Another beautiful arctan integral $\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx$ Do you think we can express the closed form of the integral below in a very nice and short way?
As you already know, your opinions weighs much to me, so I need them!
Calculate in closed-form 
$$\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx.$$
I'm looking forward to your precious feedback!
Mathematica tells us the closed form is 
$$\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}+\frac{2 i}{5}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}+\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{8}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{8}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}+\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}-\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}+\frac{8 i}{15}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}-\frac{8 i}{15}\right)+\frac{1}{4} \log (4) \tan ^{-1}(6)-\frac{1}{4} \log (4) \tan ^{-1}(9)+\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{6}\right)-\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{9}\right)-\frac{1}{4} \log (9) \tan ^{-1}(2)+\frac{1}{4} \log (9) \tan ^{-1}(3)-\frac{1}{4} \log (9) \tan ^{-1}(6)+\frac{1}{4} \log (9) \tan ^{-1}(9)-\frac{1}{4} \log (9) \tan ^{-1}\left(\frac{3}{55}\right)+\frac{1}{4} \log (16) \tan ^{-1}(2)-\frac{1}{4} \log (16) \tan ^{-1}(3).$$
 A: Exploiting integration by parts, the problem boils down to computing:
$$ \int_{1/2}^{1}\left(\log(1+x)-\log(1-x)\right)\left(\frac{3}{5x^2+8x+5}-\frac{4}{5x^2+6x+5}\right)\,dx $$
and by partial fraction decomposition that is equivalent to computing:
$$ I_{\pm}(\zeta)=\int_{1/2}^{1}\frac{\log(1\pm x)}{x-\zeta}\,dx $$
with $\zeta\in\left\{-\frac{3}{5}-\frac{4i}{5},-\frac{3}{5}+\frac{4i}{5},-\frac{4}{5}-\frac{3i}{5},-\frac{4}{5}+\frac{3i}{5}\right\} $. We may check that:
$$ \int \frac{\log(1+x)}{x-a} = \log(1+x)\log\left(1-\frac{1+x}{1+a}\right)+\text{Li}_2\left(\frac{1+x}{1-a}\right) $$
holds by differentiation to compute the closed form of the original integral in terms of dilogarithms and products of logarithms only.
A: Substituting $x\to\frac{1-x}{1+x}$ is usually a first reaction of mine. Here it works surprisingly well: the integral is equivalent to 
$$I=\int_{1/2}^1 \frac{\tan^{-1}\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx=\frac12\int_0^{\frac13}\frac{\tan^{-1}\left(\frac{x}{x^2+6}\right)}{x} dx.$$
Integrating by parts, 
$$I=-\frac12\ln3 \tan^{-1}\left(\frac{3}{55}\right)-\frac12\int_0^{\frac13} \frac{6-x^2}{x^4+13x^2+36}\,\ln x \, dx$$
This looks managable. We can decompose $\displaystyle \, \frac{6-x^2}{x^4+13x^2+36}=\frac{2}{x^2+4}-\frac{3}{x^2+9} $
so that 
$$\int_0^{\frac13} \frac{6-x^2}{x^4+13x^2+36}\,\ln x \, dx=2\int_0^{\frac13} \frac{\ln x}{x^2+2^2}\,dx-3\int_0^{\frac13}\frac{\ln x}{x^2+3^2}\,dx$$
Now these integrals may also be done by partial fraction (whence emerge the complex stuff) and the antiderivative of the remaining parts is easily found in terms of logarithms and dilogarithms..
So the logarithm and $tan^{-1}$ cancel at the end, leaving:

$$ I=\frac12\Im\operatorname{Li_2}\left(\frac1{9i}\right)-\frac12\Im\operatorname{Li_2}\left(\frac1{6i}\right)$$

Edit 
The final cancellation and Chris' comment below suggest a detour to the integration by parts and partial fractions decomposition: 
We just notice that $\displaystyle \tan^{-1}\left(\frac{x}{x^2+6}\right)=\tan^{-1}\left(\frac{x}{2}\right)-\tan^{-1}\left(\frac{x}{3}\right)$, whence 
$$I=\frac12\int_0^{\frac13}\frac{\tan^{-1}\left(\frac{x}{x^2+6}\right)}{x} dx=\frac12\int_0^{\frac13}\frac{\tan^{-1}\left(\frac{x}{2}\right)-\tan^{-1}\left(\frac{x}{3}\right)}{x}dx\\=\frac12\int_0^{\frac16}\frac{\tan^{-1}x}{x}dx-\frac12\int_0^{\frac19}\frac{\tan^{-1}x}{x}dx=\frac{1}{2}\operatorname{Ti}_2\left(\frac{1}{6}\right)-\frac12\operatorname{Ti}‌​_2\left(\frac{1}{9}\right).$$
A: It can be easily checked by differentiation that:
$${\large\int}\,\frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx=\\
\frac i4\left[\vphantom{\Large|}\operatorname{Li}_2\left(\left(-\tfrac12+\tfrac i6\right)(x-1)\right)-\operatorname{Li}_2\left(\left(-\tfrac12-\tfrac i6\right)(x-1)\right)\\
\,+\operatorname{Li}_2\left(\left(-\tfrac12-\tfrac i4\right)(x-1)\right)-\operatorname{Li}_2\left(\left(-\tfrac12+\tfrac i4\right)(x-1)\right)\\
\!\!\!\!\!\!+\operatorname{Li}_2\left(\left(\tfrac12+i\right)(x+1)\right)\,\,\,\,\,\,-\operatorname{Li}_2\left(\left(\tfrac12-i\right)(x+1)\right)\\
\,\,\,\,\,\,\,\,\,\,+\operatorname{Li}_2\left(\left(\tfrac12-\tfrac{3i}2\right)(x+1)\right)\,\,\,-\operatorname{Li}_2\left(\left(\tfrac12+\tfrac{3i}2\right) (x+1)\right)\right]\color{gray}{+C}$$
This enables us to evaluate a definite integral over any interval.
If you are only interested in real values of $x$ then the antiderivative can be simplified to:
$$\frac12\Im\left[\vphantom{\Large|}\operatorname{Li}_2\left(\left(\tfrac12+\tfrac{3i}2\right)(x+1)\right)-\operatorname{Li}_2\left(\left(\tfrac12+i\right)(x+1)\right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\operatorname{Li}_2\left(\left(\tfrac i4-\tfrac12\right)(x-1)\right)-\operatorname{Li}_2\left(\left(\tfrac i6-\tfrac12\right)(x-1)\right)\right]\color{gray}{+C}$$
