Define the function $\mathcal{I}:\mathbb{R}_{>0}^{4}\rightarrow\mathbb{R}$ via the quadruple integral
$$\mathcal{I}{\left(a,b,c,d\right)}:=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\int_{0}^{z}\mathrm{d}w\,\exp{\left(-\frac{ax^{2}}{2}-\frac{by^{2}}{2}-\frac{cz^{2}}{2}-\frac{dw^{2}}{2}\right)}.\tag{1}$$
If we look at how this integral transforms under the substitution $\left(x,y,z,w\right)\mapsto\left(px,py,pz,pw\right)$ for some fixed but arbitrary positive real $p$, we obtain the following scaling relation:
$$\forall\left(a,b,c,d,p\right)\in\mathbb{R}_{>0}^{5}:\mathcal{I}{\left(a,b,c,d\right)}=p^{4}\mathcal{I}{\left(ap^{2},bp^{2},cp^{2},dp^{2}\right)}.\tag{2}$$
As such, in our general evaluation of $\mathcal{I}{\left(a,b,c,d\right)}$ it will suffice to consider the $a=1$ case.
To begin with, we derive some quick integration formulas that will be helpful below.
For any $p\in\mathbb{R}_{>0}$,
$$\begin{align}
\int_{0}^{\infty}\mathrm{d}x\,x^{3}\exp{\left(-\frac{px^{2}}{2}\right)}
&=\int_{0}^{\infty}\mathrm{d}y\,\frac{y}{2}\exp{\left(-\frac{py}{2}\right)};~~~\small{\left[x=\sqrt{y}\right]}\\
&=\int_{0}^{\infty}\mathrm{d}z\,\frac{2}{p}\cdot\frac{z}{p}\exp{\left(-z\right)};~~~\small{\left[y=\frac{2z}{p}\right]}\\
&=\frac{2}{p^{2}}\int_{0}^{\infty}\mathrm{d}z\,z\exp{\left(-z\right)}\\
&=\frac{2}{p^{2}}.\tag{3a}\\
\end{align}$$
Next, given $p\in\mathbb{R}_{>0}$ and setting $q:=\sqrt{p}$,
$$\begin{align}
\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left(1+pt^{2}\right)^{2}}
&=\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left(1+q^{2}t^{2}\right)^{2}}\\
&=\frac{1}{q^{3}}\int_{0}^{q}\mathrm{d}x\,\frac{2x^{2}}{\left(1+x^{2}\right)^{2}};~~~\small{\left[qt=x\right]}\\
&=\frac{1}{q^{3}}\int_{0}^{q}\mathrm{d}x\,\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan{\left(x\right)}-\frac{x}{1+x^{2}}\right]\\
&=\frac{1}{q^{3}}\left[\arctan{\left(q\right)}-\frac{q}{1+q^{2}}\right]\\
&=\left[\frac{\arctan{\left(q\right)}}{q^{3}}-\frac{1}{q^{2}\left(1+q^{2}\right)}\right]\\
&=\left[\frac{\arctan{\left(\sqrt{p}\right)}}{p\sqrt{p}}-\frac{1}{p\left(1+p\right)}\right].\tag{3b}\\
\end{align}$$
Defining the function $\mathcal{J}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ by
$$\mathcal{J}{\left(p,q,x\right)}:=\int_{0}^{x}\mathrm{d}y\,\left[\frac{\arctan{\left(\sqrt{p+qy^{2}}\right)}}{\left(p+qy^{2}\right)^{3/2}}-\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\right],\tag{3c}$$
we then find that for any $\left(p,q,x\right)\in\mathbb{R}_{>0}^{3}$,
$$\begin{align}
\mathcal{J}{\left(p,q,x\right)}
&=\int_{0}^{x}\mathrm{d}y\,\left[\frac{\arctan{\left(\sqrt{p+qy^{2}}\right)}}{\left(p+qy^{2}\right)^{3/2}}-\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\right]\\
&=\int_{0}^{x}\mathrm{d}y\,\frac{\arctan{\left(\sqrt{p+qy^{2}}\right)}}{\left(p+qy^{2}\right)^{3/2}}-\int_{0}^{x}\mathrm{d}y\,\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\\
&=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\int_{0}^{x}\mathrm{d}y\,\frac{qy^{2}}{p\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)};~~~\small{I.B.P.}\\
&~~~~~-\int_{0}^{x}\mathrm{d}y\,\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\\
&=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\int_{0}^{x}\mathrm{d}y\,\frac{1}{p\left(1+p+qy^{2}\right)}\\
&=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\frac{1}{p\sqrt{q\left(1+p\right)}}\int_{0}^{x\sqrt{\frac{q}{1+p}}}\mathrm{d}t\,\frac{1}{\left(1+t^{2}\right)};~~~\small{\left[y=t\sqrt{\frac{1+p}{q}}\right]}\\
&=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\frac{\arctan{\left(\frac{x\sqrt{q}}{\sqrt{1+p}}\right)}}{p\sqrt{q\left(1+p\right)}}.\tag{3d}\\
\end{align}$$
Suppose $\left(b,c,d\right)\in\mathbb{R}_{>0}^{3}$.
We begin by reducing $\mathcal{I}$ to a single-variable integral in the following way:
$$\begin{align}
\mathcal{I}{\left(1,b,c,d\right)}
&=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\int_{0}^{z}\mathrm{d}w\,\exp{\left(-\frac{x^{2}}{2}-\frac{by^{2}}{2}-\frac{cz^{2}}{2}-\frac{dw^{2}}{2}\right)}\\
&=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,x^{3}\\
&~~~~~\times\exp{\left(-\frac{x^{2}}{2}-\frac{bx^{2}t^{2}}{2}-\frac{cx^{2}u^{2}}{2}-\frac{dx^{2}v^{2}}{2}\right)};~~~\small{\left[\left(y,z,w\right)=\left(xt,xu,xv\right)\right]}\\
&=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,x^{3}e^{-\frac{\left(1+bt^{2}+cu^{2}+dv^{2}\right)x^{2}}{2}}\\
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\int_{0}^{\infty}\mathrm{d}x\,x^{3}e^{-\frac{\left(1+bt^{2}+cu^{2}+dv^{2}\right)x^{2}}{2}}\\
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\frac{2}{\left(1+bt^{2}+cu^{2}+dv^{2}\right)^{2}},\\
\end{align}$$
and then,
$$\begin{align}
\mathcal{I}{\left(1,b,c,d\right)}
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\frac{2}{\left(1+bt^{2}+cu^{2}+dv^{2}\right)^{2}}\\
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{2t^{2}}{\left(1+bt^{2}+ct^{2}x^{2}+dt^{2}y^{2}\right)^{2}};~~~\small{\left[\left(u,v\right)=\left(tx,ty\right)\right]}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left(1+bt^{2}+ct^{2}x^{2}+dt^{2}y^{2}\right)^{2}}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left[1+\left(b+cx^{2}+dy^{2}\right)t^{2}\right]^{2}}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\bigg{[}\frac{\arctan{\left(\sqrt{b+cx^{2}+dy^{2}}\right)}}{\left(b+cx^{2}+dy^{2}\right)^{3/2}}\\
&~~~~~-\frac{1}{\left(b+cx^{2}+dy^{2}\right)\left(1+b+cx^{2}+dy^{2}\right)}\bigg{]}\\
&=\int_{0}^{1}\mathrm{d}x\,\mathcal{J}{\left(b+cx^{2},d,x\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\left[\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}-\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}\right].\tag{4}\\
\end{align}$$
Define the auxiliary functions $\mathcal{F}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ and $\mathcal{G}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ by the respective integrals
$$\begin{align}
\mathcal{F}{\left(b,c,d\right)}
&:=\int_{0}^{1}\mathrm{d}x\,\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}\tag{5a}\\
\end{align}$$
and
$$\begin{align}
\mathcal{G}{\left(b,c,d\right)}
&:=\int_{0}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}.\tag{5b}\\
\end{align}$$
Suppose $\left(b,c,d\right)\in\mathbb{R}_{>0}^{3}$, and set $p:=\sqrt{\frac{b}{c}}\land q:=\sqrt{\frac{1+b}{c}}\land r:=\sqrt{\frac{d}{c}}$. Next, noting that $0<p<q$, set $s:=\frac{p}{q}\land z:=q^{-1}$. We then obtain the following expressions for $\mathcal{G}$ and $\mathcal{F}$:
$$\begin{align}
\mathcal{G}{\left(b,c,d\right)}
&=\int_{0}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}\\
&=\frac{1}{c\sqrt{cd}}\int_{0}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{rx}{\sqrt{q^{2}+x^{2}}}\right)}}{\left(p^{2}+x^{2}\right)\sqrt{q^{2}+x^{2}}}\\
&=\frac{1}{c\sqrt{cd}}\int_{0}^{q^{-1}}\mathrm{d}y\,\frac{q\arctan{\left(\frac{rqy}{\sqrt{q^{2}+q^{2}y^{2}}}\right)}}{\left(p^{2}+q^{2}y^{2}\right)\sqrt{q^{2}+q^{2}y^{2}}};~~~\small{\left[x=qy\right]}\\
&=\frac{1}{c\sqrt{cd}}\int_{0}^{q^{-1}}\mathrm{d}y\,\frac{\arctan{\left(\frac{ry}{\sqrt{1+y^{2}}}\right)}}{\left(p^{2}+q^{2}y^{2}\right)\sqrt{1+y^{2}}}\\
&=\frac{1}{\left(1+b\right)\sqrt{cd}}\int_{0}^{z}\mathrm{d}y\,\frac{\arctan{\left(\frac{ry}{\sqrt{1+y^{2}}}\right)}}{\left(s^{2}+y^{2}\right)\sqrt{1+y^{2}}}\\
&=\frac{1}{\left(1+b\right)\sqrt{cd}}\int_{0}^{\frac{z}{\sqrt{1+z^{2}}}}\mathrm{d}t\,\frac{\arctan{\left(rt\right)}}{s^{2}+\left(1-s^{2}\right)t^{2}};~~~\small{\left[\frac{y}{\sqrt{1+y^{2}}}=t\right]}\\
&=\frac{1}{\left(1+b\right)\sqrt{cd}}\int_{0}^{\frac{rz}{\sqrt{1+z^{2}}}}\mathrm{d}u\,\frac{r\arctan{\left(u\right)}}{r^{2}s^{2}+\left(1-s^{2}\right)u^{2}};~~~\small{\left[rt=u\right]}\\
&=\int_{0}^{\sqrt{\frac{d}{1+b+c}}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}},\\
\end{align}$$
and
$$\begin{align}
\mathcal{F}{\left(b,c,d\right)}
&=\int_{0}^{1}\mathrm{d}x\,\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x\arctan{\left(\sqrt{b+\left(c+d\right)x^{2}}\right)}}{2\left(b+cx^{2}\right)\sqrt{b+\left(c+d\right)x^{2}}}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{\arctan{\left(\sqrt{b+\left(c+d\right)y}\right)}}{2\left(b+cy\right)\sqrt{b+\left(c+d\right)y}};~~~\small{\left[x^{2}=y\right]}\\
&=\int_{b}^{b+c+d}\mathrm{d}t\,\frac{\arctan{\left(\sqrt{t}\right)}}{2\left(bd+ct\right)\sqrt{t}};~~~\small{\left[y=\frac{t-b}{c+d}\right]}\\
&=\int_{\sqrt{b}}^{\sqrt{b+c+d}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}};~~~\small{\left[\sqrt{t}=u\right]}.\\
\end{align}$$
Hence, we can express $\mathcal{I}$ as
$$\begin{align}
\mathcal{I}{\left(1,b,c,d\right)}
&=\int_{0}^{1}\mathrm{d}x\,\left[\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}-\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}\right]\\
&=\mathcal{F}{\left(b,c,d\right)}-\mathcal{G}{\left(b,c,d\right)}\\
&=\int_{\sqrt{b}}^{\sqrt{b+c+d}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}}-\int_{0}^{\sqrt{\frac{d}{1+b+c}}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}}\\
&=\frac{1}{\sqrt{bcd}}\int_{\sqrt{\frac{c}{d}}}^{\sqrt{\frac{(b+c+d)c}{bd}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}\\
&~~~~~-\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{c}{b(1+b+c)}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}};~~~\small{\left[u=x\sqrt{\frac{bd}{c}}\right]}\\
&=\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{(b+c+d)c}{bd}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}-\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{c}{d}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}\\
&~~~~~-\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{c}{b(1+b+c)}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}.\tag{6}\\
\end{align}$$
Finally, it can be shown (see Appendix) that the following integration formula holds for all $\left(p,z\right)\in\mathbb{R}_{>0}^{2}$:
$$\int_{0}^{z}\mathrm{d}x\,\frac{2\arctan{\left(px\right)}}{1+x^{2}}=\arctan^{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(-\frac{1-p}{1+p}\right)}-\operatorname{Li}_{2}{\left(\frac{1-p}{1+p},\pi-2\arctan{\left(z\right)}\right)},\tag{7}$$
where the two-variable variant of the dilogarithm is defined by the integral representation
$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$
Since each of the three remaining integrals in the last line of $(6)$ above can be evaluated with formula $(7)$, this in principle completes the derivation. I don't see much point in going through the tedium of actually writing out the explicit expression though.
Appendix.
Define the function $\mathcal{K}:\mathbb{R}_{>0}^{2}\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{K}{\left(a,z\right)}:=\int_{0}^{z}\mathrm{d}x\,\frac{\arctan{\left(ax\right)}}{1+x^{2}}.$$
In the special case $a=1$ the integral is elementary, and we have
$$\mathcal{K}{\left(1,z\right)}=\int_{0}^{z}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{1+x^{2}}=\frac12\arctan^{2}{\left(z\right)};~~~\small{z\in\mathbb{R}_{>0}}.$$
Suppose $\left(a,z\right)\in\mathbb{R}_{>0}^{2}$, and assume $a\neq1$. Then, $-1<\frac{1-a}{1+a}<1\land\frac{1-a}{1+a}\neq0$.
Set $\omega:=2\arctan{\left(z\right)}$, and note that $0<\omega<\pi\land z=\tan{\left(\frac{\omega}{2}\right)}$.
$$\begin{align}
\mathcal{K}{\left(a,z\right)}
&=\int_{0}^{z}\mathrm{d}x\,\frac{\arctan{\left(ax\right)}}{1+x^{2}}\\
&=\int_{0}^{z}\mathrm{d}x\,\frac{1}{1+x^{2}}\int_{0}^{a}\mathrm{d}t\,\frac{x}{1+x^{2}t^{2}}\\
&=\int_{0}^{z}\mathrm{d}x\int_{0}^{a}\mathrm{d}t\,\frac{x}{\left(1+x^{2}\right)\left(1+t^{2}x^{2}\right)}\\
&=\int_{0}^{a}\mathrm{d}t\int_{0}^{z}\mathrm{d}x\,\frac{x}{\left(1+x^{2}\right)\left(1+t^{2}x^{2}\right)}\\
&=\frac12\int_{0}^{a}\mathrm{d}t\int_{0}^{z^{2}}\mathrm{d}y\,\frac{1}{\left(1+y\right)\left(1+t^{2}y\right)};~~~\small{\left[x^{2}=y\right]}\\
&=\frac12\int_{0}^{a}\mathrm{d}t\int_{0}^{z^{2}}\mathrm{d}y\,\frac{d}{dy}\left[\frac{\ln{\left(1+y\right)}-\ln{\left(1+t^{2}y\right)}}{1-t^{2}}\right]\\
&=\frac12\int_{0}^{a}\mathrm{d}t\,\frac{\ln{\left(1+z^{2}\right)}-\ln{\left(1+z^{2}t^{2}\right)}}{1-t^{2}}\\
&=-\frac12\int_{0}^{a}\mathrm{d}t\,\frac{\ln{\left(\frac{1+z^{2}t^{2}}{1+z^{2}}\right)}}{1-t^{2}}\\
&=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\ln{\left(\frac{1+z^{2}\left(\frac{1-x}{1+x}\right)^{2}}{1+z^{2}}\right)};~~~\small{\left[t=\frac{1-x}{1+x}\right]}\\
&=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\ln{\left(\frac{\left(1+x\right)^{2}+z^{2}\left(1-x\right)^{2}}{\left(1+z^{2}\right)\left(1+x\right)^{2}}\right)}\\
&=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\ln{\left(\frac{1+2\left(\frac{1-z^{2}}{1+z^{2}}\right)x+x^{2}}{\left(1+x\right)^{2}}\right)}\\
&=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\left[\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}-2\ln{\left(1+x\right)}\right]\\
&=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}+\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{x}\\
&=-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}+\frac12\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}\\
&~~~~~-\frac12\operatorname{Li}_{2}{\left(-1\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}\\
&=+\frac12\operatorname{Li}_{2}{\left(-1\right)}+\frac18\omega^{2}+\frac12\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}\\
&~~~~~-\frac12\operatorname{Li}_{2}{\left(-1\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}\\
&=\frac12\arctan^{2}{\left(z\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}\\
&~~~~~-\frac12\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}x\,\frac{(-1)\ln{\left(1-2x\cos{\left(\pi-\omega\right)}+x^{2}\right)}}{2x}\\
&=\frac12\arctan^{2}{\left(z\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}-\frac12\operatorname{Li}_{2}{\left(\frac{1-a}{1+a},\pi-\omega\right)}.\blacksquare\\
\end{align}$$
