Prove $$\int_{0}^{1}\frac{1}{\sqrt{1-x^2}}\arctan\left(\frac{88\sqrt{21}}{36x^2+215}\right)dx=\frac{\pi^2}{6}.\tag1$$
My attempts are as follows.
We define for $a,b>0$ $$f(a,b)=\int_0^1\frac{\arctan(ax^2+b)}{\sqrt{1-x^2}}dx,\tag2$$ so that $$\int_0^1\frac{1}{\sqrt{1-x^2}}\arctan\left(\frac1{ax^2+b}\right)dx=\frac{\pi^2}{4}-f(a,b).$$
The only sensible thing I could think to do was substitute $t=ax^2+b$ so that $f(a,b)=r(b,a+b)$, where $$r(p,q)=\int_{p}^{q}\frac{\arctan x}{\sqrt{(x-p)(q-x)}}dx,$$ which seems only slightly more manageable. I decided to give this approach up and return to $f(a,b)$. We see that $$\frac1{\sqrt{1-x^2}}=\frac12\sqrt{\frac{1-x}{1+x}}+\frac12\sqrt{\frac{1+x}{1-x}},\tag3$$ so that $$f(a,b)=\frac12\int_{-1}^{1}\sqrt{\frac{1-x}{1+x}}\arctan(ax^2+b)dx,\tag4$$ as we can split up $(2)$ into two integrals using $(3)$, preform the change of variable $x\mapsto-x$ in the second, then add the two integrals back together. As it stands, $(4)$ may be the easiest form of the integral to work with, but I am not sure. Another idea is to set $t=\sqrt{\frac{1-x}{1+x}}$ in $(4)$ so that $$f(a,b)=2\int_0^\infty \frac{x^2}{(1+x^2)^2}\arctan(ax+b)dx.\tag5$$ From $ax+b\mapsto x$ followed by integration by parts with $u=\arctan x$, we have $$f(a,b)=\frac{\pi^2}{4a^2}-\frac2a\int_b^\infty \left[\frac1a\arctan\left(\frac{x-b}{a}\right)-\frac{x-b}{(x-b)^2+a^2}\right]\frac{dx}{1+x^2}.\tag6$$ The first integral $$J_1(a,b)=\int_b^\infty \frac{\arctan\left(\frac{x-b}{a}\right)}{x^2+1}dx$$ does not seem very easy at all and I do not know how to go about evaluating it. However, the second integral $$J_2(a,b)=\int_b^\infty \frac{x-b}{(x-b)^2+a^2}\frac{dx}{1+x^2}$$ can be evaluated explicitly as $$J_2(a,b)=M\pi b(a^2+b^2+2a+1)-M\left[(a^2-b^2-1)\ln\frac{a^2}{b^2+1}+2b(a^2+b^2+1)\arctan b\right],$$ where $$M=-\frac{1}{2(a^2+(b-i)^2)(a^2+(b+i)^2)}.$$ I feel, however, that there is an easy way to evaluate the integral $f(a,b)$, but as of now it is beyond me. Any help is appreciated.
