$I = \int_0^k z^{m_1 - 1} \ln(1 + z) \left(\frac{m_1 z}{a} + \frac{m_2}{b} \right)^{-(m_1 + m_2)} \mathrm dz.$ Question: How to find the closed-form solution for the given integral?
$$I = \int_0^k z^{m_1 - 1} \ln(1 + z) \left(\dfrac{m_1 z}{a} + \dfrac{m_2}{b} \right)^{-(m_1 + m_2)} \mathrm dz,$$
where $k, a, b, m_1, m_2 \in \mathbb R_+$. Any leads appreciated. 
Attempt: I find it difficult because of the limits. In case the upper limit is $\infty$, the integral can be solved as 
\begin{align}
I < & \int_0^\infty z^{m_1 - 1} \ln(1 + z) \left(\dfrac{m_1 z}{a} + \dfrac{m_2}{b} \right)^{-(m_1 + m_2)} \mathrm dz \\
= & \left( \dfrac{b}{m_2}\right)^{m_1 + m_2}\int_0^\infty z^{m_{1} - 1} G_{2, 2}^{1, 2} \left( z  \left\vert \begin{smallmatrix} 1, & 1 \\[0.6em] 1, & 0\end{smallmatrix} \right.\right) G_{1, 1}^{1, 1} \left( \left. \dfrac{b m_1}{a m_2} z\right\vert \begin{smallmatrix} 1 - m_{1} - m_{2}\\[0.6em] 0\end{smallmatrix}\right) \mathrm dz \\
= & \left( \dfrac{b}{m_2}\right)^{m_1 + m_2} G_{3, 3}^{3, 2} \left( \left. \dfrac{b m_1}{a m_2} \right\vert \begin{smallmatrix} 1- m_{1} - m_{2}, & -m_{1}, & 1 - m_{1} \\[0.6em] 0, & -m_{1}, & -m_{1}\end{smallmatrix}\right),
\end{align}
where the integration above is solved using [1, eqns. (7, 10, 11, & 21)] and $G_{p, q}^{m, n}(\cdot)$ is Meijer's G-function. But this results in a (very) loose bound.
 A: I suggest the following simplifications: First we introduce the abbreviations:
$$\delta =k^{m_1} \left(\frac{m_2}{b}\right){}^{-\left(m_1+m_2\right)}$$
$$\nu =-\frac{b k m_1}{a m_2}$$
$$\mu =m_1+m_2$$
to write:
$$I=\delta  \int_0^1 z^{-1+m_1} \text{Log}[1+k z] (1-\nu  z)^{-\mu } \
\, dz$$ Second we have to do a partial integration with:
$$\nu ^{-m_1} \partial _z\left(\text{Beta}\left[\nu  z,m_1,1-\mu \right] \text{Log}[1+kz]\right)=\frac{k \nu ^{-m_1} \text{Beta}\left[z \nu ,m_1,1-\mu
\right]}{1+k z}+z^{-1+m_1} (1-z \nu )^{-\mu } \text{Log}[1+k z]$$
to reduce the integral to:
$$J=\int_0^1 \frac{\text{Beta}[z \nu ,m_1,1-\mu ]}{1+k z} \, dz$$
Now we see that the solution of the integral is a function of two variables. This integral appears in a more general form in the calculation of the porosity of an electrode of a lead acid battery. The solution is needed to calculate the coverage with lead sulfate and have information of the state of charge of the battery. If $k=\nu$ the solution is given by Wolfram. For the general the strategy is to express the numerator and denominator as a hypergeometric function:
$$\text{Beta}[z \nu ,\alpha ,1-\mu ]=\frac{(z \nu )^{m_1} \text{HypergeometricPFQ}\left[\left\{m_1,\mu \right\},\left\{m_1+1\right\},z \nu \right]}{m_1}$$
$$\frac{1}{1+k z}=\text{Hypergeometric2F1}[1,1,1,-z k]$$
As a result one has to solve the integral over two hypergeometric functions. This is done by SRIVASTAVA and leads to a Kampe de Feriet Function. The generalization is a H-Fox-Function with two variables. If I have a email address I can share the notebook with you and also the paper. 
EDIT
I found the solution of the integral in the paper Sanchis-Lozano. It is formula (4) or Expression 2.
