Integrating a Ratio of Elliptic Integrals Can anyone help evaluate $$\int dx\frac{\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-k^{2}\sin^{2}\left(\theta\right)}}}{x\int_{0}^{\frac{\pi}{2}}d\theta\sqrt{1-k^{2}\sin^{2}\left(\theta\right)}}$$
where $$k = \frac{x}{R}$$
I'm not exactly sure if this is possible, though someone may know a trick...
Explanation
I'm trying to use the method of reduction of order to find the second solution to
$$
\frac{d^{2}}{d\epsilon^{2}}g^{\star}+\frac{\left(R^{2}+3\epsilon^{2}\right)}{\epsilon\left(R^{2}-\epsilon^{2}\right)}\frac{d}{d\epsilon}g^{\star}+\frac{\left(5R^{2}+3\epsilon^{2}\right)}{\left(R^{2}-\epsilon^{2}\right)^{2}}g^{\star}=0
$$
where $g^{\star}$ is a function of $\epsilon$, given $$g_1 = \frac{1}{\pi R^{2}}\left(2\left(R^{2}-\epsilon^{2}\right)E\left(\frac{\epsilon}{R}\right)\right)$$ is a solution; $E(x)$ is the complete elliptic integral of the second kind:
$$
E\left(k\right)=\int_{0}^{\frac{\pi}{2}}d\theta\sqrt{1-k^{2}\sin^{2}\left(\theta\right)}
$$
The method of order of reduction involves this integrating factor $\mu(\epsilon)$
$$
\log(\mu(\epsilon)) = \int{dx\frac{3 E\left(\frac{x}{R}\right)-2 K\left(\frac{x}{R}\right)}{x E\left(\frac{x}{R}\right)}}
$$
where
$$
K\left(k\right)=\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-k^{2}\sin^{2}\left(\theta\right)}}
$$
is the complete elliptic integral of the first kind.

Notice
$$k\frac{dE(k) }{dk}
= k\int_0^{\pi/2} \frac{\partial}{\partial k}\sqrt{1-k^2\sin^2\theta} d\theta
= \int_0^{\pi/2} \frac{-2k^2\sin^2\theta}{2\sqrt{1-k^2\sin^2\theta}} d\theta
= E(k) - K(k)$$
This leads to
$$\frac{K(k)}{E(k)} = 1 - k\frac{d}{dk}\log E(k) = 1 - x\frac{d}{dx}\log E(k)$$
and hence
$$\int\frac{dx}{x}\frac{K(k)}{E(k)}
= \int\left[\frac{1}{x} - \frac{d}{dx}\log E(k)\right] dx = \log\frac{x}{E(k)} + \text{const.}
$$
