Evaluate $\int_{0}^{\infty}\text{Ei}(-x)^3\left(\text{Ei}(-x)^2-5\text{Ei}(x)\text{Ei}(-x)+10\text{Ei}(x)^2\right)\text{d}x$ I want to evaluate an integral
$$I=\int_{0}^{\infty}\operatorname{Ei}(-x)^3\left(\text{Ei}(-x)^2-5\operatorname{Ei}(x)\operatorname{Ei}(-x)+10\operatorname{Ei}(x)^2\right)\text{d}x.$$
$\operatorname{Ei}(.)$ is the exponential integral, which is defined by
$$
\operatorname{Ei}(x)=\int_{0}^{x} \frac{e^{t}-1}{t}\text{d}t
+\ln\left ( \left | x \right |  \right )+\gamma
$$
with Euler-Mascheroni constant $\gamma=0.5772156649015932...$.
Actually,
$$
I=16\int_{0}^{\infty}\operatorname{si}(x)^5\text{d}x-30\pi^2\operatorname{Li}_2\left(\frac14\right)-60\pi^2\ln(2)^2.
$$
Where $\operatorname{si}(x)=-\int_{x}^{\infty} \frac{\sin(t)}{t}\text{d}t$ and $\operatorname{Li}_2(.)$ is the dilogarithm.

Question: How can we compute $I$ explicitly? Thanks for your help.
 A: Not an answer, but too long for a comment.
For what it's worth, $I$ is equivalent to the following expression:
\begin{align*}I &=-\frac{5\pi ^2}{3} \operatorname{Li}_2\left(\frac{1}{16}\right)+30 \operatorname{Li}_2\left(\frac{1}{4}\right)^2-\frac{105}{2}\operatorname{Li}_4\left(\frac{1}{9}\right)+60
   \operatorname{Li}_4\left(\frac{1}{4}\right)+780 \operatorname{Li}_4\left(\frac{1}{3}\right)\\ &-\frac{1120}{3}\operatorname{Li}_4\left(\frac{1}{2}\right)-480 \operatorname{Li}_4\left(\frac{2}{3}\right)+120
   \operatorname{Li}_2\left(\frac{1}{4}\right) \ln^2(2)+120 \operatorname{Li}_3\left(\frac{1}{4}\right) \ln(3)+10 \ln(3) \zeta (3)\\ &-\frac{34 \pi^4}{27}+\frac{1300}{9}\ln^4(2)-\frac{45}{2}\ln^4(3)-160\ln^3(2)\ln(3)+80\ln(2)\ln^3(3)+\frac{200\pi^2}{9}\ln^2(2)\\ &+5 \pi ^2 \ln^2(3)-40\pi^2\ln(2)\ln(3)-40\int_{1}^{\infty}\ln\left(\frac{t+2}{t+4}\right)\frac{\left(\ln(t) \ln(t+1)+\operatorname{Li}_2(-t)\right)}{t+3}\, dt\end{align*}
which this integral may be potentially more tractable, however, I do not have much hope. It is likely that this polylogarithm expression can be simplified, however, I have not spent much thought on that matter.
