How to prove $$\int_{\sqrt{3/5}}^1 \frac{\arctan (x)}{\sqrt{2 x^2-1} \left(3 x^2-1\right)} \, dx=\frac{3\pi^2}{160}$$ I found the integral neat enough but also tough. Is it somehow related to the Ahmed integral$?$ Any help will be appreciated.

Update: Please see the link under @pisco's answer for further reference.

  • 1
    $\begingroup$ Try the change of variable $y=\sqrt{2x^2-1}$ $\endgroup$
    – FDP
    Aug 22, 2019 at 11:19
  • 2
    $\begingroup$ This answer seems to be relevant: math.stackexchange.com/a/580736/44121 $\endgroup$ Aug 22, 2019 at 11:27
  • 3
    $\begingroup$ Looks like it is a Coxeter integral. As a start let: $\sqrt{2x^2-1}=xt$ to get: $\int_\frac{1}{\sqrt 3}^1\frac{\arctan\left(\frac{1}{\sqrt{2-t^2}}\right)}{1+t^2}dt$. Now try Feynman's trick maybe. $\endgroup$
    – Zacky
    Aug 22, 2019 at 20:33
  • 1
    $\begingroup$ Just a comment for now. It seems that $$ I =\frac{5\pi^2}{96} - \int_0^{\sqrt 3} \frac{\arctan (\sqrt 5 x)}{1+x^2} dx +2 \int_0^{\frac 1 {\sqrt 3}} \frac{\arctan (\sqrt 5 x)}{1+x^2} dx$$ I'm not sure this will help, but does anyone have an idea about how to to evaluate integrals of the form $\int_0^a \frac{\arctan(bx)}{1+x^2} dx$? My thought is that it is related to the dilogarithm. $\endgroup$ Aug 24, 2019 at 2:26
  • 1
    $\begingroup$ One can prove that:\begin{align}\int_{\sqrt{\frac{3}{5}}}^{\sqrt{3}} \frac{\arctan x}{\sqrt{2 x^2-1} \left(3 x^2-1\right)} \, dx=\frac{\pi^2}{10}-\frac{1}{3}\pi\arctan\left(\sqrt{\frac{3}{5}}\right)\end{align} $\endgroup$
    – FDP
    Aug 24, 2019 at 17:04

3 Answers 3


Two proofs will be given. One proof is the remaining of this answer, along lines of Schläfli and Coxeter. Second proof, more direct, is given at this answer's remark.

Note that $$\int_{\sqrt {3/5} }^1 {\frac{{\arctan x}}{{\sqrt {2{x^2} - 1} (3{x^2} - 1)}}dx} = \int_1^{\sqrt {5/3} } {\frac{{x(\frac{\pi }{2} - \arctan x)}}{{\sqrt {2 - {x^2}} (3 - {x^2})}}dx} = \frac{{{\pi ^2}}}{{24}} - \frac{1}{2}\int_1^{5/3} {\frac{{\arctan \sqrt x }}{{\sqrt {2 - x} (3 - x)}}dx}$$ therefore your integral is equivalent to $$\int_1^{5/3} {\frac{{\arctan \sqrt x }}{{\sqrt {2 - x} (3 - x)}}dx} = \frac{{11{\pi ^2}}}{{240}}$$ I will prove this by establishing $$\tag{1}\int_0^1 {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {2 - t} (3 - t)}}dt} = \frac{{{\pi ^2}}}{{48}}$$ $$\tag{2} \int_0^{5/3} {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {2 - t} (3 - t)}}dt} = \frac{{{\pi ^2}}}{{15}}$$

The following ideas are an expounded version of Polylogarihm and Associated Functions by Leonard Lewin page 115-117, arguments there are very terse.

Let $$S(\alpha ,\beta ,\gamma ) = \sum\limits_{n = 1}^\infty {\frac{{{k^n}}}{{{n^2}}}(\cos 2n\alpha - \cos 2n\beta + \cos 2n\gamma - 1)} - {\alpha ^2} + {\beta ^2} - {\gamma ^2}$$ where $0\leq \alpha,\gamma \leq \pi/2, 0\leq \beta < \pi$ and $k$ is a function of $\alpha,\beta,\gamma$ which will be determined soon. It is easy to check that $$\frac{{\partial S}}{{\partial \alpha }} = - 2\sum\limits_{n = 1}^\infty {\frac{{{k^n}}}{n}\sin 2n\alpha } - 2\alpha = - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \alpha } \right)$$ Therefore the differential $dS$ is $$\begin{aligned}dS &= \frac{1}{k}\sum\limits_{n = 1}^\infty {\frac{{{k^n}}}{n}(\cos 2n\alpha - \cos 2n\beta + \cos 2n\gamma - 1)} dk - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \alpha } \right)d\alpha+\\ & 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \beta } \right)d\beta - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \gamma } \right)d\gamma \end{aligned}$$ Now choose $k$ such that the coefficient of $dk$ vanishes, using $\sum\limits_{n = 1}^\infty {\frac{{{a^n}\cos nx}}{n}} = - \frac{1}{2}\ln ({a^2} - 2a\cos x + 1)$ one easily see such $k$ is $$ k = \frac{{\sqrt {{{\cos }^2}\alpha {{\cos }^2}\gamma - {{\cos }^2}\beta } - \sin \alpha \sin \gamma }}{{\sqrt {{{\cos }^2}\alpha {{\cos }^2}\gamma - {{\cos }^2}\beta } + \sin \alpha \sin \gamma }}$$ This completes the definition of $S(\alpha,\beta,\gamma)$. Note that in order for $k$ to be real, we need to assume the term inside radical is always $\geq 0$, we confine ourselves exclusive to this case. Now $dS$ becomes $$\tag{3}dS = - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \alpha } \right)d\alpha+ 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \beta } \right)d\beta - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \gamma } \right)d\gamma $$

Four observations:

  • $S(0,\beta,\gamma) = \pi(\beta-\gamma)$
  • When $\sin^2 \alpha + \sin^2 \gamma = \sin^2 \beta$, $S(\alpha,\beta,\gamma) = -\alpha^2+\beta^2-\gamma^2$
  • When $\cos \alpha \cos\gamma = \cos\beta$, $S(\alpha,\beta,\gamma)= 0$.
  • $S(\alpha,\pi - 2\alpha,\alpha) = 6S(\alpha,\pi/3,\pi/6)$

Proof. For the first one, $\alpha = 0$ implies $k=1$, so $$S(0,\beta ,\gamma ) = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}( - \cos 2n\beta + \cos 2n\gamma )} + {\beta ^2} - {\gamma ^2} = \pi (\beta - \gamma )$$ For second one, the condition implies $k=0$. For third one, the condition implies $k=-1$, then just evaluate the series. The fourth assertion is more crucial. We prove it carefully. By first bullet, it suffices to prove their derivative with respect to $\alpha$ are equal. When $\beta = \pi/3, \gamma = \pi/6$, one sees that $$\frac{{1 + k}}{{1 - k}}\tan \alpha = \sqrt{3 \cos^2 \alpha-1} \sec \alpha $$ Since $\beta,\gamma$ are constants, $d\gamma = d\beta = 0$, $(3)$ shows $$\tag{A}-\frac{1}{2}\frac{dS(\alpha,\pi/3,\pi/6)}{d\alpha} = \arctan(\sqrt{3 \cos^2 \alpha-1} \sec \alpha)$$

Now consider $S(\alpha,\pi - 2\alpha,\alpha)$, with $\beta = \pi - 2\alpha, \gamma = \alpha, d\beta = -2d\alpha$, $d\gamma = d\alpha$, one computes via $(3)$, $$\tag{B}-\frac{1}{2}\frac{dS(\alpha,\pi - 2\alpha,\alpha)}{d\alpha} = 2 \arctan \left(\sqrt{\cos ^4\alpha-\cos ^2 2\alpha} \csc \alpha \sec \alpha\right)-2 \arctan \left(\sqrt{\cos ^4 \alpha-\cos ^2 2 \alpha} \tan 2 \alpha \csc ^2 \alpha \right)+2\pi $$ To complete the proof, it suffices to differentiate RHS of $6\times (A), (B)$, and see whether they are equal. This becomes a trivial but computational heavy task.

Now let $a>b$, consider $$\sqrt {a - b} \int_0^x {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {b - t} (a - t)}}dt} = - 2\int_0^x {{{\tan }^{ - 1}}\sqrt t d({{\tan }^{ - 1}}\sqrt {\frac{{b - t}}{{a - b}}} )} $$ fix $\alpha = {\tan ^{ - 1}}\sqrt {\frac{{b - t}}{{a - b}}}$, we find $\beta,\gamma$ (depends on $a,b$ but not on $t$) such that $$\frac{{1 + k}}{{1 - k}}\tan \alpha = \sqrt t $$ One easily verifies one such pair $\beta,\gamma$ is $$\gamma = {\tan ^{ - 1}}\frac{1}{{\sqrt a }}\qquad \beta = {\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} $$ Since $\beta,\gamma$ are independent of $t$, $(3)$ implies $$dS(\alpha ,{\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} ,{\tan ^{ - 1}}\frac{1}{{\sqrt a }}) = - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \alpha } \right)d\alpha = - 2{\tan ^{ - 1}}\sqrt t d\alpha $$ Hence $$\int_0^x { - 2{{\tan }^{ - 1}}\sqrt t d\alpha } = S({\tan ^{ - 1}}\sqrt {\frac{{b - x}}{{a - b}}} ,{\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} ,{\tan ^{ - 1}}\frac{1}{{\sqrt a }}) + C$$ for a constant $C$ independent of $x$. Set $x=0$, then one checks the third bullet point applies for $$S({\tan ^{ - 1}}\sqrt {\frac{b}{{a - b}}} ,{\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} ,{\tan ^{ - 1}}\frac{1}{{\sqrt a }})$$ therefore it is $0$, hence $C=0$. Thus we proved

$$\tag{4}\int_0^x {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{(a - t)\sqrt {b - t} }}dt} = \frac{1}{{\sqrt {a - b} }}S({\tan ^{ - 1}}\sqrt {\frac{{b - x}}{{a - b}}} ,{\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} ,{\tan ^{ - 1}}\frac{1}{{\sqrt a }})$$

Now let $a=3, b=2$, we deduce from $(4)$ $$\int_0^1 {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {2 - t} (3 - t)}}dt} = S(\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{6})\qquad \int_0^{5/3} {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {2 - t} (3 - t)}}dt} = S(\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{6})$$ Now the second bullet applies for the former one, so $S(\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{6}) = \frac{{{\pi ^2}}}{{48}}$, this is $(1)$. For latter one, note that fourth bullet implies $$6S(\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{6}) = S(\frac{\pi }{6},\frac{{2\pi }}{3},\frac{\pi }{6})$$ but directly from definition (the $k$ associated with these two pairs are equal), one sees that $$S(\frac{\pi }{6},\frac{{2\pi }}{3},\frac{\pi }{6}) - S(\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{6}) = \sum\limits_{n = 1}^\infty {\frac{{{k^n}}}{{{n^2}}}(\underbrace{\cos \frac{{2\pi n}}{3} - \cos \frac{{4\pi n}}{3}}_{=0})} + {(\frac{{2\pi }}{3})^2} - {(\frac{\pi }{3})^2} = \frac{\pi^2}{3}$$ thus $S(\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{6}) = \frac{{{\pi ^2}}}{{15}}$, this is $(2)$. The integral claimed by OP is now established.


This is a partial solution which transformed the original integral into an Ahmed-like integral. Firstly, substitute $y\to \sqrt{2 x^2-1}$ and introduce a parameter $a$ into $\tan ^{-1}\left(a \sqrt{\frac{1}{2} \left(x^2+1\right)}\right)$. Differentiate w.r.t $a$, the integrand will be a rational function. After integration w.r.t $y$, we have $I=-A+B+\frac{C \left(\pi -3 \tan ^{-1}\left(\sqrt{\frac{3}{5}}\right)\right)}{\sqrt{3}}$ where $\small A=\int_0^1 \frac{x \tan ^{-1}\left(\frac{x}{\sqrt{x^2+2}}\right)}{\sqrt{x^2+2} \left(x^2+3\right)} \, dx=\frac{\pi ^2}{288},$ $\small B=\int_0^1 \frac{x \tan ^{-1}\left(\frac{x}{\sqrt{5} \sqrt{x^2+2}}\right)}{\sqrt{x^2+2} \left(x^2+3\right)} \, dx,$ $ C=\int_0^1 \frac{1}{x^2+3} \, dx=\frac{\pi }{6 \sqrt{3}}$. For $A$, integrate by parts using $\small \int \frac{x}{\sqrt{x^2+2} \left(x^2+3\right)} \, dx=\tan ^{-1}\left(\sqrt{x^2+2}\right)$ it reduces to the original Ahmed integral. Apply the same method to $B$, I arrive at corresponding $\small B'=\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(3 x^2+5\right)} \, dx$ whose value should be $\small\frac{1}{5} \sqrt{5} \left(\frac{\pi ^2}{30}-\frac{1}{6} \pi \tan ^{-1}\left(\sqrt{\frac{3}{5}}\right)+\frac{1}{3} \pi \tan ^{-1}\left(\sqrt{\frac{1}{15}}\right)\right)$ due to the conjectural result. So now we only need to justify the value of $B'$.

Update: According to the link FDP provided under pisco's answer, we are able to evaluate more nontrivial Coxeter integrals such as $\int_0^{\frac{\pi }{5}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{71 \pi ^2}{900}$. Based on $S$ function's representation as well as Feynman's trick, an elegant formula is found:

  • $\small \int_{\frac{1}{\sqrt{y}}}^{\frac{1}{\sqrt{x}}} \frac{\arctan (x)}{\sqrt{2 x^2-1} \left(3 x^2-1\right)} \, dx=\frac{1}{2} \pi \left(\tan ^{-1}\left(\sqrt{2-y}\right)-\tan ^{-1}\left(\sqrt{2-x}\right)\right)+\frac{1}{2}\left(S\left(\tan ^{-1}\left(\sqrt{2-y}\right),\frac{\pi }{3},\frac{\pi }{6}\right)-S\left(\tan ^{-1}\left(\sqrt{2-x}\right),\frac{\pi }{3},\frac{\pi }{6}\right)\right)=\left(\sqrt{y (2-y)} A(y)-\sqrt{x (2-x)} A(x)\right)+\frac{1}{6} \pi \left(\tan ^{-1}\left(\sqrt{\frac{3 (2-y)}{y}}\right)-\tan ^{-1}\left(\sqrt{\frac{3 (2-x)}{x}}\right)\right)-\frac{1}{3} \pi \left(\tan ^{-1}\left(\sqrt{\frac{2-y}{3 y}}\right)-\tan ^{-1}\left(\sqrt{\frac{2-x}{3 x}}\right)\right)$

Where function $S$ is the one defined in pisco's answer, $A$ the generalized Ahmed integral:

  • $A(t)=\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(t+x^2\right)} \, dx,A(1)=\frac{5 \pi ^2}{96}$

Using the original Ahmed integral and special value of $S$, one may let $x\to1$ and assign special values to $y$ to evaluate, say

  • $\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(x^2+\frac{2 \sqrt{5}}{5}+1\right)} \, dx=\sqrt{5} \left(\frac{71 \pi ^2}{3600}+\frac{1}{3} \pi \tan ^{-1}\left(\sqrt{\frac{1}{3} \left(9-4 \sqrt{5}\right)}\right)-\frac{1}{6} \pi \tan ^{-1}\left(\sqrt{27-12 \sqrt{5}}\right)\right)$
  • $\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(x^2-\frac{2 \sqrt{5}}{5}+1\right)} \, dx=\sqrt{5} \left(\frac{241 \pi ^2}{3600}+\frac{1}{3} \pi \tan ^{-1}\left(\sqrt{\frac{1}{3} \left(4 \sqrt{5}+9\right)}\right)-\frac{1}{6} \pi \tan ^{-1}\left(\sqrt{12 \sqrt{5}+27}\right)\right)$

From which we deduce the final one (via PFD), a remarkable quartic Ahmed integral:

  • $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(5 x^4+10 x^2+1\right)} \, dx=\frac{37 \pi ^2}{1440}$
  • $\begingroup$ Same negative phenomenon remains for $B^\prime$ if you consider\begin{align}F(a)=\int_0^1 \frac{\tan ^{-1}\left(\frac{a}{\sqrt{x^2+2}}\right)}{\sqrt{x^2+2} \left(3 x^2+5\right)} \, dx\end{align} If you consider \begin{align}F(a)=\int_0^1 \frac{\tan ^{-1}\left(a\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(3 x^2+5\right)} \, dx\end{align}things are even worse. $\endgroup$
    – FDP
    Aug 23, 2019 at 10:34
  • $\begingroup$ i know this book thank you anyway. I think i have collected all articles/books about Ahmed and Coxeter integrals publicly available. For Ahmed's integral there is a few known reference: arxiv.org/ftp/arxiv/papers/1505/1505.03314.pdf it's an amazing proof. $\endgroup$
    – FDP
    Aug 27, 2019 at 15:26
  • $\begingroup$ Xion: parametric differentiation or double integral are the same thing. $\endgroup$
    – FDP
    Aug 27, 2019 at 15:30

Some substitutions which I think make the integral more simple.

$$I=\int_{\sqrt{\frac{3}{5}}}^1 \frac{\arctan (x)}{\sqrt{2 x^2-1} \left(3 x^2-1\right)} \, dx$$


$$I=\int_1^{\sqrt{\frac{5}{3}}} \frac{y \arctan \frac{1}{y}}{\sqrt{2 -y^2} \left(3-y^2\right)} \, dy$$

$$y=\sqrt{2} z$$

$$I=\frac{\sqrt{2}}{3} \int_{\frac{1}{\sqrt{2}}}^{\sqrt{\frac{5}{6}}} \frac{z \arctan \frac{1}{\sqrt{2} z}}{\sqrt{1 -z^2} \left(1-\frac23 z^2\right)} \, dz$$


$$I=\frac{\sqrt{2}}{6} \int_{\frac{1}{2}}^{\frac{5}{6}} \frac{\arctan \frac{1}{\sqrt{2 u} }}{\sqrt{1 -u} \left(1-\frac23 u\right)} \, du$$

$$u= \frac{1+s}{2}$$

$$I=\frac{1}{4} \int_{0}^{\frac{2}{3}} \frac{\arctan \frac{1}{\sqrt{1+s} }}{\sqrt{1 -s} \left(1-\frac12 s\right)} \, ds$$

So we need to prove that:

$$J=\int_{0}^{\frac{2}{3}} \frac{\arctan \frac{1}{\sqrt{1+s} }}{\sqrt{1 -s} \left(1-\frac12 s\right)} \, ds= \frac{3 \pi^2}{40}$$

Let's try integration by parts. Turns out that:

$$ \int \frac{ds}{\sqrt{1 -s} \left(1-\frac12 s\right)}=-4 \arctan \sqrt{1-s}$$

$$\frac{d}{ds} \arctan \frac{1}{\sqrt{1+s} }=-\frac{1}{4} \frac{ds}{\sqrt{1 +s} \left(1+\frac12 s\right)}$$

So our integral is equal to:

$$J=-4 \arctan\frac{1}{\sqrt{1+s} } \arctan \sqrt{1-s} \bigg|_0^{2/3}-\int_{0}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

$$J=\frac{\pi^2}{4}-\frac{2 \pi}{3} \arctan \sqrt{\frac{3}{5}} -\int_{0}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

Maybe this symmetry could help.

Substituting $s \to -s$ we have:

$$J=\int_{-\frac{2}{3}}^0 \frac{\arctan \frac{1}{\sqrt{1-s} }}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

$$\arctan \frac{1}{\sqrt{1-s} }= \frac{\pi}{2}-\arctan \sqrt{1-s}$$

$$J= \frac{\pi}{2}\int_{-\frac{2}{3}}^0 \frac{ds}{\sqrt{1 +s} \left(1+\frac12 s\right)} -\int_{-\frac{2}{3}}^0 \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

$$J= \frac{\pi^2}{6} -\int_{-\frac{2}{3}}^0 \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

Adding the two expressions for $J$ we obtain:

$$2J=\frac{5\pi^2}{12}-\frac{2 \pi}{3} \arctan \sqrt{\frac{3}{5}} -\int_{-\frac{2}{3}}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

$$J=\frac{5\pi^2}{24}-\frac{\pi}{3} \arctan \sqrt{\frac{3}{5}} -\int_{-\frac{2}{3}}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(2+s\right)} \, ds$$

Again, the symmetry might help with the last integral.

So now we need to show:

$$Y=\int_{-\frac{2}{3}}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(2+s\right)} \, ds=\frac{2\pi^2}{15}-\frac{\pi}{3} \arctan \sqrt{\frac{3}{5}}$$

Note that a related integral (from numerical results):

$$\int_{-1}^{1} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(2+s\right)} \, ds= \frac{\pi^2}{6}$$


Turning arctangent into an integral, substituting $s=\sin \theta= \frac{2t}{1+t^2}$ and then integrating a rational integral w.r.t. $t$ (with Mathematica's help), I have made yet another form of the conjecture:

Prove that: $$\int_0^1 \frac{\arctan \frac{2 \sqrt{1+2 p^2}}{\sqrt{5} (1+p^2)}}{\sqrt{1+2 p^2} (1+3 p^2)} dp= \frac{\pi}{2} \arctan \sqrt{\frac{3}{5}}- \frac{\pi^2}{15}$$

This one looks more complicated, but at least the limits are nice.

The integral is similar to $B$ from Fengshan Xiong's solution.

  • $\begingroup$ Thank you, @ClaudeLeibovici, now if only someone managed to finish the proof, since I'm fresh out of tricks $\endgroup$
    – Yuriy S
    Aug 23, 2019 at 14:56
  • $\begingroup$ I bet that if i try to compute $Y$ (Feynman's trick) i will get the same problem we have already faced. $\endgroup$
    – FDP
    Aug 23, 2019 at 15:14
  • $\begingroup$ @FDP, you are right, see the last edit. I get almost the same complicated integral as the OP. In any case, an interesting problem $\endgroup$
    – Yuriy S
    Aug 23, 2019 at 17:35
  • $\begingroup$ Probably the solution is to write the integral to compute as a sum of two nice integrals doable using Feynman's trick. i don't know yet how to do this. $\endgroup$
    – FDP
    Aug 23, 2019 at 19:32

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