$ \int_0^\infty dx\, x^{-3/4} \ln(1+x) \operatorname{Li}_2 \left( -\frac 1 x \right)$ How to show the following identity from Wolfram?
$$
\int_0^\infty dx\, x^{-3/4} \ln(1+x) \operatorname{Li}_2 \left( -\frac 1 x \right) =-2\pi \sqrt 2\left[\frac{5 \pi^2}{3} + 16 \left(3 \ln 2 + G - 4 \right) \right].
$$
Here $G$ is Catalan's constant.
I thought of using $\operatorname{Li}_2(-x) + \operatorname{Li}_2(-1/x) = -\zeta(2) - \frac 1 2 \ln(x)^ 2$, but the result is not much easier.

We start from the integral 
$$
\int_{0}^{\infty} \frac{x^u }{(1+s x)(1+t x) }dx= \frac{\pi}{\sin \pi u} \cdot \frac{s^u - t^u}{s^u t^u (s-t)}
$$
for $s$, $t>0$ and $0 < u < 1$.  
Integrating  with respect to $t$ we get 
\begin{eqnarray}
\int_{0}^{\infty} \frac{x^u \log ( 1 + \frac{1}{t x})}{(1+s x)}= - \frac{\pi}{\sin \pi u}\cdot \frac{1}{s^{u+1}} \cdot \left(\, B(s/t;1+u, 0) + \log( 1 - s/t\,)\,\right)
\end{eqnarray}
for $u>0$ and $0 < s < t$, where $B(z;a,b) \colon = \int_0^z v^{a-1}(1-v)^{b-1} d v$ is the  incomplete Beta function.  Note that we have $B(z;a,0) = z^a \cdot \Phi(z, 1, a)$, where $\Phi(z,s,a) \colon = \sum_{n \ge 0} \frac{z^n}{(n+ a)^s}$ is the  Lerch transcendent. 
Integrating again with respect to $t$ we get 
\begin{eqnarray}
\int_0^{\infty}\frac{x^u}{1 + s x}\, Li_2(-\frac{1}{t x}) = \frac{\pi}{\sin \pi u \cdot (s t)^{1+u}} \cdot \left( s^{1+u} \Phi(s/t, 2, 1+u) - t^{1+u} Li_2(s/t)\right)
\end{eqnarray}
again for $u>0$ and $0 < s < t$. In particular, for $u=1/4$ we get
$$\int_0^{\infty}\frac{x^{1/4}}{1 + s x}\, Li_2(-\frac{1}{t x}) = \frac{\pi \sqrt{2}}{(s t)^{5/4}} \cdot \left( s^{5/4} \Phi(s/t, 2, 5/4) - t^{5/4} Li_2(s/t)\right)$$
Now integrate with respect to $s$ and get, with $t=1$ and $s = v^4$:
\begin{eqnarray}
\int_0^{\infty}x^{-3/4}\log(1 + v^4 x)\, Li_2(-\frac{1}{ x}) =-\frac{4 \pi \sqrt{2}}{v}\left( v^5 \phi(v^4,2,5/4) -Li_2(v^4) - 16 v + 8(1-v) \arctan v + 4 (1+v) \log [(1+v)(1+v^2)] +\\+ 4 (1+v)\log(1-v) + 8(1+v) \textrm{arctanh} (v)\right) 
\end{eqnarray}
for $0 < v<1$ and taking $v \to 1$ we get 
$$\int_0^{\infty}x^{-3/4}\log(1 +  x)\, Li_2(-\frac{1}{ x})=
 \frac{2}{3} \sqrt{2} \pi \,( \, 96 + \pi^2 -144 \log 2 - 6\, \zeta(2, 5/4)\,)= -3.339758...$$ 
and now use that $\zeta(2, 5/4) = \pi^2 + 8 G - 16$, where  $\zeta(z, a)$ is the Hurwitz zeta function, and $G$ is the Catalan's constant. 
