1, & 1 \\ 1, & 0\end{smallmatrix} \right.\right) G_{2, 2}^{1, 2} \left( \left.\dfrac{\Omega_3}{\Omega_4}x \right\vert \begin{smallmatrix} -N_b, & 1 - N_b \\ 0, & -N_b\end{smallmatrix}\right)dx, \tag{1}
\end{align}
where the second term inside the integral is represented as a Meijer's G function using [1, Section IV-C], the log function is represented as a Meijer's G function using [2, below Fig. 1] and the Gauss hypergeometric function is represented as a Meijer's G function using [3, eqn. (17)]. The closed-form solution to (1) is now straightforward to represent using 4, in terms of extended generalized bivariate Meijer’s G function (EGBMGF).
Note: Looking for some expert comments/opinions.
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[2]. P.S. Bithas, "Digital Communications over Generalized-K Fading Channels"
[3]. V.S. Adamchik, et.al. "The algorithm for calculating integrals of Hypergeometric type functions and its realization in REDUCE system"
[4]. http://functions.wolfram.com/07.34.21.0081.01
