I came across the following integral identity $$I = \int^1_0 \mathrm{d}x \frac{x^2}{\sqrt{1-x^4}} = \frac{\sqrt{\pi}}{3}\frac{\Gamma\left(\frac{7}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}$$ involving the gamma function, but I cannot seem to derive the result myself.
My take is to expand the denominator and do term-by-term integration: $$ \begin{align} I &= \int^1_0 \mathrm{d}x \frac{x^2}{\sqrt{1-x^4}} \\ &= \int^1_0 \mathrm{d}x \text{ }x^2\left(1-x^4\right)^{-1/2} \\ &= \int^1_0 \mathrm{d}x \sum^\infty_{n=0}\binom{-\frac{1}{2}}{n}(-1)^n x^{4n+2}. \\ \end{align} $$ Here we can exchange the sum and the integral to get $$ \begin{align} I &= \sum^\infty_{n=0}\binom{-\frac{1}{2}}{n}(-1)^n\int^1_0 \mathrm{d}x\text{ }x^{4n+2} \\ &= \sum^\infty_{n=0}\binom{-\frac{1}{2}}{n}\frac{(-1)^n}{4n+3} \\ &= \sum^\infty_{n=0}\frac{1}{4n+3}\frac{(2n-1)!!}{2^n n!}, \end{align} $$ where we define $(-1)!!$ to be $1$. I think this method reaches a dead end, since showing that this expansion is equivalent to the desired expression feels just as cumbersome as the original problem.
The substitution $$x = u^{1/4}, \quad dx = \frac{1}{4} u^{-3/4} \, du,$$ immediately gives $$\int_{x=0}^1 x^2 (1-x^4)^{-1/2} \, dx = \frac{1}{4} \int_{u=0}^1 u^{-1/4} (1-u)^{-1/2} \, du.$$ Now apply the beta function identity $$\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = \int_{x=0}^1 x^{a-1} (1-x)^{b-1} \, dx$$ and also recall that $$\Gamma(\tfrac{1}{2}) = \sqrt{\pi}.$$