# Integral ${\large\int}_0^{\pi/2}\arctan^2\!\left(\frac{\sin x}{\sqrt3+\cos x}\right)dx$

I need to evaluate this integral: $$I=\int_0^{\pi/2}\arctan^2\!\left(\frac{\sin x}{\sqrt3+\cos x}\right)dx$$ Maple and Mathematica cannot evaluate it in this form. Its numeric value is $$I\approx0.156371391375711701230837603266631522020409597791339398428...$$ that is not recognized by WolframAlpha and Inverse Symbolic Calculator+.

Is it possible to evaluate this integral in a closed form?

I found similar questions here, here and here, but approaches shown in the answers do not seem to be directly applicable here.

• Please see this post for the similar problem: math.stackexchange.com/questions/1593334/…. Dec 31, 2015 at 19:21
• Since $$\arctan^2\left(\frac{r\sin x}{1+r\cos x}\right)=-\frac14\ln^2\left(\frac{1+re^{ix}}{1+re^{-ix}}\right)$$ one may simply find the antiderivative of $$\frac i{4z}\ln^2\left(\frac{1+rz}{1+rz^{-1}}\right)$$ either by hand or with Mathematica and plug in the integration limits at $z=1$ and $z=i$ (the integrand is analytic in the first quadrant) . Cleo's answer should follow from some simplifications through the standard polylogarithm identities. Jan 1, 2016 at 7:44
• @M.N.C.E. That should work, but the antiderivative looks scary. I hope there is a different approach to this problem. Jan 2, 2016 at 4:08
• By the way, if you want, write \mathrm dx to generate $\mathrm dx$ as opposed to $dx$. Apr 25, 2018 at 4:58

$$I=\frac\pi{20}\ln^23+\frac\pi4\operatorname{Li_2}\left(\tfrac13\right)-\frac15\operatorname{Ti}_3\left(\sqrt3\right),$$ where $$\operatorname{Ti}_3\left(\sqrt3\right)=\Im\Big[\operatorname{Li}_3\left(i\sqrt3\right)\Big]=\frac{\sqrt3}8\Phi\left(-3,3,\tfrac12\right)=\frac{5\sqrt3}4\,{_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac34\right)-\frac{5\pi^3}{432}-\frac\pi{16}\ln^23.$$

• This is not an answer. Dec 31, 2015 at 20:44
• @CarlMummert: And this is not a pipe. :-$)$ Dec 31, 2015 at 20:57
• @BenLongo I doubt very much it could be made any simpler than the forms already shown here. See this answer and comments below it. Jan 1, 2016 at 4:15
• Is there a table somewhere of known hypergeometric forms? Jan 1, 2016 at 19:42
• @BenLongo The most extensive list I know (but still far from being complete): for $_2F_1$, for $_pF_q$. You also can download corresponding Mathematica notebooks from there. Jan 1, 2016 at 23:00

With the same approach I took in this answer, we have:

$$\begin{eqnarray*}\arctan\left(\frac{\sin x}{\sqrt{3}+\cos x}\right) &=& \text{Im}\log\left(\sqrt{3}+e^{ix}\right)\\&=&\text{Im}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n\sqrt{3}^n}\,e^{inx}\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n\sqrt{3}^n}\,\sin(nx)\tag{1}\end{eqnarray*}$$ and by computing $\int_{0}^{\pi/2}\sin(nx)\sin(mx)\,dx$ it follows that: $$\begin{eqnarray*}I&=&\int_{0}^{\pi/2}\arctan^2\left(\frac{\sin x}{\sqrt{3}+\cos x}\right)\,dx\\&=&\frac{\pi}{4}\,\text{Li}_2\left(\frac{1}{3}\right)+\sum_{m\neq n}\frac{(-1)^{n+m}}{nm\sqrt{3}^{n+m}}\int_{0}^{\pi/2}\sin(nx)\sin(mx)\,dx\tag{2}\end{eqnarray*}$$ but the last integral is zero if $n$ and $m$ have the same parity.

It follows that the last series in $(2)$ can be written as:

$$\begin{eqnarray*}-\sqrt{3}\sum_{a\geq 1}\sum_{b\geq 1}\frac{1}{(2b-1)3^{a+b}}\cdot\frac{(-1)^{a+b}}{(2a)^2-(2b-1)^2}\tag{3}\end{eqnarray*}$$ and by reindexing the last double series on $a+b=s$ we get:

$$\begin{eqnarray*}-\sqrt{3}\sum_{s\geq 2}\frac{(-1)^s}{3^s}\sum_{b=1}^{s-1}\frac{1}{2b-1}\cdot\frac{1}{(2s-2b)^2-(2b-1)^2}\tag{3}\end{eqnarray*}$$ than can be dealt through partial fraction decomposition and dilogarithm identities.

I would like to actually take the time to solve a more general problem because a) we can, and b) the final result is much simpler than you might expect (or a CAS might lead you believe by offering you an antiderivative with an ungodly number of terms).

Define the function $$\mathcal{I}:\left(0,1\right)\rightarrow\mathbb{R}$$ via the definite integral

$$\mathcal{I}{\left(a\right)}:=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\arctan^{2}{\left(\frac{a\sin{\left(\varphi\right)}}{1+a\cos{\left(\varphi\right)}}\right)}.$$

We will show below that $$\mathcal{I}{\left(a\right)}$$ has the following closed-form expression in terms of polylogarithms for all $$0:

\begin{align} \mathcal{I}{\left(a\right)} &=\Im{\left[\operatorname{Li}_{3}{\left(-iae^{2i\arctan{\left(a\right)}}\right)}\right]}+\Im{\left[\operatorname{Li}_{3}{\left(ia\right)}\right]}\\ &~~~~~+\left[\frac{\pi}{2}-2\arctan{\left(a\right)}\right]\operatorname{Li}_{2}{\left(a,\frac{\pi}{2}-2\arctan{\left(a\right)}\right)}\\ &~~~~~+\frac{\pi}{8}\operatorname{Li}_{2}{\left(-a^{2}\right)}+\frac43\arctan^{3}{\left(a\right)}-\frac{\pi}{2}\arctan^{2}{\left(a\right)}.\\ \end{align}

Suppose $$a\in\left(0,1\right)$$.

\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\arctan^{2}{\left(\frac{a\sin{\left(\varphi\right)}}{1+a\cos{\left(\varphi\right)}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\arctan^{2}{\left(\frac{a\sin{\left(2\arctan{\left(x\right)}\right)}}{1+a\cos{\left(2\arctan{\left(x\right)}\right)}}\right)};~~~\small{\left[\varphi=2\arctan{\left(x\right)}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\arctan^{2}{\left(\frac{a\left(\frac{2x}{1+x^{2}}\right)}{1+a\left(\frac{1-x^{2}}{1+x^{2}}\right)}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\arctan^{2}{\left(\frac{2ax}{\left(1+x^{2}\right)+a\left(1-x^{2}\right)}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\arctan^{2}{\left(\frac{2ax}{1+a+\left(1-a\right)x^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\arctan^{2}{\left(\frac{\left(\frac{2a}{1+a}\right)x}{1+\left(\frac{1-a}{1+a}\right)x^{2}}\right)}.\\ \end{align}

Set $$\frac{1-a}{1+a}=:p$$. Then, $$0, and hence

\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\arctan^{2}{\left(\frac{\left(\frac{2a}{1+a}\right)x}{1+\left(\frac{1-a}{1+a}\right)x^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\arctan^{2}{\left(\frac{\left(1-p\right)x}{1+px^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\left[\arctan{\left(\frac{x-px}{1+px^{2}}\right)}\right]^{2}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\left[\arctan{\left(x\right)}-\arctan{\left(px\right)}\right]^{2},\\ \end{align}

where in the last line above we've used the following formula for the difference of two arctangents:

$$\arctan{\left(z\right)}-\arctan{\left(w\right)}=\arctan{\left(\frac{z-w}{1+zw}\right)};~~~\small{\left(z,w\right)\in\mathbb{R}^{2}\land zw>-1}.$$

Then,

\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\left[\arctan{\left(x\right)}-\arctan{\left(px\right)}\right]^{2}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{1+x^{2}}\left[\arctan^{2}{\left(x\right)}-2\arctan{\left(x\right)}\arctan{\left(px\right)}+\arctan^{2}{\left(px\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2\arctan^{2}{\left(x\right)}}{1+x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{4\arctan{\left(x\right)}\arctan{\left(px\right)}}{1+x^{2}}+\int_{0}^{1}\mathrm{d}x\,\frac{2\arctan^{2}{\left(px\right)}}{1+x^{2}}\\ &=\frac23\arctan^{3}{\left(1\right)}\\ &~~~~~-2\arctan^{2}{\left(1\right)}\arctan{\left(p\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{2p\arctan^{2}{\left(x\right)}}{1+p^{2}x^{2}};~~~\small{I.B.P.}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2\arctan^{2}{\left(px\right)}}{1+x^{2}}\\ &=\frac{\pi^{3}}{96}-\frac{\pi^{2}}{8}\arctan{\left(p\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{2p\arctan^{2}{\left(x\right)}}{1+p^{2}x^{2}}+\int_{0}^{1}\mathrm{d}x\,\frac{2\arctan^{2}{\left(px\right)}}{1+x^{2}}.\\ \end{align}

Lengthy aside on the evaluation of $$\int_{0}^{z}\mathrm{d}x\,\frac{2b\arctan^{2}{\left(ax\right)}}{1+b^{2}x^{2}}$$:

Define the function $$\mathcal{J}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$$

$$\mathcal{J}{\left(a,b,z\right)}:=\int_{0}^{z}\mathrm{d}x\,\frac{2b\arctan^{2}{\left(ax\right)}}{1+b^{2}x^{2}}.$$

It's a simple matter to show by rescaling the integral that

$$\forall\left(a,b,z\right)\in\mathbb{R}_{>0}^{3}:\mathcal{J}{\left(a,b,z\right)}=\mathcal{J}{\left(az,bz,1\right)},$$

so we may assume WLOG that $$z=1$$ in the general evaluation of $$\mathcal{J}{(a,b,z)}$$.

Suppose $$\left(a,b\right)\in\mathbb{R}_{>0}^{2}$$, and set $$\frac{b}{a}=:c\in\mathbb{R}_{>0}\land\arctan{\left(a\right)}=:\alpha\in\left(0,\frac{\pi}{2}\right)$$. Then,

\begin{align} \mathcal{J}{\left(a,b,1\right)} &=\int_{0}^{1}\mathrm{d}x\,\frac{2b\arctan^{2}{\left(ax\right)}}{1+b^{2}x^{2}}\\ &=\int_{0}^{a}\mathrm{d}y\,\frac{2ab\arctan^{2}{\left(y\right)}}{a^{2}+b^{2}y^{2}};~~\small{\left[x=\frac{y}{a}\right]}\\ &=\int_{0}^{a}\mathrm{d}y\,\frac{2c\arctan^{2}{\left(y\right)}}{1+c^{2}y^{2}}\\ &=\int_{0}^{\arctan{\left(a\right)}}\mathrm{d}\varphi\,\frac{2c\varphi^{2}\sec^{2}{\left(\varphi\right)}}{1+c^{2}\tan^{2}{\left(\varphi\right)}};~~\small{\left[\arctan{\left(y\right)}=\varphi\right]}\\ &=\int_{0}^{\alpha}\mathrm{d}\varphi\,\frac{2c\varphi^{2}\sec^{2}{\left(\varphi\right)}}{1+c^{2}\tan^{2}{\left(\varphi\right)}}\\ &=\left[2\varphi^{2}\arctan{\left(c\tan{\left(\varphi\right)}\right)}\right]_{\varphi=0}^{\varphi=\alpha}-\int_{0}^{\alpha}\mathrm{d}\varphi\,4\varphi\arctan{\left(c\tan{\left(\varphi\right)}\right)};~~~\small{I.B.P.}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\varphi\,\varphi\arctan{\left(c\tan{\left(\varphi\right)}\right)}.\\ \end{align}

Next, by rewriting the integral as a multiple integral and changing the order of integration in the appropriate way, we obtain the following:

\begin{align} \mathcal{J}{\left(a,b,1\right)} &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\varphi\,\varphi\arctan{\left(c\tan{\left(\varphi\right)}\right)}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\varphi\int_{0}^{\varphi}\mathrm{d}\vartheta\,\arctan{\left(c\tan{\left(\varphi\right)}\right)}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\arctan{\left(c\tan{\left(\varphi\right)}\right)}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\int_{0}^{c}\mathrm{d}y\,\frac{d}{dy}\arctan{\left(y\tan{\left(\varphi\right)}\right)}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\int_{0}^{c}\mathrm{d}y\,\frac{\tan{\left(\varphi\right)}}{1+y^{2}\tan^{2}{\left(\varphi\right)}}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{0}^{c}\mathrm{d}y\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\frac{\tan{\left(\varphi\right)}}{1+y^{2}\tan^{2}{\left(\varphi\right)}}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{0}^{c}\mathrm{d}y\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\frac{\sin{\left(\varphi\right)}\cos{\left(\varphi\right)}}{\cos^{2}{\left(\varphi\right)}+y^{2}\sin^{2}{\left(\varphi\right)}}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{0}^{c}\mathrm{d}y\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\frac{\sin{\left(\varphi\right)}\cos{\left(\varphi\right)}}{1+\left(y^{2}-1\right)\sin^{2}{\left(\varphi\right)}}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-4\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{0}^{c}\mathrm{d}y\int_{\sin{\left(\vartheta\right)}}^{\sin{\left(\alpha\right)}}\mathrm{d}t\,\frac{t}{1+\left(y^{2}-1\right)t^{2}};~~~\small{\left[\sin{\left(\varphi\right)}=t\right]}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-2\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{0}^{c}\mathrm{d}y\int_{\sin^{2}{\left(\vartheta\right)}}^{\sin^{2}{\left(\alpha\right)}}\mathrm{d}u\,\frac{1}{1+\left(y^{2}-1\right)u};~~~\small{\left[t^{2}=u\right]}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}\\ &~~~~~-2\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{0}^{c}\mathrm{d}y\,\frac{\ln{\left(1+\left(y^{2}-1\right)\sin^{2}{\left(\alpha\right)}\right)}-\ln{\left(1+\left(y^{2}-1\right)\sin^{2}{\left(\vartheta\right)}\right)}}{\left(y^{2}-1\right)}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}+\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{0}^{c}\mathrm{d}y\,\frac{2\ln{\left(\frac{1-\left(1-y^{2}\right)\sin^{2}{\left(\alpha\right)}}{1-\left(1-y^{2}\right)\sin^{2}{\left(\vartheta\right)}}\right)}}{\left(1-y^{2}\right)}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}\\ &~~~~~+\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{\frac{1-c}{1+c}}^{1}\mathrm{d}x\,\frac{1}{x}\ln{\left(\frac{x^{2}+2x+1-4x\sin^{2}{\left(\alpha\right)}}{x^{2}+2x+1-4x\sin^{2}{\left(\vartheta\right)}}\right)};~~~\small{\left[y=\frac{1-x}{1+x}\right]}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}\\ &~~~~~+\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{r}^{1}\mathrm{d}x\,\frac{1}{x}\ln{\left(\frac{1+2x\cos{\left(2\alpha\right)}+x^{2}}{1+2x\cos{\left(2\vartheta\right)}+x^{2}}\right)};~~~\small{\left[r:=\frac{1-c}{1+c}\in\left(-1,1\right)\right]}.\\ \end{align}

At this point it will be helpful to introduce the following two-variable extension of the dilogarithm:

$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\int_{0}^{r}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{2x};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$

This function gives the real part of the dilogarithm of complex argument inside the unit circle:

$$\Re{\left(\operatorname{Li}_{2}{\left(re^{i\theta}\right)}\right)}=\operatorname{Li}_{2}{\left(r,\theta\right)};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}\land|r|<1}.$$

The function $$\operatorname{Li}_{2}{\left(r,\theta\right)}$$ can be shown to have the following special cases:

$$\operatorname{Li}_{2}{\left(1,\theta\right)}=\frac14\left(\pi-\theta\right)^{2}-\frac{\pi^{2}}{12};~~~\small{0\le\theta\le2\pi},$$

$$\operatorname{Li}_{2}{\left(r,\frac{\pi}{2}\right)}=\frac14\operatorname{Li}_{2}{\left(-r^{2}\right)};~~~\small{r\in\mathbb{R}}.$$

Continuing with our evaluation of $$\mathcal{J}$$, we obtain

\begin{align} \mathcal{J}{\left(a,b,1\right)} &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}+\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{r}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+2x\cos{\left(2\alpha\right)}+x^{2}}{1+2x\cos{\left(2\vartheta\right)}+x^{2}}\right)}}{x}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}+\int_{0}^{\alpha}\mathrm{d}\vartheta\,\bigg{[}2\operatorname{Li}_{2}{\left(r,\pi-2\alpha\right)}-2\operatorname{Li}_{2}{\left(1,\pi-2\alpha\right)}\\ &~~~~~-2\operatorname{Li}_{2}{\left(r,\pi-2\vartheta\right)}+2\operatorname{Li}_{2}{\left(1,\pi-2\vartheta\right)}\bigg{]}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}+\int_{0}^{\alpha}\mathrm{d}\vartheta\,\bigg{[}2\vartheta^{2}-2\alpha^{2}+2\operatorname{Li}_{2}{\left(r,\pi-2\alpha\right)}\\ &~~~~~-2\operatorname{Li}_{2}{\left(r,\pi-2\vartheta\right)}\bigg{]}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-\frac43\alpha^{3}+2\alpha\operatorname{Li}_{2}{\left(r,\pi-2\alpha\right)}\\ &~~~~~-2\int_{0}^{\alpha}\mathrm{d}\vartheta\,\operatorname{Li}_{2}{\left(r,\pi-2\vartheta\right)}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-\frac43\alpha^{3}+2\alpha\operatorname{Li}_{2}{\left(r,\pi-2\alpha\right)}\\ &~~~~~-2\int_{0}^{\alpha}\mathrm{d}\vartheta\,\Re{\left[\operatorname{Li}_{2}{\left(re^{i\left(\pi-2\vartheta\right)}\right)}\right]}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-\frac43\alpha^{3}+2\alpha\operatorname{Li}_{2}{\left(r,\pi-2\alpha\right)}\\ &~~~~~-2\Re\int_{0}^{\alpha}\mathrm{d}\vartheta\,\operatorname{Li}_{2}{\left(re^{i\left(\pi-2\vartheta\right)}\right)}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-\frac43\alpha^{3}+2\alpha\operatorname{Li}_{2}{\left(r,\pi-2\alpha\right)}\\ &~~~~~-\Re\int_{\pi-2\alpha}^{\pi}\mathrm{d}\vartheta\,\operatorname{Li}_{2}{\left(re^{i\vartheta}\right)}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-\frac43\alpha^{3}+2\alpha\operatorname{Li}_{2}{\left(r,\pi-2\alpha\right)}\\ &~~~~~-\Re{\left[\frac{1}{i}\operatorname{Li}_{3}{\left(re^{i\pi}\right)}-\frac{1}{i}\operatorname{Li}_{3}{\left(re^{i\left(\pi-2\alpha\right)}\right)}\right]}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-\frac43\alpha^{3}+2\alpha\operatorname{Li}_{2}{\left(r,\pi-2\alpha\right)}\\ &~~~~~-\Im{\left[\operatorname{Li}_{3}{\left(-r\right)}-\operatorname{Li}_{3}{\left(re^{i\left(\pi-2\alpha\right)}\right)}\right]}\\ &=2\alpha^{2}\arctan{\left(c\tan{\left(\alpha\right)}\right)}-\frac43\alpha^{3}+2\alpha\operatorname{Li}_{2}{\left(r,\pi-2\alpha\right)}+\Im{\left[\operatorname{Li}_{3}{\left(-re^{-2i\alpha}\right)}\right]}.\\ \end{align}

The following pair of integration formulas then follow from special cases of the previous result:

$$\int_{0}^{1}\mathrm{d}x\,\frac{2p\arctan^{2}{\left(x\right)}}{1+p^{2}x^{2}}=\Im{\left[\operatorname{Li}_{3}{\left(i\frac{1-p}{1+p}\right)}\right]}+\frac{\pi}{8}\operatorname{Li}_{2}{\left(-\left(\frac{1-p}{1+p}\right)^{2}\right)}+\frac{\pi^{2}}{8}\arctan{\left(p\right)}-\frac{\pi^{3}}{48},$$

\begin{align} \int_{0}^{1}\mathrm{d}x\,\frac{2\arctan^{2}{\left(px\right)}}{1+x^{2}} &=\Im{\left[\operatorname{Li}_{3}{\left(\frac{1-p}{1+p}e^{-2i\arctan{\left(p\right)}}\right)}\right]}+2\arctan{\left(p\right)}\operatorname{Li}_{2}{\left(\frac{1-p}{1+p},2\arctan{\left(p\right)}\right)}\\ &~~~~~-\frac43\arctan^{3}{\left(p\right)}+\frac{\pi}{2}\arctan^{2}{\left(p\right)},\\ \end{align}

where $$0.

We finally have what we need to complete our evaluation of $$\mathcal{I}$$. We find

\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{1}\mathrm{d}x\,\frac{2\arctan^{2}{\left(px\right)}}{1+x^{2}}+\int_{0}^{1}\mathrm{d}x\,\frac{2p\arctan^{2}{\left(x\right)}}{1+p^{2}x^{2}}-\frac{\pi^{2}}{8}\arctan{\left(p\right)}+\frac{\pi^{3}}{96}\\ &=\Im{\left[\operatorname{Li}_{3}{\left(\frac{1-p}{1+p}e^{-2i\arctan{\left(p\right)}}\right)}\right]}+2\arctan{\left(p\right)}\operatorname{Li}_{2}{\left(\frac{1-p}{1+p},2\arctan{\left(p\right)}\right)}\\ &~~~~~-\frac43\arctan^{3}{\left(p\right)}+\frac{\pi}{2}\arctan^{2}{\left(p\right)}\\ &~~~~~+\Im{\left[\operatorname{Li}_{3}{\left(i\frac{1-p}{1+p}\right)}\right]}+\frac{\pi}{8}\operatorname{Li}_{2}{\left(-\left(\frac{1-p}{1+p}\right)^{2}\right)}+\frac{\pi^{2}}{8}\arctan{\left(p\right)}-\frac{\pi^{3}}{48}\\ &~~~~~-\frac{\pi^{2}}{8}\arctan{\left(p\right)}+\frac{\pi^{3}}{96}\\ &=\Im{\left[\operatorname{Li}_{3}{\left(\frac{1-p}{1+p}e^{-2i\arctan{\left(p\right)}}\right)}\right]}+\Im{\left[\operatorname{Li}_{3}{\left(i\frac{1-p}{1+p}\right)}\right]}\\ &~~~~~+2\arctan{\left(p\right)}\operatorname{Li}_{2}{\left(\frac{1-p}{1+p},2\arctan{\left(p\right)}\right)}+\frac{\pi}{8}\operatorname{Li}_{2}{\left(-\left(\frac{1-p}{1+p}\right)^{2}\right)}\\ &~~~~~-\frac43\arctan^{3}{\left(p\right)}+\frac{\pi}{2}\arctan^{2}{\left(p\right)}-\frac{\pi^{3}}{96}\\ &=\Im{\left[\operatorname{Li}_{3}{\left(ae^{-2i\arctan{\left(\frac{1-a}{1+a}\right)}}\right)}\right]}+\Im{\left[\operatorname{Li}_{3}{\left(ia\right)}\right]}\\ &~~~~~+2\arctan{\left(\frac{1-a}{1+a}\right)}\operatorname{Li}_{2}{\left(a,2\arctan{\left(\frac{1-a}{1+a}\right)}\right)}+\frac{\pi}{8}\operatorname{Li}_{2}{\left(-a^{2}\right)}\\ &~~~~~-\frac43\arctan^{3}{\left(\frac{1-a}{1+a}\right)}+\frac{\pi}{2}\arctan^{2}{\left(\frac{1-a}{1+a}\right)}-\frac{\pi^{3}}{96}\\ &=\Im{\left[\operatorname{Li}_{3}{\left(-iae^{2i\arctan{\left(a\right)}}\right)}\right]}+\Im{\left[\operatorname{Li}_{3}{\left(ia\right)}\right]}\\ &~~~~~+\left[\frac{\pi}{2}-2\arctan{\left(a\right)}\right]\operatorname{Li}_{2}{\left(a,\frac{\pi}{2}-2\arctan{\left(a\right)}\right)}\\ &~~~~~+\frac{\pi}{8}\operatorname{Li}_{2}{\left(-a^{2}\right)}+\frac43\arctan^{3}{\left(a\right)}-\frac{\pi}{2}\arctan^{2}{\left(a\right)}.\blacksquare\\ \end{align}

• More than impressive ! Aug 3, 2021 at 9:01
• @ClaudeLeibovici Thanks. Always nice to be noticed. =) Aug 3, 2021 at 12:13