Evaluating $\int_0^{\pi/2} \log(1 - x \cot x) \, dx\;$ (Is there a closed form?) I am interested in knowing if it is possible to find a closed-form solution to the following challenging log-cotangent integral

$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx$$

I very much doubt a closed form in terms of known mathematical constants can be found and have no reason to suspect one exists.
In the absence of a simple closed-form value being found, maybe the integral can be evaluated in terms of an infinite series. In this direction, since $0 < x \cot x < 1$ for all $x \in (0,\frac{\pi}{2})$ the log term appearing in the integrand can be expanded. Doing so produces
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx = -\sum_{n = 1}^\infty \frac{a_n}{n},$$
where
$$a_n = \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx. \tag1$$
Noting that
\begin{align*}
a_1 &= \int_0^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log (2)\\
a_2 &= \int_0^{\frac{\pi}{2}} (x \cot x)^2 \, dx = \pi \log (2) - \frac{\pi^3}{24}\\
a_3 &= \int_0^{\frac{\pi}{2}} (x \cot x)^3 \, dx = \frac{9\pi}{16} \zeta (3) - \frac{\pi^3}{16} + \frac{3\pi}{2} \log (2) - \frac{\pi^3}{8} \log (2),
\end{align*}
perhaps it is possible to find a general expression for $a_n$. Indeed, an attempt at finding (1) can be found here.
Any other approaches or suggestions to this challenging integral would be welcome.

Approximating the integral using rational approximants
Using integration by parts, we easily get
$$I = \int_0^{\pi/2} \ln(1 - x\cot x) \mathrm{d} x
= - \frac{\pi^3}{24} + \frac{\pi}{2}\ln 2 - \int_0^{\pi/2} \frac{x^3}{\tan x - x} \mathrm{d} x.$$
$I \approx -3.353337262889472...$
We focus on
$$I_1 = \int_0^{\pi/2} \frac{x^3}{\tan x - x} \mathrm{d} x.$$
We may approximate $I_1$ by rational approximants to $\tan x$ at $0$.

*

*Pade $(5, 4)$ approximant $\tan x \approx \frac{x(x^4 - 105x^2 + 945)}{15x^4 - 420x^2 + 945}
\triangleq g(x)$
We can bound the error by
$$0 \le \frac{x^3}{g(x) - x} - \frac{x^3}{\tan x - x} \le g(\pi/2) \approx 0.000045.$$
We have
$$I_1 \approx \int_0^{\pi/2} \frac{x^3}{g(x) - x} \mathrm{d} x = 
- \frac{5}{112}\pi^3 + \frac{165}{56}\pi - \frac{243}{56}\sqrt{10} 
\operatorname{arctanh} \frac{\pi\sqrt{10}}{30}.$$
Then we have $I \approx -3.353344689$ with relative error $< 0.000003$.
