A closed form for $\int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:\mathrm{d}x$ We already know that 

\begin{align} 
\displaystyle & \int_{0}^{\pi/2} x \ln(2 \cos x)\:\mathrm{d}x  = -\frac{7}{16} \zeta(3), 
\\\\ &  \int_{0}^{\pi/2} x^2 \ln^2(2 \cos x)\:\mathrm{d}x  = \frac{11 \pi}{16} \zeta(4).  \end{align} 

Does the following integral admit a closed form?

\begin{align} \displaystyle &  \int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:\mathrm{d}x  \end{align} 


We can try the harmonic analysis path. Since:
$$\log(2\cos x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}\cos(2nx),$$
$$\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n},\tag{1}$$
we have, as an example:
$$\int_{0}^{\pi/2}\log^3(2\sin x)dx=-\frac{3\pi}{4}\zeta(3)$$
since 
$$\int_{0}^{\pi/2}\cos(2n_1 x)\cos(2n_2 x)\cos(2n_3 x)\,dx = \frac{\pi}{8}\delta_{2\cdot\max n_i=(n_1+n_2+n_3)}\tag{2}$$
Now:
$$\pi/2-x = \frac{\pi}{4}+\frac{2}{\pi}\sum_{m=0}^{+\infty}\frac{\cos((4m+2)x)}{(2m+1)^2}\tag{3}$$
hence by multiplying $(1)$ and $(3)$ we can write the Fouries cosine series of $(\pi/2-x)\log(2\sin x)$ over $(0,\pi/2)$ and grab from $(2)$ a combinatorial equivalent for
$$\int_{0}^{\pi/2}\left((\pi/2-x)\log(2\sin x)\right)^3\,dx.$$
With the aid of Mathematica I got:
$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(-(2n+1)\left(\psi'\left(\frac{2n+1}{2}\right)-\psi'\left(-\frac{2n+1}{2}\right)\right)+2\left(2\gamma+\psi\left(\frac{2n+1}{2}\right)+\psi\left(-\frac{2n+1}{2}\right)\right)\right),$$
$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(2 H_{\frac{2n+1}{2}}-(2n+1)\left(\psi'\left(\frac{2n+1}{2}\right)-\psi'\left(-\frac{2n+1}{2}\right)\right)\right).$$

$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(-4\log 2+\sum_{j=0}^n\frac{4}{2j+1}+\frac{4}{2n+1}+(2n+1)\sum_{j=0}^{n}\frac{8}{(2j+1)^2}\right).\tag{1}$$

So we have the Fourier cosine series of $(\pi/2-x)\log(2\sin x)$ but the path does not look promising from here. However, if we replace $\pi/2-x$ with a periodic continuation we get the way nicer identity:

$$(\pi/2-x)\log(2\sin x)=-\left(\sum_{n=1}^{+\infty}\frac{\sin(2nx)}{n}\right)\left(\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n}\right)\tag{2}$$

that directly leads to:

$$f(x)=(\pi/2-x)\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{H_{n-1}}{n}\sin(2nx).\tag{3}$$

Now since $\int_{0}^{\pi/2}\sin(2mx)dx=\frac{\mathbb{1}_{m\equiv 1\pmod{2}}}{m}$ and $\int_{0}^{\pi/2}\sin(2ax)\sin(2bx)dx=\frac{\pi}{4}\delta_{a,b}$, the first two identites are easily proven. Now the three-terms integral
$$\int_{0}^{\pi/2}\sin(2ax)\sin(2bx)\sin(2cx)dx$$
is a linear combination of $\frac{1}{a+b+c},\frac{1}{-a+b+c},\frac{1}{a-b+c},\frac{1}{a+b-c}$ depending on the parity of $a,b,c$, so it is quite difficult to find, explicitly, the Fourier cosine series of $f(x)^2$ or the integral $\int_{0}^{\pi/2}f(x)^3\,dx$, but still not impossible. In particular, by this answer we know that the Taylor coefficients of the powers of $\log(1-x)$ depends on the generalized harmonic numbers. In our case,
$$-\log(1-x)=\sum_{n=1}^{+\infty}\frac{1}{n}x^n,$$
$$\log(1-x)^2 = \sum_{n=2}^{+\infty}\frac{2H_{n-1}}{n}x^n,$$
$$-\log(1-x)^3 = \sum_{n=3}^{+\infty}\frac{3H_{n-1}^2-3H_{n-1}^{(2)}}{n}x^n,$$
$$\log(1-x)^4 = \sum_{n=4}^{+\infty}\frac{4H_{n-1}^3+8H_{n-1}^{(3)}-12 H_{n-1}H_{n-1}^{(2)}}{n}x^n\tag{4}$$
hence we can just find a closed form for
$$\int_{0}^{\pi/2}x^3 (1-2\cos x)^n dx$$
and sum everything through the third previous identity. Ugh.
