Family of definite integrals involving Dedekind eta function of a complex argument, $\int_0^{\infty} \eta^k(ix)dx$

Let $q = e^{-\pi x}$ and then $$\eta(ix) = q^{1/12}\prod_{n = 1}^{\infty}(1 - q^{2n}) = 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}$$ and $x = K'/K$. Then $$\eta^{4}(ix) = 2^{-4/3}\left(\frac{2K}{\pi}\right)^{2}(kk')^{2/3}$$ Now we can see that $x = 0$ then $k = 1$ and $x \to \infty$ implies $k \to 0$ and $$\frac{dx}{dk} = -\frac{\pi}{2kk'^{2}K^{2}}$$ and hence the integral $$I(4) = \int_{0}^{\infty}\eta^{4}(ix)\,dx = \int_{0}^{1}2^{-4/3}\left(\frac{2K}{\pi}\right)^{2}(kk')^{2/3}\cdot\frac{\pi}{2kk'^{2}K^{2}}\,dk$$ and then $$I(4) = \frac{2^{-1/3}}{\pi}\int_{0}^{1}k^{-1/3}(1 - k^{2})^{-2/3}\,dk$$ and putting $k^{2} = m$ we get $dk = \dfrac{dm}{2\sqrt{m}}$ and hence $$I(4) = \frac{2^{-4/3}}{\pi}\int_{0}^{1}m^{-2/3}(1 - m)^{-2/3}\,dm = \frac{2^{-4/3}}{\pi}\frac{\Gamma^{2}(1/3)}{\Gamma(2/3)}$$ and since $\Gamma(1/3)\Gamma(2/3) = 2\pi/\sqrt{3}$ we get the desired value of $I(4)$. I was lucky to get $I(4)$ because the $K^{2}$ term got cancelled in the evaluation of integral.

Incidentally I get the evaluation of a nice sum $$\sum_{n = 0}^{\infty} (-1)^n \dfrac{\sinh\dfrac{\pi}{\sqrt{3}}(2n + 1)}{\cosh\dfrac{\sqrt{3}\pi}{2}(2n + 1)} = \frac{\pi}{3\sqrt[3]{2}\Gamma^{3}(2/3)}$$

I feel like an update is due.

As of today, I possess the (remarkably beautiful) closed forms of $\, I(k)\,$ corresponding to each $\,k\,$ for which $\,\eta^k\,$ is lacunary (Serre 1985): $$k \in \mathcal{S} := \{1,2,3,4,6,8,10,14,26\}.$$ See also Records on the vanishing of Fourier coefficients of Powers Of the Dedekind Eta Function (2018).

I originally found these closed forms (excluding $k=26$) experimentally, by some educated guessing.
In this answer by @pisco, he proves the $\,k=8\,$ case and with similar methods proves the latter cases $\,k= 10,14,26\,\,$, thereby completing the list.

As of now, it remains unknown whether a closed form exists for $\,k$'s not in $\mathcal{S}$, and any new information is welcome.

Enjoy:

$$\begin{align*} &\int_0^\infty \eta( i x) \, \mathrm{d} x = \frac{2 \pi}{\sqrt{3}}\\[0.1em] \\& \int_0^\infty \eta^2( i x) \, \mathrm{d} x = \ln\left( 1+ \sqrt{3} +\sqrt{3+2\sqrt{3}}\right) \\[0.1em] \\& \int_0^\infty \eta^3( i x) \, \mathrm{d} x = 1 \\[0.1em] \\& \int_0^\infty \eta^4( i x) \, \mathrm{d} x = \frac{3^{1/2}}{2^{7/3} \, \pi^2} \,\Gamma\left(\frac13\right)^3 \\[0.1em] \\& \int_0^\infty \eta^6( i x) \, \mathrm{d} x = \frac{1}{2^{4} \, \pi^3} \,\Gamma\left(\frac14\right)^4 \\[0.1em] \\& \int_0^\infty \eta^8( i x) \, \mathrm{d} x = \frac{3^{1/2}}{2^{6} \, \pi^6} \,\Gamma\left(\frac13\right)^9 \\[0.1em] \\& \int_0^\infty \eta^{10}( i x) \, \mathrm{d} x = \frac{3^{1/4} \, \left(\sqrt{3}-1\right)^2}{2^{15/2} \, \pi^6} \,\Gamma\left(\frac14\right)^8 \\[0.1em] \\& \int_0^\infty \eta^{14}( i x) \, \mathrm{d} x = \frac{3^{13/4} \, \left(\sqrt{3}-1\right)^4}{2^{27/2} \, \pi^{12}} \,\Gamma\left(\frac13\right)^{18} \\[0.1em] \\& \int_0^\infty \eta^{26}( i x) \, \mathrm{d} x = \frac{3^{31/4} \, 5^2 \, \left(1+\sqrt{3}\right)^8 \, \left(35+58\sqrt{3}\right)}{2^{55/2}\cdot 11\cdot 13 \, \pi^{24}} \,\Gamma\left(\frac13\right)^{36} \\&\qquad \qquad\qquad\quad+ \frac{3^{1/4} \cdot 7 \, \left(1+\sqrt{3}\right)^8 \, \left(194-215\sqrt{3}\right)}{2^{35/2}\cdot 11\cdot 13 \, \pi^{18}} \,\Gamma\left(\frac14\right)^{24} \end{align*}$$