Our goal here is the general evaluation of the following six-parameter class of integrals:
$$\int_{t_{0}}^{t_{1}}\mathrm{d}t\,\frac{s\arctan{\left(p+qt\right)}}{\left(t-r\right)^{2}+s^{2}};~~~\small{(p,q,r,s,t_{0},t_{1})\in\mathbb{R}^{6}\land0<s}.$$
Such an integral can be rewritten and split up as
$$\begin{align}
\int_{t_{0}}^{t_{1}}\mathrm{d}t\,\frac{s\arctan{\left(p+qt\right)}}{\left(t-r\right)^{2}+s^{2}}
&=\int_{\frac{t_{0}-r}{s}}^{\frac{t_{1}-r}{s}}\mathrm{d}u\,\frac{\arctan{\left(p+qr+qsu\right)}}{1+u^{2}};~~~\small{\left[t=r+su\right]}\\
&=\int_{\arctan{\left(\frac{t_{0}-r}{s}\right)}}^{\arctan{\left(\frac{t_{1}-r}{s}\right)}}\mathrm{d}\tau\,\arctan{\left(p+qr+qs\tan{\left(\tau\right)}\right)};~~~\small{\left[u=\tan{\left(\tau\right)}\right]}\\
&=\int_{0}^{\arctan{\left(\frac{t_{1}-r}{s}\right)}}\mathrm{d}\tau\,\arctan{\left(p+qr+qs\tan{\left(\tau\right)}\right)}\\
&~~~~~-\int_{0}^{\arctan{\left(\frac{t_{0}-r}{s}\right)}}\mathrm{d}\tau\,\arctan{\left(p+qr+qs\tan{\left(\tau\right)}\right)}\\
&=\mathcal{I}{\left(p+qr,qs,\arctan{\left(\frac{t_{1}-r}{s}\right)}\right)}-\mathcal{I}{\left(p+qr,qs,\arctan{\left(\frac{t_{0}-r}{s}\right)}\right)},\\
\end{align}$$
where in the last line we've introduced the function $\mathcal{I}:\mathbb{R}\times\mathbb{R}\times(-\frac{\pi}{2},\frac{\pi}{2})\rightarrow\mathbb{R}$ given by the three-parameter definite integral
$$\mathcal{I}{\left(a,b,\theta\right)}:=\int_{0}^{\theta}\mathrm{d}\tau\,\arctan{\left(a+b\tan{\left(\tau\right)}\right)}.$$
Our focus then shifts naturally to the general evaluation of $\mathcal{I}$, which fortunately has significantly less parameter clutter than what we started with.
Observe that for $(a,b,\theta)\in\mathbb{R}\times\mathbb{R}\times(-\frac{\pi}{2},\frac{\pi}{2})$, we can show that
We can quickly derive a couple of basic functional relations for $\mathcal{I}$ that essentially allow us to choose the signs of up to two of the three parameters: for $(a,b,\theta)\in\mathbb{R}\times\mathbb{R}\times(-\frac{\pi}{2},\frac{\pi}{2})$,
$$\begin{align}
\mathcal{I}{\left(-a,-b,\theta\right)}
&=\int_{0}^{\theta}\mathrm{d}\tau\,\arctan{\left(-a-b\tan{\left(\tau\right)}\right)}\\
&=-\int_{0}^{\theta}\mathrm{d}\tau\,\arctan{\left(a+b\tan{\left(\tau\right)}\right)}\\
&=-\mathcal{I}{\left(a,b,\theta\right)},\\
\end{align}$$
and
$$\begin{align}
\mathcal{I}{\left(-a,b,-\theta\right)}
&=\int_{0}^{-\theta}\mathrm{d}\tau\,\arctan{\left(-a+b\tan{\left(\tau\right)}\right)}\\
&=\int_{0}^{\theta}\mathrm{d}\tau\,(-1)\arctan{\left(-a+b\tan{\left(-\tau\right)}\right)};~~~\small{\left[\tau\mapsto-\tau\right]}\\
&=\int_{0}^{\theta}\mathrm{d}\tau\,(-1)\arctan{\left(-a-b\tan{\left(\tau\right)}\right)}\\
&=\int_{0}^{\theta}\mathrm{d}\tau\,\arctan{\left(a+b\tan{\left(\tau\right)}\right)}\\
&=\mathcal{I}{\left(a,b,\theta\right)}.\\
\end{align}$$
Going forward, we will assume without loss of generality that both $b$ and $\theta$ are nonnegative. In fact, we might as well go ahead and say $b$ and $\theta$ are strictly positive since the $\theta=0$ or $b=0$ cases can be excluded as basically trivial:
$$\mathcal{I}{\left(a,b,0\right)}=0,$$
$$\mathcal{I}{\left(a,0,\theta\right)}=\theta\arctan{\left(a\right)}.$$
We consider the $a=0$ case first. Suppose $b\in\mathbb{R}_{>0}\land\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$.
$$\begin{align}
\mathcal{I}{\left(0,b,\theta\right)}
&=\int_{0}^{\theta}\mathrm{d}\tau\,\arctan{\left(b\tan{\left(\tau\right)}\right)}\\
&=\int_{0}^{\theta}\mathrm{d}\tau\int_{0}^{b}\mathrm{d}x\,\frac{\tan{\left(\tau\right)}}{1+x^{2}\tan^{2}{\left(\tau\right)}}\\
&=\int_{0}^{b}\mathrm{d}x\int_{0}^{\theta}\mathrm{d}\tau\,\frac{\tan{\left(\tau\right)}}{1+x^{2}\tan^{2}{\left(\tau\right)}}\\
&=\int_{0}^{b}\mathrm{d}x\int_{0}^{\theta}\mathrm{d}\tau\,\frac{\cos{\left(\tau\right)}\sin{\left(\tau\right)}}{\cos^{2}{\left(\tau\right)}+x^{2}\sin^{2}{\left(\tau\right)}}\\
&=\int_{0}^{b}\mathrm{d}x\int_{0}^{\theta}\mathrm{d}\tau\,\frac{\cos{\left(\tau\right)}\sin{\left(\tau\right)}}{1-\left(1-x^{2}\right)\sin^{2}{\left(\tau\right)}}\\
&=\int_{0}^{b}\mathrm{d}x\int_{0}^{\sin{\left(\theta\right)}}\mathrm{d}t\,\frac{t}{1-\left(1-x^{2}\right)t^{2}};~~~\small{\left[\sin{\left(\tau\right)}=t\right]}\\
&=\int_{0}^{b}\mathrm{d}x\int_{0}^{\sin^{2}{\left(\theta\right)}}\mathrm{d}u\,\frac{1}{2\left[1-\left(1-x^{2}\right)u\right]};~~~\small{\left[t^{2}=u\right]}\\
&=\int_{0}^{b}\mathrm{d}x\int_{0}^{\sin^{2}{\left(\theta\right)}}\mathrm{d}u\,\frac{d}{du}\left[-\frac{\ln{\left(1-\left(1-x^{2}\right)u\right)}}{2\left(1-x^{2}\right)}\right]\\
&=\int_{0}^{b}\mathrm{d}x\,\left[-\frac{\ln{\left(1-\left(1-x^{2}\right)\sin^{2}{\left(\theta\right)}\right)}}{2\left(1-x^{2}\right)}\right]\\
&=-\frac12\int_{0}^{b}\mathrm{d}x\,\frac{1}{1-x^{2}}\ln{\left(1-\left(1-x^{2}\right)\sin^{2}{\left(\theta\right)}\right)}\\
&=-\frac12\int_{1}^{\frac{1-b}{1+b}}\mathrm{d}y\,\frac{(-2)}{\left(1+y\right)^{2}}\cdot\frac{\left(1+y\right)^{2}}{4y}\ln{\left(1-\frac{4y}{\left(1+y\right)^{2}}\sin^{2}{\left(\theta\right)}\right)};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\
&=-\frac12\int_{\frac{1-b}{1+b}}^{1}\mathrm{d}y\,\frac{1}{2y}\ln{\left(\frac{\left(1+y\right)^{2}-4y\sin^{2}{\left(\theta\right)}}{\left(1+y\right)^{2}}\right)}\\
&=-\frac12\int_{\frac{1-b}{1+b}}^{1}\mathrm{d}y\,\frac{1}{2y}\ln{\left(\frac{1+2y\left[1-2\sin^{2}{\left(\theta\right)}\right]+y^{2}}{\left(1+y\right)^{2}}\right)}\\
&=-\frac12\int_{\frac{1-b}{1+b}}^{1}\mathrm{d}y\,\frac{1}{2y}\ln{\left(\frac{1+2y\cos{\left(2\theta\right)}+y^{2}}{\left(1+y\right)^{2}}\right)}\\
&=\frac12\int_{\frac{1-b}{1+b}}^{1}\mathrm{d}y\,\frac{1}{2y}\left[2\ln{\left(1+y\right)}-\ln{\left(1+2y\cos{\left(2\theta\right)}+y^{2}\right)}\right]\\
&=\frac12\left[\int_{\frac{1-b}{1+b}}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}-\int_{\frac{1-b}{1+b}}^{1}\mathrm{d}y\,\frac{\ln{\left(1+2y\cos{\left(2\theta\right)}+y^{2}\right)}}{2y}\right]\\
&=\frac12\left[-\operatorname{Li}_{2}{\left(-1\right)}+\operatorname{Li}_{2}{\left(-\frac{1-b}{1+b}\right)}+\operatorname{Li}_{2}{\left(1,\pi-2\theta\right)}-\operatorname{Li}_{2}{\left(\frac{1-b}{1+b},\pi-2\theta\right)}\right]\\
&=\frac12\left[\theta^{2}+\operatorname{Li}_{2}{\left(-\frac{1-b}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{1-b}{1+b},\pi-2\theta\right)}\right],\\
\end{align}$$
where
$$\operatorname{Li}_{2}{\left(z\right)}:=-\int_{0}^{z}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x};~~~\small{z\in(-\infty,1]},$$
$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}y\,\frac{\ln{\left(1-2y\cos{\left(\theta\right)}+y^{2}\right)}}{y};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$
Note: It can be shown (see appendix below for proof) that
$$\operatorname{Li}_{2}{\left(1,\theta\right)}=\operatorname{Li}_{2}{\left(-1\right)}+\left(\frac{\theta}{2}-\frac{\pi}{2}\right)^{2};~~~\small{0\le\theta\le2\pi}.$$
We now consider the $a\neq0$ case.
Suppose $a\in\mathbb{R}_{\neq0}\land b\in\mathbb{R}_{>0}\land\theta\in(0,\frac{\pi}{2})$. The trick here is to find a set of values $(A,B,\varphi)\in\mathbb{R}^{3}$ such that
$$\forall\tau\in[0,\theta]:\arctan{\left(a+b\tan{\left(\tau\right)}\right)}=\arctan{\left(A\right)}+\arctan{\left(B\tan{\left(\tau-\varphi\right)}\right)}.$$
Note: In order for the RHS of $(1)$ to be singularity-free over the integration interval (like the LHS is), we require that
$$\forall\tau\in[0,\theta]:-\frac{\pi}{2}<\tau-\varphi<\frac{\pi}{2}.$$
Also note that we must have $B\neq0$ in order for the RHS of $(1)$ to be non-constant with respect to $\tau$.
Recall the arctangent addition formula
$$\arctan{\left(z\right)}+\arctan{\left(w\right)}=\arctan{\left(\frac{z+w}{1-zw}\right)};~~~\small{(z,w)\in\mathbb{R}^{2}\land zw<1},$$
as well as the angle difference formula for tangent
$$\tan{\left(x-y\right)}=\frac{\tan{\left(x\right)}-\tan{\left(y\right)}}{1+\tan{\left(x\right)}\tan{\left(y\right)}};~~~\small{(x,y)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)^{2}}.$$
Assuming $0<1-AB\tan{\left(\tau-\varphi\right)}$, we then have
$$\arctan{\left(A\right)}+\arctan{\left(B\tan{\left(\tau-\varphi\right)}\right)}=\arctan{\left(\frac{A+B\tan{\left(\tau-\varphi\right)}}{1-AB\tan{\left(\tau-\varphi\right)}}\right)}.$$
Then, given $B\in\mathbb{R}_{>0}\land B\neq1\land\tau\in\left[0,\frac{\pi}{2}\right)\land\varphi\in\left(0,\frac{\pi}{2}\right)$, and setting $A:=\frac{\tan{\left(\varphi\right)}}{B}\land\beta:=2\arctan{\left(B\right)}$, we have $0<\beta<\pi\land\beta\neq\frac{\pi}{2}\land B=\tan{\left(\frac{\beta}{2}\right)}$, and
$$1-AB\tan{\left(\tau-\varphi\right)}=\frac{1+\tan^{2}{\left(\varphi\right)}}{1+\tan{\left(\tau\right)}\tan{\left(\varphi\right)}}>0,$$
and then
$$\begin{align}
\frac{A+B\tan{\left(\tau-\varphi\right)}}{1-AB\tan{\left(\tau-\varphi\right)}}
&=\frac{A+B\left[\frac{\tan{\left(\tau\right)}-\tan{\left(\varphi\right)}}{1+\tan{\left(\tau\right)}\tan{\left(\varphi\right)}}\right]}{1-AB\left[\frac{\tan{\left(\tau\right)}-\tan{\left(\varphi\right)}}{1+\tan{\left(\tau\right)}\tan{\left(\varphi\right)}}\right]}\\
&=\frac{A\left[1+\tan{\left(\tau\right)}\tan{\left(\varphi\right)}\right]+B\left[\tan{\left(\tau\right)}-\tan{\left(\varphi\right)}\right]}{\left[1+\tan{\left(\tau\right)}\tan{\left(\varphi\right)}\right]-AB\left[\tan{\left(\tau\right)}-\tan{\left(\varphi\right)}\right]}\\
&=\frac{\left[A-B\tan{\left(\varphi\right)}\right]+\left[B+A\tan{\left(\varphi\right)}\right]\tan{\left(\tau\right)}}{\left[1+AB\tan{\left(\varphi\right)}\right]-\left[AB-\tan{\left(\varphi\right)}\right]\tan{\left(\tau\right)}}\\
&=\frac{\left[B^{-1}\tan{\left(\varphi\right)}-B\tan{\left(\varphi\right)}\right]+\left[B+B^{-1}\tan^{2}{\left(\varphi\right)}\right]\tan{\left(\tau\right)}}{\left[1+\tan^{2}{\left(\varphi\right)}\right]};~~~\small{AB=\tan{\left(\varphi\right)}}\\
&=\frac{\left(B^{-1}-B\right)\tan{\left(\varphi\right)}+\left[B+B^{-1}\tan^{2}{\left(\varphi\right)}\right]\tan{\left(\tau\right)}}{\sec^{2}{\left(\varphi\right)}}\\
&=\left(B^{-1}-B\right)\tan{\left(\varphi\right)}\cos^{2}{\left(\varphi\right)}+\left[B+B^{-1}\tan^{2}{\left(\varphi\right)}\right]\cos^{2}{\left(\varphi\right)}\tan{\left(\tau\right)}\\
&=\left(\frac{1-B^{2}}{B}\right)\sin{\left(\varphi\right)}\cos{\left(\varphi\right)}+\frac{B^{2}\cos^{2}{\left(\varphi\right)}+\sin^{2}{\left(\varphi\right)}}{B}\tan{\left(\tau\right)}\\
&=\frac{1-B^{2}}{2B}\sin{\left(2\varphi\right)}+\frac{B^{2}\left[1+\cos{\left(2\varphi\right)}\right]+\left[1-\cos{\left(2\varphi\right)}\right]}{2B}\tan{\left(\tau\right)}\\
&=\frac{1-B^{2}}{2B}\sin{\left(2\varphi\right)}+\left[\frac{1+B^{2}}{2B}-\frac{1-B^{2}}{2B}\cos{\left(2\varphi\right)}\right]\tan{\left(\tau\right)}\\
&=\cot{\left(\beta\right)}\sin{\left(2\varphi\right)}+\left[\csc{\left(\beta\right)}-\cot{\left(\beta\right)}\cos{\left(2\varphi\right)}\right]\tan{\left(\tau\right)}\\
&=a+b\tan{\left(\tau\right)},\\
\end{align}$$
provided $\beta,\varphi$ satisfy the system of equations
$$\begin{cases}
&a=\cot{\left(\beta\right)}\sin{\left(2\varphi\right)},\\
&b=\csc{\left(\beta\right)}-\cot{\left(\beta\right)}\cos{\left(2\varphi\right)}.\\
\end{cases}$$
We find
$$\begin{cases}
&a=\cot{\left(\beta\right)}\sin{\left(2\varphi\right)},\\
&b=\csc{\left(\beta\right)}-\cot{\left(\beta\right)}\cos{\left(2\varphi\right)},\\
\end{cases}$$
$$\implies\begin{cases}
&\sin{\left(\beta\right)}=\frac{2b}{1+a^{2}+b^{2}},\\
&\cot{\left(2\varphi\right)}=\frac{1+a^{2}-b^{2}}{2ab},\\
\end{cases}$$
$$\implies\begin{cases}
&\frac{2B}{1+B^{2}}=\frac{2b}{1+a^{2}+b^{2}},\\
&\tan{\left(\frac{\pi}{2}-2\varphi\right)}=\frac{1+a^{2}-b^{2}}{2ab},\\
\end{cases}$$
$$\implies\begin{cases}
&B=\frac{1+a^{2}+b^{2}-\operatorname{sgn}{\left(a\right)}\sqrt{\left[a^{2}+\left(b-1\right)^{2}\right]\left[a^{2}+\left(b+1\right)^{2}\right]}}{2b},\\
&\varphi=\frac{\pi}{4}-\frac12\arctan{\left(\frac{1+a^{2}-b^{2}}{2ab}\right)},\\
&\varphi=\arctan{\left(\frac{-1-a^{2}+b^{2}+\operatorname{sgn}{\left(a\right)}\sqrt{\left[a^{2}+\left(b-1\right)^{2}\right]\left[a^{2}+\left(b+1\right)^{2}\right]}}{2ab}\right)}.\\
\end{cases}$$
Hence, for $a\in\mathbb{R}_{\neq0}\land b\in\mathbb{R}_{>0}\land\theta\in(0,\frac{\pi}{2})$ with
$$B:=\frac{1+a^{2}+b^{2}-\operatorname{sgn}{\left(a\right)}\sqrt{\left[a^{2}+\left(b-1\right)^{2}\right]\left[a^{2}+\left(b+1\right)^{2}\right]}}{2b},$$
$$\varphi:=\frac{\pi}{4}-\frac12\arctan{\left(\frac{1+a^{2}-b^{2}}{2ab}\right)},$$
we obtain
$$\begin{align}
\mathcal{I}{\left(a,b,\theta\right)}
&=\int_{0}^{\theta}\mathrm{d}\tau\,\arctan{\left(a+b\tan{\left(\tau\right)}\right)}\\
&=\int_{0}^{\theta}\mathrm{d}\tau\,\left[\arctan{\left(\frac{\tan{\left(\varphi\right)}}{B}\right)}+\arctan{\left(B\tan{\left(\tau-\varphi\right)}\right)}\right]\\
&=\int_{0}^{\theta}\mathrm{d}\tau\,\arctan{\left(\frac{\tan{\left(\varphi\right)}}{B}\right)}+\int_{0}^{\theta}\mathrm{d}\tau\,\arctan{\left(B\tan{\left(\tau-\varphi\right)}\right)}\\
&=\theta\arctan{\left(\frac{\tan{\left(\varphi\right)}}{B}\right)}+\int_{0}^{\theta}\mathrm{d}\tau\,\arctan{\left(B\tan{\left(\tau-\varphi\right)}\right)}\\
&=\theta\arctan{\left(\frac{\tan{\left(\varphi\right)}}{B}\right)}+\int_{-\varphi}^{\theta-\varphi}\mathrm{d}\tau\,\arctan{\left(B\tan{\left(\tau\right)}\right)};~~~\small{\left[\tau\mapsto\tau+\varphi\right]}\\
&=\theta\arctan{\left(\frac{\tan{\left(\varphi\right)}}{B}\right)}+\int_{0}^{\theta-\varphi}\mathrm{d}\tau\,\arctan{\left(B\tan{\left(\tau\right)}\right)}\\
&~~~~~-\int_{0}^{-\varphi}\mathrm{d}\tau\,\arctan{\left(B\tan{\left(\tau\right)}\right)}\\
&=\theta\arctan{\left(\frac{\tan{\left(\varphi\right)}}{B}\right)}\\
&~~~~~+\frac12\left(\theta-\varphi\right)^{2}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-B}{1+B}\right)}-\frac12\operatorname{Li}_{2}{\left(\frac{1-B}{1+B},\pi-2\theta+2\varphi\right)}\\
&~~~~~-\frac12\varphi^{2}-\frac12\operatorname{Li}_{2}{\left(-\frac{1-B}{1+B}\right)}+\frac12\operatorname{Li}_{2}{\left(\frac{1-B}{1+B},\pi+2\varphi\right)}\\
&=\frac12\left(\theta-\varphi\right)^{2}-\frac12\varphi^{2}+\theta\arctan{\left(\frac{\tan{\left(\varphi\right)}}{B}\right)}\\
&~~~~~+\frac12\operatorname{Li}_{2}{\left(\frac{1-B}{1+B},\pi+2\varphi\right)}-\frac12\operatorname{Li}_{2}{\left(\frac{1-B}{1+B},\pi-2\theta+2\varphi\right)}.\\
\end{align}$$
Appendix. For $0<\theta<2\pi$,
$$\begin{align}
\operatorname{Li}_{2}{\left(1,\theta\right)}
&=\operatorname{Li}_{2}{\left(1,\pi\right)}+\int_{\pi}^{\theta}\mathrm{d}\varphi\,\frac{d}{d\varphi}\operatorname{Li}_{2}{\left(1,\varphi\right)}\\
&=-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-2y\cos{\left(\pi\right)}+y^{2}\right)}}{y}\\
&~~~~~+\int_{\pi}^{\theta}\mathrm{d}\varphi\,\frac{d}{d\varphi}\left[-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-2y\cos{\left(\varphi\right)}+y^{2}\right)}}{y}\right]\\
&=-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+2y+y^{2}\right)}}{2y}\\
&~~~~~+\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}y\,\frac{\partial}{\partial\varphi}\left[-\frac{\ln{\left(1-2y\cos{\left(\varphi\right)}+y^{2}\right)}}{2y}\right]\\
&=-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left((1+y)^{2}\right)}}{2y}+\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}y\,\left[-\frac{\sin{\left(\varphi\right)}}{1-2y\cos{\left(\varphi\right)}+y^{2}}\right]\\
&=-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}+\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}y\,\frac{\left[-\sin{\left(\varphi\right)}\right]}{1-2y\cos{\left(\varphi\right)}+y^{2}}\\
&=\operatorname{Li}_{2}{\left(-1\right)}+\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}y\,\frac{\left[-\sin{\left(\varphi\right)}\right]}{1-2y\cos{\left(\varphi\right)}+y^{2}}\\
&=\operatorname{Li}_{2}{\left(-1\right)}+\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{1}^{0}\mathrm{d}t\,\frac{(-2)}{\left(1+t\right)^{2}}\cdot\frac{\left[-\sin{\left(\varphi\right)}\right]}{1-2\left(\frac{1-t}{1+t}\right)\cos{\left(\varphi\right)}+\left(\frac{1-t}{1+t}\right)^{2}};~~~\small{\left[y=\frac{1-t}{1+t}\right]}\\
&=\operatorname{Li}_{2}{\left(-1\right)}+\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}t\,\frac{2\left[-\sin{\left(\varphi\right)}\right]}{\left(1+t\right)^{2}-2\left(1-t^{2}\right)\cos{\left(\varphi\right)}+\left(1-t\right)^{2}}\\
&=\operatorname{Li}_{2}{\left(-1\right)}+\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}t\,\frac{\left[-\sin{\left(\varphi\right)}\right]}{\left(1+t^{2}\right)-\left(1-t^{2}\right)\cos{\left(\varphi\right)}}\\
&=\operatorname{Li}_{2}{\left(-1\right)}+\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}t\,\frac{\left[-\sin{\left(\varphi\right)}\right]}{\left[1-\cos{\left(\varphi\right)}\right]+t^{2}\left[1+\cos{\left(\varphi\right)}\right]}\\
&=\operatorname{Li}_{2}{\left(-1\right)}+\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}t\,\frac{\left[-2\sin{\left(\frac{\varphi}{2}\right)}\cos{\left(\frac{\varphi}{2}\right)}\right]}{2\sin^{2}{\left(\frac{\varphi}{2}\right)}+2t^{2}\cos^{2}{\left(\frac{\varphi}{2}\right)}}\\
&=\operatorname{Li}_{2}{\left(-1\right)}-\int_{\pi}^{\theta}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}t\,\frac{\cot{\left(\frac{\varphi}{2}\right)}}{1+t^{2}\cot^{2}{\left(\frac{\varphi}{2}\right)}}\\
&=\operatorname{Li}_{2}{\left(-1\right)}-\int_{\pi}^{\theta}\mathrm{d}\varphi\,\arctan{\left(\cot{\left(\frac{\varphi}{2}\right)}\right)}\\
&=\operatorname{Li}_{2}{\left(-1\right)}-\int_{\pi}^{\theta}\mathrm{d}\varphi\,\arctan{\left(\tan{\left(\frac{\pi}{2}-\frac{\varphi}{2}\right)}\right)}\\
&=\operatorname{Li}_{2}{\left(-1\right)}-\int_{\pi}^{\theta}\mathrm{d}\varphi\,\left(\frac{\pi}{2}-\frac{\varphi}{2}\right)\\
&=\operatorname{Li}_{2}{\left(-1\right)}+\int_{\pi}^{\theta}\mathrm{d}\varphi\,\left(\frac{\varphi}{2}-\frac{\pi}{2}\right)\\
&=\operatorname{Li}_{2}{\left(-1\right)}+\left(\frac{\theta}{2}-\frac{\pi}{2}\right)^{2}.\\
\end{align}$$
