Proving a closed form for an integral with nested radicals Is there a simple way to prove the following identity?
$$\int_0^1\sqrt{\frac{u^2-2-2 \sqrt{u^4-u^2+1}}{4 u^6-8 u^4+8 u^2-4}}\mathrm du=\frac{\sqrt{3+2 \sqrt{3}}}{2^{10/3}\pi}\Gamma\left(\frac13\right)^3$$

Context:
This integral came up in trying to evaluate the complete elliptic integral of the first kind $K(m)$ ($m$ is the parameter),
$$K\left(\exp\left(\frac{i\pi}{3}\right)\right)$$
in terms of simpler functions. In particular,
$$K\left(\exp\left(\frac{i\pi}{3}\right)\right)=C\left(1+i \left(2-\sqrt{3}\right)\right)$$
and $C$ is the integral mentioned in the first part.
I was able to show this through an indirect route, but I am hoping my messy method can be easily outdone.
 A: Here is another approach that doesn’t requires Hypergeometric Functions.
$$I=\int_{0}^{1}\sqrt{\frac{u^2-2-2\sqrt{u^4-u^2+1}}{4u^6-8u^4+8u^2-4}}du=\int_{0}^{1}\underbrace{\sqrt{\frac{2-u^2+2\sqrt{{u^4-u}^2+1}}{4\left(1-u^2\right)\left({u^4-u}^2+1\right)}}}_{u\rightarrow\sqrt{x}}du$$
$$I=\frac{1}{4}\int_{0}^{1}\underbrace{\sqrt{\frac{2-x+2\sqrt{x^2-x+1}}{x\left(1-x\right)\left(x^2-x+1\right)}}}_{x\rightarrow y+\frac{1}{2}}dx=\frac{1}{4}\int_{-\frac{1}{2}}^{\frac{1}{2}}\underbrace{\sqrt{\frac{\frac{3}{2}-y+2\sqrt{y^2+\frac{3}{4}}}{\left(\frac{1}{4}-y^2\right)\left(y^2+\frac{3}{4}\right)}}}_{f(y)}dy$$
$$I=\frac{1}{4}\int_{0}^{\frac{1}{2}}\left(f\left(y\right)+f\left(-y\right)\right)dy=\frac{\sqrt{2+\sqrt3}}{2\sqrt2}\int_{0}^{\frac{1}{2}}\underbrace{\sqrt{\frac{\frac{\sqrt3}{2}+\sqrt{y^2+\frac{3}{4}}}{\left(\frac{1}{4}-y^2\right)\left(y^2+\frac{3}{4}\right)}}}_{y=\sqrt{z^2-\frac{3}{4}}}dy$$
$$I=\frac{\sqrt{2+\sqrt3}}{2\sqrt2}\int_{\frac{\sqrt3}{2}}^{1}\underbrace{\frac{dz}{\sqrt{\left(1-z^2\right)\left(z-\frac{\sqrt3}{2}\right)}}}_{z=\cos{\left(\theta\right)}}=\frac{\sqrt{2+\sqrt3}}{2\sqrt2}\int_{0}^{\frac{\pi}{6}}\underbrace{\frac{dz}{\sqrt{\cos{\left(\theta\right)}-\cos{\left(\frac{\pi}{6}\right)}}}}_{\sin{\left(\frac{\theta}{2}\right)}=\sin{\left(\phi\right)}\sin{\left(\frac{\pi}{12}\right)}}$$
$$I=\frac{\sqrt{2+\sqrt3}}{2}\int_{0}^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-\sin^2{\left(\frac{\pi}{12}\right)\sin^2{\left(\phi\right)}}}}=\frac{\sqrt{2+\sqrt3}}{2}K\left(\sin{\left(\frac{\pi}{12}\right)}\right)$$
$$I=\frac{\sqrt{2+\sqrt3}}{2}\left(\frac{1}{2\ 3^\frac{3}{4}}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{4}{6}\right)}\right)=\frac{\sqrt{3+2\sqrt3}}{2^\frac{10}{3}\pi}{\Gamma\left(\frac{1}{3}\right)}^3$$
Some explanation:
1)  I’ve started factorizing the denominator to check if there was any common term that could be canceled.
2)  Then, I applied a couple of substitutions. The first one had the objective of decreasing the degree of the variable, and the second was aiming to eliminate the first-degree term of the polynomial inside the inner root.
3)  Then I rewrote the integral exploiting the symmetry of its integration limits, which result in a sum of two square roots ($f(y)+f(-y)$) that could be rewritten after some algebra as a single square root.
4)  After that, some other substitutions were applied to make calculations easier. The last one may look kind trick, but it’s intuitive after rewriting the expression using sines of half angle.
5)  Finally, a well-known representation of the Complete Elliptic Integral of the First Kind was found, and its value was taken from a table and then the result was simplified using both duplication and reflection formulas of Gamma.
