<s>[INST] 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 
$$\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}+\frac{2 i}{5}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}+\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{8}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{8}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}+\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}-\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}+\frac{8 i}{15}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}-\frac{8 i}{15}\right)+\frac{1}{4} \log (4) \tan ^{-1}(6)-\frac{1}{4} \log (4) \tan ^{-1}(9)+\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{6}\right)-\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{9}\right)-\frac{1}{4} \log (9) \tan ^{-1}(2)+\frac{1}{4} \log (9) \tan ^{-1}(3)-\frac{1}{4} \log (9) \tan ^{-1}(6)+\frac{1}{4} \log (9) \tan ^{-1}(9)-\frac{1}{4} \log (9) \tan ^{-1}\left(\frac{3}{55}\right)+\frac{1}{4} \log (16) \tan ^{-1}(2)-\frac{1}{4} \log (16) \tan ^{-1}(3).$$
 [/INST] 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.
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