About the integral $\int_{0}^{+\infty}\frac{\text{arcsinh}(x)\,\text{arcsinh}(\lambda x)}{x^2}\,dx$ Context: I am working on a project involving integral means, algebraic inequalities and hypergeometric functions. Today I was studying the integral over the region $A<b<a<A+1$ of $$\int_{0}^{+\infty}\frac{dt}{\sqrt{(t^2+a^2)(t^2+b^2)}}$$
which is well-known to be related with the complete elliptic integral of the first kind and the AGM mean. My train of thoughts led me to the parametric integral
$$ I(\lambda) = \int_{0}^{+\infty}\frac{\text{arcsinh}(x)\,\text{arcsinh}(\lambda x)}{x^2}\,dx,\qquad \lambda>0 $$
which my version of Mathematica returns as a Meijer G function, and by other ways I know to be related with a series involving squared central binomial coefficients and values of the incomplete Beta function. $I(1)=\frac{\pi^2}{2}$ is straightforward to prove.

Question: Is it possible to find a closed form for $I(\lambda)$ in terms of "usual" functions, or at least a manageable representation as a fast-convergent series?

My first temptation was to apply Feyman's trick, but $\int_{0}^{+\infty}\frac{\text{arcsinh}(x)}{x\sqrt{1+\lambda^2 x^2}}\,dx $ does not seem really easier to tackle or to integrate. I hope that I am wrong, of course. There also is a similar integral which has a simple closed form:
$$ \int_{0}^{+\infty}\frac{\log(1+x^2)\log(\lambda^2+x^2)}{x^2}\,dx = 2\pi\left(1+\tfrac{1}{\lambda}\right)\log(\lambda+1).$$
 A: From the generalized Parseval relation for the Mellin transform, we can use the result
$$
\int_0^\infty g(\lambda x)g(x)x^{-2}\,dx=\frac{1}{2i\pi}\int_{\delta-i\infty}^{\delta+i\infty}\tilde{g}(s)\tilde{g}(-1-s)\lambda^{-s}\,ds
$$
Here, from Ederlyi table (6.6.13) p.323, for $g(x)=\text{arcsinh} (x)$, one has 
$$\tilde{g}(s)=-\frac{1}{2s}B\left( \frac{s+1}{2},\frac{-s}{2} \right)$$
for $-1<s<0$. Then
\begin{equation}
I(\lambda)=\frac{-1}{8i\pi^2}\int_{\delta-i\infty}^{\delta+i\infty}\Gamma^2\left( \frac{1+s}{2} \right)\Gamma^2\left( -\frac{s}{2} \right)\frac{\lambda^{-s}}{s(s+1)}\,ds
\end{equation}
with $-1<\delta<0$. Poles are at $s=-1,-3,-5...$ and $0,2,4...$. Poles at $s=-1,0$ are of order 3 the other are double. For $s\to\infty$, the function to be integrated is $\sim \left|\lambda\right|^{-s}s^{-3}\csc^{-2}(\pi s/2)$.
Then, for $\left|\lambda\right|<1$, we close the contour with a large half-circle on the left of the vertical line $\Re(s)=\delta$. With the help of Maple to compute the residues, one gets
\begin{align}
I(\lambda)=\frac{1}{2}&\left( \left( \ln \frac{\lambda}{4}-1 \right)^2 +1+\frac{\pi^2}{3}\right)\lambda
-\frac{1}{2\pi}\sum_{n=2}^\infty\left( \frac{\Gamma\left( n-\frac{1}{2} \right)}{\Gamma(n)} \right)^2.\\
&.\left[-\ln(\lambda)+
\psi(n+1)-\psi(n+\frac{1}{2})+\frac{4n^2-n-2}{2n(n-1)(2n-1)}
\right]\frac{\lambda^{2n-1}}{(n-1)(2n-1)}
\end{align} 
The series converges for $\left|\lambda\right| < 1$.
For $\left|\lambda\right| >1$, we may use the relation
$$ I(\lambda)=\lambda I\left(\frac{1}{\lambda}\right)$$ which is obvious from the integral expression. This functional relation may be retrieved by closing the contour by a half-circle on the right of the line.
A: This is not going to be the full answer to this question because I believe that this integral cannot be represented in terms of elementary functions plus poly-logarithms. I will simply extract the "simple" part out of this integral and leave the undigested part as it is in hope that some genius might tackle it in the future. We start by computing the anti-derivative of the integrand. Firstly we note the following result:
\begin{eqnarray}
(1) \int \frac{\log(x)}{\sqrt{1+\lambda^2 x^2}}dx=\frac{1}{2 \lambda} \left[\mbox{arcsinh}(\lambda x) \left( \mbox{arcsinh}(\lambda x) - 2 \log(2 \lambda)\right) + Li_2(e^{-2\mbox{arcsinh}(\lambda x)})\right]
\end{eqnarray}
We have:
\begin{eqnarray}
&&\int \frac{\mbox{arcsinh}(x) \mbox{arcsinh}(\lambda x)}{x} dx=\\
&& -\frac{1}{x} \mbox{arcsinh}(x) \mbox{arcsinh}(\lambda x) +\\
&& (\log(x)-\log(1+\sqrt{1+x^2})) \mbox{arcsinh}(\lambda x) + (\log(x)-\log(1+\sqrt{1+\lambda^2 x^2})) \mbox{arcsinh}( x) \lambda +\\
&&-\lambda \int\frac{\log(x)-\log(1+\sqrt{1+x^2})}{\sqrt{1+\lambda^2 x^2}}dx -\lambda \int\frac{\log(x)-\log(1+\sqrt{1+\lambda^2 x^2})}{\sqrt{1+ x^2}}dx=\\
&&\frac{1}{2} \left(\right.\\
&&\left.-\lambda \text{Li}_2\left(e^{-2 \mbox{arcsinh}(x)}\right)-\text{Li}_2\left(e^{-2 \mbox{arcsinh}(\lambda x)}\right)\right.\\
&&\left.-2 \log \left(\sqrt{x^2+1}+1\right) \mbox{arcsinh}(\lambda
   x)-2 \lambda \mbox{arcsinh}(x) \log \left(\sqrt{\lambda^2 x^2+1}+1\right)-\lambda \mbox{arcsinh}(x)^2\right.\\
&&\left.-\frac{2 \mbox{arcsinh}(x) \mbox{arcsinh}(\lambda x)}{x}-\mbox{arcsinh}(\lambda x)^2\right.\\
&&\left.+2 \lambda \log (x) \mbox{arcsinh}(x)+\lambda \log (4) \mbox{arcsinh}(x)+2 \log (2 \lambda) \mbox{arcsinh}(\lambda x)+2 \log (x) \mbox{arcsinh}(\lambda x)\right.\\
&&\left.\right)+\\
&&\lambda \underbrace{\int \frac{\log(1+\sqrt{1+x^2})}{\sqrt{1+\lambda^2 x^2}} dx}_{{\mathcal I}_1(x)}+\lambda \underbrace{\int \frac{\log(1+\sqrt{1+\lambda^2 x^2})}{\sqrt{1+ x^2}} dx}_{{\mathcal I}_2(x)}
\end{eqnarray}
In the first line we integrated by parts twice and in the second line we used the identity(1). Now, we deal with the remaining integrals by expanding them in a Taylor series about $\lambda=1$. We have:
\begin{eqnarray}
&&{\mathcal I}_1(x) = -\frac{1}{2} [\log(2 z)]^2 - 2 Li_2(-z) +\\
&& (\lambda-1) \cdot \left( 2 \arctan(z)+2 Li_2(-z)+\frac{1}{2} \log(z)^2+\right.\\
&&\left.\frac{\left(2 \log (2)-z^2 (4-2 \log (2))\right) \log (z)+4 \left(z^2-1\right) \log (z+1)+4 \log (2)}{2 \left(z^2+1\right)}\right)+O((\lambda-1)^2)\\
&&{\mathcal I}_2(x) = -\frac{1}{2} [\log(2 z)]^2 - 2 Li_2(-z) +\\
&& (\lambda-1) \cdot \left( -2 \arctan(z)+\log(z)\right) + O((\lambda-1)^2)
\end{eqnarray}
where $z:= \exp(\mbox{arcsinh}(x))$.
