Closed form for $\frac{\int_0^1 Ei(x)^5 \ln(x) dx}{\int_0^1 Ei(x)^3\ln(x) dx}$? 
Consider the expression
$$\frac{\int_0^1 Ei(x)^5 \ln(x) dx}{\int_0^1 Ei(x)^3 \ln(x) dx} $$
A) Can we rewrite this with a single integral sign?
B) Do we have a closed form for this expression in terms of hypergeometric functions?
C) Is there a closed form without hypergeometric functions for this value?

They told me A,B,C are all true.
But that was on April 1.
So I'm not sure.
I think A is true and B and C false.
What is the truth?
It seems the value is about $29.6900899$.

A partial answer.
The identity $$\text{Ei}(x)=\gamma+\log(x)+\sum_{n\geq 1}\frac{x^n}{n\cdot n!} $$
leads to
$$ \int_{0}^{1}\text{Ei}(x)\log(x)\,dx =-\gamma+2-\sum_{n\geq 1}\frac{1}{n(n+1)(n+1)!}=-\gamma+2-3+e=e-\gamma-1\tag{1}$$
and similarly
$$ \int_{0}^{1}\text{Ei}^2(x)\log(x)\,dx $$
can be computed in terms of $\gamma,e$, the hypergeometric function $\phantom{}_3 F_3\left(1,1,1;2,2,2;1\right)$ related to $\sum_{n\geq 1}\frac{1}{n\cdot n!(n+1)^3}$ and the double series
$$ \begin{eqnarray*}\sum_{n,m\geq 1}\frac{1}{n\cdot n!\cdot m\cdot m!\cdot (m+n+1)^2}&=&\sum_{s=2}^{+\infty}\frac{1}{s!(s+1)^2}\sum_{n=1}^{s-1}\frac{\binom{s}{n}}{n(s-n)}\\&=&\sum_{s=2}^{+\infty}\frac{2}{s\cdot s!(s+1)^2}\sum_{n=1}^{s-1}\binom{s}{n}\frac{1}{n}.\end{eqnarray*}$$
It turns out that the integrals $\int_{0}^{1}\text{Ei}^k(x)\log(x)\,dx $ are related to series with a nice hypergeometric structure. They are a sort of generalization of Stirling numbers of the first kind. 
So, summarizing: I am not sure about $A)$ and $C)$, but $B)$ is true for sure. 

Another chance might be to play with the inverse function of $\text{Ei}(z)$ over $(0,1)$. By denoting this function as $\text{Ie}(z)$ we have $\lim_{z\to -\infty}\text{Ie}(z)=0$, $\lim_{z\to\text{Ei}(1)}\text{Ie}(z)=1$ and $\frac{d}{dz}\log\text{Ie}(z)=e^{-\text{Ie}(z)}$:
$$ \int_{0}^{1}\text{Ei}(x)^k\log(x)\,dx=\int_{-\infty}^{\text{Ei}(1)}z^k \text{Ie}(z)\log\text{Ie}(z)\frac{d}{dz}\log\text{Ie}(z)\,dz $$
the RHS now appears to be manageable through integration by parts, and we may wonder about applications of the Lagrange inversion theorem for deriving an explicit series representation for $\text{Ie}(z)$ or its logarithm
$$ \log\text{Ie}(z) = \frac{(z-\text{Ei}(1))}{e}-\frac{(z-\text{Ei}(1))^2}{2e^2}+\frac{(z-\text{Ei}(1))^3}{6e^3}-\frac{(z-\text{Ei}(1))^5}{24e^5}+\frac{(z-\text{Ei}(1))^6}{40e^6}-\frac{(z-\text{Ei}(1))^7}{720e^7}+\frac{169(z-\text{Ei}(1))^8}{20160 e^8}-\ldots$$
