Is there a closed form for this sum?

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.

The OP already gave the series expansion for $\arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) $ in here. Using that expansion, we have

\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 \\ &\quad = 16 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \sum_{p=0}^{\infty} \sum_{q=0}^{\infty} \frac{x^{2m+1}y^{2n+1}z^{2p+1}w^{2q+1}}{(2m+1)(2n+1)(2p+1)(2q+1)} \times \\ & \qquad \times \int_{0}^{\frac{\pi}{2}} \sin(2m+1)\theta \sin(2n+1)\theta \sin(2p+1)\theta \sin(2q+1)\theta \, d\theta \quad \quad \rm{\bf (1)} \end{align*}

Now we use the following somewhat lengthy trigonometric equality for general $x,y,z,w$ (which can be shown by Euler formulae, or more readily, by combining triple and double product formulae as listed in Wikipedia here) \begin{align*} & \quad 8 \sin(w) \sin(x) \sin(y) \sin(z) = \\ & - \cos(w+x+y-z) - \cos(w+y+z-x) - \cos(w+z+x-y) + \cos(w+x+y+z)+\\ &\cos(-w+x+y-z) + \cos(-w+y+z-x) + \cos(-w+z+x-y) - \cos(-w+x+y+z) \end{align*}

Integrals arise, where 4, 3, or 2 arguments have a positive sign. Let's look at these type of integrals. In the following, $\delta_n = 1$ for $n=0$, $\delta_n =0$ else.

a) 4 positive signs: $$ \int_{0}^{\frac{\pi}{2}} \cos(2m+2n+2p+2q+4)\theta \, d\theta = \frac{\pi}{2} \delta_{m+n+p+q +2} $$ Since $m,n,p,q \geq 0$, this term will never be met in the sum.

b) 3 positive signs: $$ \int_{0}^{\frac{\pi}{2}} \cos(2m+2n+2p-2q +2)\theta \, d\theta = \frac{\pi}{2} \delta_{m+n+p-q +1} $$

c) 2 positive signs: $$ \int_{0}^{\frac{\pi}{2}} \cos(2m+2n-2p-2q)\theta \, d\theta = \frac{\pi}{2} \delta_{m+n-p-q} $$

So the full integral (1) becomes

\begin{align*} & \frac{\pi}{16}\Big[ \delta_{m+n+p+q +2} - \delta_{-m+n+p+q +1}- \delta_{m-n+p+q +1}- \delta_{m+n-p+q +1}- \delta_{m+n+p-q +1} +\\ & \delta_{m+n-p-q} +\delta_{-m+n-p+q} +\delta_{m-n-p+q} \Big] \end{align*}

So the conjecture is proved if we establish

\begin{align*} & \frac12 (-1)^{m+n+p+q} d(m,n,p,q) = \quad \quad \rm{\bf (2)}\\ & \delta_{m+n+p+q +2} - \delta_{-m+n+p+q +1}- \delta_{m-n+p+q +1}- \delta_{m+n-p+q +1}- \delta_{m+n+p-q +1} +\\ & \delta_{m+n-p-q} +\delta_{-m+n-p+q} +\delta_{m-n-p+q} \end{align*}

Since $d(m,n,p,q)$ denotes the number of choices of signatures so that $$ \pm(2m+1) \pm(2n+1) \pm (2p+1) \pm(2q+1) = 0 $$

we can convince ourselves of the various cases. This is best seen with the case description given by the OP. (Note my comment above, there appears to be a typo in the one but last case.) Here is a list of terms (with corresponding signs) which apply. Note that we need half as many terms as given by the OP, due to the factor $\frac12$ in (2):

$$ \begin{cases} \delta_{m+n-p-q} +\delta_{-m+n-p+q} +\delta_{m-n-p+q} & \text{if } m = n = p = q, \\ \delta_{-m+n-p+q} +\delta_{m-n-p+q} & \text{else if } m = n < p = q, \\ \delta_{m+n-p-q} & \text{else if } m + n = p + q, \\ - \delta_{m+n+p-q +1} & \text{else if } m + n + p = q-1, \\ 0 & \text{else}. \end{cases} $$

This holds as well for all permutations of the indices. Done. $\quad \quad \Box$