Closed-form for $\int_0^\infty {\frac{{\ln \left( {1 + x} \right)}}{{1 + ax}}{e^{ - bx}}{x^n}{\rm{d}}x} $ I am trying to find the integration of the following
$$\int_0^\infty  {\frac{{\ln \left( {1 + x} \right)}}{{1 + ax}}{e^{ - bx}}{x^n}{\rm{d}}x} $$
Here $a>0, b>0$, and $n$ is an integer.
I think if we get the Meijer-G representation of 
$$\frac{{\ln \left( {1 + x} \right)}}{{1 + ax}}$$
we can use Laplace transform to get the closed-form expression.
But I don't know how to express the above function as Meijer-G function.
Thanks.
 A: Disclaimer: Not a full solution, but I've gotten as far as a system of differential equations.
First, it's enough to consider an easier integral:
$$I(a,b)=I_0(a,b)=\int_0^\infty  {\frac{{\ln \left( {1 + x} \right)}}{{1 + ax}}{e^{ - bx}}{\rm{d}}x}$$
It's obvious, that for $n \in \mathbb{N}$:
$$I_n(a,b)=\int_0^\infty  {\frac{{\ln \left( {1 + x} \right)}}{{1 + ax}}{e^{ - bx}}x^n{\rm{d}}x}=(-1)^n \frac{\partial^n I_0}{\partial b^n}$$

The first option to convert the problem into a partial differential equation is to write $b=ac$ and:
$$I(a,ac)=J(a,c)=e^c \int_0^\infty  {\frac{{\ln \left( {1 + x} \right)}}{{1 + ax}}{e^{ - c(ax+1)}}{\rm{d}}x}$$
So if we take a $c$ derivative, we obtain a much more simple integral:
$$\frac{\partial J}{\partial c}=J-e^c \int_0^\infty  \ln (1 + x) e^{-c(ax+1)}\mathrm{d} x=J-\int_0^\infty  \ln (1 + x) e^{-cax}\mathrm{d} x$$
The latter integral has a well known solution (which can be obtained integrating by parts and using the definition of exponential integral):
$$\int_0^\infty  \ln (1 + x) e^{-cax}\mathrm{d} x=I(0,ac)= \\ =\frac{1}{ac}\int_0^\infty \frac{ e^{-acx}}{1+x}\mathrm{d} x=-\frac{e^{ac}}{ac} \text{Ei}(-ac)$$
Here Ei is the exponential integral.
Finally, we obtain a differential equation:

$$\frac{\partial J(a,c)}{\partial c}-J(a,c)-\frac{e^{ac}}{ac} \text{Ei}(-ac)=0$$


To obtain another PDE we change the variable $t=ax$:
$$I(a,b)=\frac{1}{a} \int_0^\infty  {\frac{{\ln \left( 1+\frac{t}{a} \right)}}{{1 + t}}{e^{ - \frac{b}{a}t}}{\rm{d}}t}$$
Now we take the $a$ derivative:
$$a \frac{\partial I(a,b)}{\partial a}+I(a,b)=\frac{b}{a^2} \int_0^\infty  {\frac{{\ln \left( 1+\frac{t}{a} \right)}}{{1 + t}}{t~e^{ - \frac{b}{a}t}}{\rm{d}}t}-\frac{1}{a^2} \int_0^\infty e^{-\frac{b}{a} t} \frac{t dt}{(1+t)(1+\frac{t}{a})}$$
We can see by direct comparison that:
$$\frac{b}{a^2} \int_0^\infty  {\frac{{\ln \left( 1+\frac{t}{a} \right)}}{{1 + t}}{t~e^{ - \frac{b}{a}t}}{\rm{d}}t}=-b \frac{\partial I(a,b)}{\partial b}$$
As for the second part of the derivative we can use partial fractions to compute the integral:
$$\frac{1}{a}=\alpha,~~~~\frac{b}{a}=\beta$$
$$\int_0^\infty e^{-\beta t} \frac{t dt}{(1+t)(1+\alpha t)}=\frac{1}{\alpha-1} \left(\int_0^\infty e^{-\beta t} \frac{dt}{1+t}-\int_0^\infty e^{-\beta t} \frac{dt}{1+\alpha t} \right)$$
But we already know how to compute the two latter integrals (see above), so:
$$\int_0^\infty e^{-\beta t} \frac{dt}{1+t}=-e^{\beta} \text{Ei}(-\beta)$$
$$\int_0^\infty e^{-\beta t} \frac{dt}{1+\alpha t}=-\frac{e^{\beta / \alpha}}{\alpha} \text{Ei}(-\beta / \alpha)$$

Finally we obtain the system of partial differential equations, which (I think) completely define the function $I(a,b)$:

$$a \frac{\partial I}{\partial a}+b \frac{\partial I}{\partial b}+I+\frac{1}{a(a-1)} \left( e^{b / a} \text{Ei}(-b / a)-ae^b \text{Ei}(-b)  \right)=0$$
$$a \frac{\partial I}{\partial b}-I-\frac{e^{b}}{b} \text{Ei}(-b)=0$$



$$I(0,b)=-\frac{1}{b} e^b \text{Ei}(-b)$$
$$I(a,0) \to \infty$$
$$I(a,\infty) = 0$$
$$I(\infty,b) = 0$$


From the above two equations we can also obtain another with only $a$ derivative:

$$a \frac{\partial I}{\partial a}+\left(1+\frac{b}{a} \right) I+\frac{1}{a(a-1)} \left( e^{b / a} \text{Ei}(-b / a)-e^b \text{Ei}(-b)  \right)=0$$


P.S. I would really appreciate if someone checks my post for mistakes. I'll be checking myself, but just to be sure.
