Solving the integral $\int_0^{\pi/4}\frac{\tan^3(x)}{\sqrt{1-x^2}}dx$

I've come across this integral recently (on a Facebook group): $$\int_0^{\pi/4}\frac{\tan^3(x)}{\sqrt{1-x^2}}dx$$ and have little to no clue how to solve it. Wolfram Alpha gives a numerical answer of approximately 0.209, but could not evaluate it 'manually'. My gut feeling is that it will involve some sort of contour integral, but I cannot see what might work. If anyone has any insight, it would be much appreciated.

• What sort of answer are you looking for? Do you have any reason to believe the value is not a transcendental number (i.e., requires numerical methods, similar to the one(s) used by W.Alpha to get "0.209", to evaluate)? – Eric Towers Jul 4 at 22:12
• I'm not really expecting any answer in particular. This was posed as a particularly difficult and interesting problem though, so I'm relatively confident it has a transcendental answer, else I feel this problem wouldn't be posed as 'interesting'. I thought I'd post it here in case anyone had any insight into it, as I am also confident this problem is beyond my level of education, but I'm always keen to learn of methods I haven't come across before. – KB399 Jul 4 at 22:28
• May be you could avoid tking problems from Facebook. What about McDonald as a source of inspiration ? (I am just kidding, be sure). Cheers :-) – Claude Leibovici Jul 5 at 5:01
• If the upper limit was $1$, Chebyshev quadrature would have come in handy – Yuriy S 12 hours ago

I am skeptical about a closed form for this definite integral. Its numerical value is $$0.20852064462921058579012825612057983198834121981775$$ which has not been identified by any inverse symbolic calculator. The closest found is $$\frac 1 {10}\Gamma \left(\frac{1}{6}\right) \Big(\log(2)\Big) ^{\Gamma \left(\frac{1}{3}\right)}=0.20852064115$$
What I suppose is that this could be done by series expansion to high order $$\frac{\tan ^3(x)}{\sqrt{1-x^2}}=x^3 \left( 1+\frac{3 x^2}{2}+\frac{193 x^4}{120}+\frac{22979 x^6}{15120}+\frac{275197 x^8}{201600}+\frac{48228793 x^{10}}{39916800}+O\left(x^{12}\right)\right)$$ and integrated termwise. Let us try with expansions up to $$O(x^{2n+3})$$ for the integrand. This would lead for the definite integral to the following values $$\left( \begin{array}{cc} n & \text{result for series expansion} \\ 10 & 0.20840782763821602389366491327101465 \\ 20 & 0.20852027116982251348225379128240811 \\ 30 & 0.20852064289873138870614412907285772 \\ 40 & 0.20852064461992536661178971837443094 \\ 50 & 0.20852064462915641583336414649616403 \\ 60 & 0.20852064462921025212195797014568471 \\ 70 & 0.20852064462921058365488869446074788 \\ 80 & 0.20852064462921058577607189472019213 \\ 90 & 0.20852064462921058579003367910543448 \\ 100 & 0.20852064462921058579012760861728633 \\ \cdots &\cdots \\ \infty & 0.208520644629210585790128256120579832 \end{array} \right)$$