While generalizing the previous result, I conjectured that the series expansion of
\begin{align*} \int_{0}^{\frac{\pi}{2}} \arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) \arctan \left( \frac{2y \sin\theta}{1-y^{2}} \right) \arctan \left( \frac{2z \sin\theta}{1-z^{2}} \right) \arctan \left( \frac{2w \sin\theta}{1-w^{2}} \right) \, d\theta \end{align*}
is equal to
$$ \frac{\pi}{2} \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \sum_{l=0}^{\infty} (-1)^{i+j+k+l} d(i,j,k,l) \frac{x^{2i+1}}{2i+1} \frac{y^{2j+1}}{2j+1} \frac{z^{2k+1}}{2k+1} \frac{w^{2l+1}}{2l+1}, \tag{1} $$
where $d(i,j,k,l)$ denotes the number of choices of signatures so that
$$ \pm(2i+1) \pm(2j+1) \pm (2k+1) \pm(2l+1) = 0. $$
I checked that the formula $\text{(1)}$ is correct up to degree 40. Compared to the egregious look of the original integral, this series representation is quite neat and tantalizing.
Assuming that $i \leq j \leq k \leq l$, we have the following algorithm:
$$ (-1)^{i+j+k+l} d(i,j,k,l) = \begin{cases} 6 & \text{if } i = j = k = l, \\ 4 & \text{else if } i = j < k = l, \\ 2 & \text{else if } i + l = j + k, \\ -2 & \text{else if } i + j + k = l-1, \\ 0 & \text{else}. \end{cases} $$
If we only sum the terms of $\text{(1)}$ corresponding to $d = 6$ or $4$, we get a combination of Legendre chi functions. But I have no idea how to simplify further.