Conjectured value of a difficult integral with Dedekind eta functions In my research on quantum groups I have the following conjecture:
\begin{equation}
\int_0^\infty\frac{\eta (2 i x)^8}{\eta (i x)^2 \eta (4 i x)^2}\,dx\,{\stackrel?=}\,\frac{K(\tfrac{1}{\sqrt{2}})}{\pi}\tag{1}
\end{equation}
where
\begin{equation}
\eta(ix)=e^{-\frac{\pi x}{12}}\prod_{n=1}^\infty (1-e^{-2n\pi x})
\end{equation}
is the Dedekind eta function and
\begin{equation}
K(\tfrac{1}{\sqrt{2}})=\frac{\Gamma^2(\tfrac14)}{4\sqrt{\pi}}
\end{equation}
is the Elliptic integral singular value. 
This value has been guessed using Inverse symbolic calculator and then checked numerically to a high precision, but I do not have a proof. 
Question: Is (1) true?
 A: The following solution uses the link between elliptic integrals and theta functions.

Let's use $q=e^{-\pi x} $ and then we define function $f$ via $$f(q) =q^{1/12}\prod_{n=1}^{\infty}(1-q^{2n})=\eta(ix)\tag{1}$$ Let $x=K(k') /K(k) $ where $k, k'$ are elliptic moduli complementary to each other and $K(k)=K $ is the complete elliptic integral of the first kind. From the theory of elliptic integrals and theta functions we have
\begin{align}
f(q)=\eta(ix)&=2^{-1/3}\sqrt{\frac{2K} {\pi}} (kk') ^{1/6}\tag{2a}\\
f(q^2)=\eta(2ix)&=2^{-2/3}\sqrt {\frac {2K}{\pi}}k^{1/3}k'^{1/12}\tag{2b}\\
f(q^4)=\eta(4ix)&=2^{-13/12}\sqrt{\frac{2K}{\pi}}\frac{k^{2/3}k'^{1/24}}{(1+k')^{1/4}}\tag{2c}
\end{align}
The integrand in question is $$\frac{f^8(q^2)} {f^2(q)f^2(q^4)}=2^{-5/2}\left(\frac{2K}{\pi}\right)^2kk'^{1/4}(1+k')^{1/2} $$ and we have $$\frac{dx} {dk} =\frac{dx}{dq}\cdot\frac {dq} {dk} =-\frac{1} {\pi q} \cdot\frac{\pi^2 q}{2kk'^2K^2} =-\frac{\pi} {2kk'^2K^2}\tag{3}$$ so that the desired integral is equal to $$\frac{1}{2\pi\sqrt{2}}\int_{0}^{1}k'^{-7/4}(1+k')^{1/2}\,dk=\frac{1}{2\pi\sqrt{2}}\int_{0}^{1}t^{-3/4}(1-t)^{-1/2}\,dt\tag{4}$$ (via substitution $k'=\sqrt{1-k^2}=t$) which is evaluated easily via Beta/Gamma functions to obtain desired result.

We were lucky that after switching from $x$ to $k$ the elliptic integral $K$ got cancelled and the resulting integral was having an algebraic function as an integrand. Thus the original integral appears to be designed to have such a nice evaluation. See this answer for another instance of this technique. 
