Integral $I(\tau_1,a,b) = \int_{\tau_1}^\infty d\tau_2\ \frac{1}{b^2 + \tau_2^2} \left(\pi - 2 \tan^{-1} \frac{\tau_2}{a} \right)^2$ I am looking at the integral:
$$I(\tau_1,a,b) = \int_{\tau_1}^\infty d\tau_2\ \frac{1}{b^2 + \tau_2^2} \left(\pi - 2 \tan^{-1} \frac{\tau_2}{a} \right)^2, \tag{1}$$
where $\tau_1$ is real and $a, b$ real positive. So far I was only able to solve the following special case:
$$I(\tau_1,a,a) = \int_{\tau_1}^\infty d\tau_2\ \frac{1}{a^2 + \tau_2^2} \left(\pi - 2 \tan^{-1} \frac{\tau_2}{a} \right)^2 = \frac{1}{6 a} \left(\pi - \tan^{-1} \frac{\tau_1}{a} \right)^3, \tag{2}$$
but I cannot find a way to crack $(1)$. I am mostly interested by the case $b=1$.
 A: 
Some work, too large for a comment.

Well, we have the following function:
$$\text{y}\left(\text{k},\text{m},\text{n},\text{p},x\right):=\frac{1}{\text{n}+x^2}\left(\text{m}-\text{k}\arctan\left(\frac{x}{\text{p}}\right)\right)^2=$$
$$\frac{\text{m}^2}{\text{n}+x^2}-\frac{2\text{k}\text{m}\arctan\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}+\frac{\text{k}^2\arctan^2\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\tag1$$
So, when we integrate:
$$\mathcal{I}_\epsilon\left(\text{k},\text{m},\text{n},\text{p}\right):=\int_\epsilon^\infty\text{y}\left(\text{k},\text{m},\text{n},\text{p},x\right)\space\text{d}x=$$
$$\underbrace{\int_\epsilon^\infty\frac{\text{m}^2}{\text{n}+x^2}\space\text{d}x}_{\text{I}_1}-\underbrace{\int_\epsilon^\infty\frac{2\text{k}\text{m}\arctan\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\space\text{d}x}_{\text{I}_2}+\underbrace{\int_\epsilon^\infty\frac{\text{k}^2\arctan^2\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\space\text{d}x}_{\text{I}_3}\tag2$$
Now, for $\text{I}_1$ we get:
$$\text{I}_1=\int_\epsilon^\infty\frac{\text{m}^2}{\text{n}+x^2}\space\text{d}x=\frac{\text{m}^2}{\text{n}}\int_\epsilon^\infty\frac{1}{1+\frac{x^2}{\text{n}}}\space\text{d}x\tag3$$
Let $\text{u}=\frac{x}{\sqrt{\text{n}}}$, so we get:
$$\text{I}_1=\frac{\text{m}^2}{\sqrt{\text{n}}}\lim_{x\to\infty}\int_\frac{\epsilon}{\sqrt{\text{n}}}^\frac{x}{\sqrt{\text{n}}}\frac{1}{1+\text{u}^2}\space\text{du}=\frac{\text{m}^2}{\sqrt{\text{n}}}\lim_{x\to\infty}\left[\arctan\left(\text{u}\right)\right]_\frac{\epsilon}{\sqrt{\text{n}}}^\frac{x}{\sqrt{\text{n}}}=$$
$$\frac{\text{m}^2}{\sqrt{\text{n}}}\lim_{x\to\infty}\left(\arctan\left(\frac{x}{\sqrt{\text{n}}}\right)-\arctan\left(\frac{\epsilon}{\sqrt{\text{n}}}\right)\right)\tag4$$
Knowing that $\text{n}>0$ implies that $\sqrt{\text{n}}>0$, so:
$$\text{I}_1=\frac{\text{m}^2}{\sqrt{\text{n}}}\left(\frac{\pi}{2}-\arctan\left(\frac{\epsilon}{\sqrt{\text{n}}}\right)\right)\tag5$$
Now, for $\text{I}_2$ we get:
$$\text{I}_2=\int_\epsilon^\infty\frac{2\text{k}\text{m}\arctan\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\space\text{d}x=2\text{k}\text{m}\int_\epsilon^\infty\frac{\arctan\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\space\text{d}x\tag6$$
Now, let's find:
$$\frac{\partial\text{I}_2}{\partial\text{p}}=-2\text{k}\text{m}\int_\epsilon^\infty\frac{x}{\left(\text{n}+x^2\right)\left(\text{p}^2+x^2\right)}\space\text{d}x\tag7$$
Using partial fractions it is not difficult to see that:
$$\frac{\partial\text{I}_2}{\partial\text{p}}=\frac{\text{km}\ln\left(\frac{\text{p}^2+\epsilon^2}{\text{n}+\epsilon^2}\right)}{\text{p}^2-\text{n}}\tag8$$
A: The given integral can be presented in the form of
$$I(\tau,a,b) = 4\int\limits_\tau^\infty \operatorname{arccot}^2\dfrac{\tau_2}a \,\dfrac{\mathrm d\tau_2}{\tau_2^2+b^2}.\tag1$$
Substitution
$$\varphi=\operatorname{arccot} \dfrac{\tau_2}a,\quad \tau_2 = a\cot\varphi,\quad\mathrm d\tau_2=-a(\cot^2\varphi+1)\,\mathrm d\varphi,$$ 
allows to write
$$I(\tau,a,b) = 4a\int\limits_0^{\operatorname{arccot}{\Large\frac\tau a}} \dfrac{\cot^2\varphi+1}{a^2\cot^2\varphi+b^2}\,\varphi^2\,\mathrm d\varphi
= 4a\int\limits_0^{\operatorname{arccot}{\Large\frac\tau a}} \dfrac{\varphi^2\,\mathrm d\varphi}{a^2\cos^2\varphi+b^2\sin^2\varphi}.\tag2$$
Then
$$I(\tau,a,a) = \dfrac4{3a}\varphi^3\bigg|_0^{{\Large\frac\pi2}-\arctan\Large\frac\tau a}
= \dfrac1{6a}\left(\pi-2\arctan\frac\tau a\right)^3\tag3$$
(see Wolfram Alpha test).
At the same time, the antiderivative of $(2)$ is
$$\color{brown}{\mathbf{\begin{align}
&J(\varphi,a.b)=4a\int \dfrac{\varphi^2}{a^2\cos^2\varphi+b^2\sin^2\varphi}\,\mathrm d\varphi =
\dfrac{2\varphi}b \big(\operatorname{Li_2}(r e^{2i\varphi})-\operatorname{Li_2}(^1\!/_{\large r}\, e^{2i\varphi})\big)\\[4pt]
&+\dfrac{i}{b}\big(\operatorname{Li_3}(re^{2i\varphi})-\operatorname{Li_3}(^1\!/_{\large r}\, e^{2i\varphi})\big)
+\dfrac{2i}{b}\varphi^2\ln\dfrac{1-re^{2i\varphi}}{1-\,^1\!/_{\large r}\,e^{2i\varphi}} +\operatorname{const},
\end{align}}}\tag4$$
(see also Wolfram Alpha calculations), where
$\operatorname{Li_n}$ is the Polylogarithm, 
$$r=\dfrac{b-a}{a+b}.\tag5$$
Therefore,
\begin{align}
&\color{brown}{\mathbf{I(\tau,a,b)= J\left(\operatorname{arccot}\frac \tau a,a,b\right) -J(0,a,b).}}\tag6
\end{align}
If $a=11,\ b=17,\ \tau = 5,$ then $r = \frac3{14},$
$$I(\tau,a,b)\approx 0.10429\,46124\,85634,$$
(see also Wolfram Alpha result),
$$J\left(\operatorname{arccot}\frac \tau a,a,b\right)\approx 0.32355\,66131\,49807
-0.26227\,19119\,51703\,i$$
(see also Wolfram Alpha result),
$$J(0,a,b)\approx0.21926\,20006\,64173 - 0.26227\,19119\,51703\,i$$
(see also Wolfram Alpha result),
$$J\left(\operatorname{arccot}\frac \tau a,a,b\right)-J(0,a,b)\approx 0.10429\,46124\,85634\approx I(\tau,a,b).$$
Test results confirm obtained closed form for the given integral.
