Closed-form solution of the integral $\int_{0}^{\infty}\frac{\ln(1+x)}{(1+ax)^{(1+m)}}\,dx$ I want to know whether there is a closed form solution for the integral $$\int_{0}^{\infty}\frac{\ln(1+x)}{(1+ax)^{(1+m)}}\,dx$$
where $a$ and $m$ are positive (not necessarily integers).
A closed-form solution can be found for the very specific case of $a=1$ for the above integral, i.e., $\int_{0}^{\infty}\frac{\ln(1+x)}{(1+x)^{(1+m)}}\,dx = B(1,m) \cdot [\psi(1+m) - \psi(m)]$, where $B(\cdot,\cdot)$ is the Beta function and $\psi(\cdot)$ is the Euler psi function. 
But I want to know whether there is a solution for the general case. If there is not, is there a way to represent the integral approximately with a closed form expression (or in terms of hypergeometric functions).
I have tried to express the nominator and denominator using Meijer-G functions and employ the convolution theorem (the product of two arbitrary G-functions integrated over the positive real axis can be represented by just another G-function), but it turns out there are too many strict constraints to employ the theorem.
Thanks a lot in advance.
 A: 
$I=\dfrac{B\bigg(1-\dfrac1a~;~m~;~1-m\bigg)}{m~a~(a-1)^m}~:~$ See incomplete beta function for more information.

In particular, for integer and half-integer values of m we have the following results:

$$\int_0^\infty\dfrac{\ln(1+x)}{(A+ax)^2}~dx~=~\dfrac{2\ln\sqrt{\dfrac Aa}}{a~(A-a)}$$
$$\int_0^\infty\dfrac{\ln(1+x)}{(A+ax)^{3/2}}~dx~=~\dfrac{4\arccos\sqrt{\dfrac Aa}}{a~(a-A)}$$

from which the others can be evaluated by repeatedly differentiating both sides with regard to A, and then setting $A=1$.
A: Suppose that $a>0$ and $a\not =1$ and $m\in \mathbb{N}, m\geq 1$. 
We use integration by parts:
$$F_m(a)=\int_0^{+\infty}\frac{\log(1+x)}{(1+ax)^{m+1}}dx=\big[-\frac{\log(1+x)}{am(1+ax)^m}\big]_0^{+\infty}+\frac{1}{ma}\int_0^{+\infty}\frac{dx}{(x+1)(1+ax)^m}$$ 
Hence
$$maF_m(a)=G_m(a)=\int_0^{+\infty}\frac{dx}{(x+1)(1+ax)^m}$$
Now put $\displaystyle R(x)=\frac{1}{(x+1)(1+ax)^m}$. We have
$$R(x)=\frac{1}{(1-a)^m(1+x)}-\frac{a}{1-a}\sum_{k=0}^{m-1}\frac{1}{(1-a)^k(1+ax)^{m-k}}$$
and hence
$$G_m(a)=-\frac{\log a}{(1-a)^m}-\sum_{k=0}^{m-2}\frac{1}{(m-k-1)(1-a)^{k+1}}$$
