Introduction
I found this definition of the Rutherford constant on Wolfram's site:
$$\mathcal{K}_R=\sqrt{2}\int_{-1}^{\infty}\frac{R(x)^2}{S(x)}\mathrm{d}x$$ Where $$R(x)=\frac{1}{2\pi}\int_{\overline{u}}^{2\pi-\overline{u}}\frac{\cos(u)}{\sqrt{x-\cos(u)}}\mathrm{d}u\qquad S(x)=\frac{1}{2\pi}\int_{\overline{u}}^{2\pi-\overline{u}}\frac{1}{\sqrt{x-\cos(u)}}\mathrm{d}u$$
and
$$\overline{u}=\begin{cases}0&\text{for }x>1\\\cos^{-1}(x)&\text{for }-1<x<1\end{cases}$$
Question
It seemed like a definition that was a bit too much to feed into a calculator (nested functions and piecewise functions), so I tried to simplify it a bit and arrived at this formulation:
$$\mathcal{K}_{R}=\frac{4}{\pi}\left[\int_{0}^{1}\left(\frac{(4E'(x)+K(x))^2}{x^{\frac{3}{2}}}+4E(x)^2\right)\frac{\mathrm{d}x}{K(x)}-\frac{10}{3}\right]$$
Where
$$K(z)\text{ and }E(z)\text{ are the complete elliptic integral of the 1st and 2nd kind}$$
$E'(x)$ is the derivative of $E(x)$
Does anyone know any formula that can further simplify this integral?
I think this might be an easier part to deal with:
$$\displaystyle \int_{0}^{1}\frac{E(x)^2}{K(x)}\mathrm{d}x$$