Integral of complete elliptic Integral of second kind I am currently wondering wether there is an indefinite integral for 
$$\int dx \;\sqrt{1-x^2} E\left (1+\frac{1}{x^2-1} \right )$$
WolframAlpha states, that there is no result in terms of analytical functions. I am not a Mathematician and don't know the current literature in the field. I wonder whether there maybe is literature more recent, which has a solution for this.
Also if there is no solution, I wonder whether there is a way to prove that there is no solution.
Looking forward to helpfull answers.
Cheers!
 A: 
Define the function $\mathcal{I}:\left[-1,1\right]\rightarrow\mathbb{R}$ via the definite integral
$$\begin{align}
\mathcal{I}{\left(z\right)}
&:=\int_{0}^{z}\mathrm{d}x\,\sqrt{1-x^{2}}\operatorname{E}{\left(1+\frac{1}{x^{2}-1}\right)},\tag{1}\\
\end{align}$$
where the complete elliptic integral of the second is defined as a function of the elliptic parameter $\mu$ by the integral representation
$$\operatorname{E}{\left(\mu\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{\sqrt{1-\mu x^{2}}}{\sqrt{1-x^{2}}};~~~\small{\mu\in\left(-\infty,1\right]}.\tag{2}$$
By the fundamental theorem of calculus, $\mathcal{I}$ is a valid antiderivative over the interval $[-1,1]$. The question is whether $\mathcal{I}$ has a closed form in terms of recognizable special functions.

The following functional relation for $E$ is very helpful here:
$$\begin{align}
\operatorname{E}{\left(\frac{z}{z-1}\right)}
&=\frac{1}{\sqrt{1-z}}\operatorname{E}{\left(z\right)};~~~\small{z\in\left(-\infty,1\right)}.\tag{3}\\
\end{align}$$
This relation is just a special case of Pfaff's transformation for hypergeometric functions.

Let $z\in\left[-1,1\right]$. Using Pfaff's transformation, the integral simplifies to
$$\begin{align}
\mathcal{I}{\left(z\right)}
&=\int_{0}^{z}\mathrm{d}x\,\sqrt{1-x^{2}}\operatorname{E}{\left(1+\frac{1}{x^{2}-1}\right)}\\
&=\int_{0}^{z}\mathrm{d}x\,\sqrt{1-x^{2}}\operatorname{E}{\left(\frac{x^{2}}{x^{2}-1}\right)}\\
&=\int_{0}^{z}\mathrm{d}x\,\operatorname{E}{\left(x^{2}\right)}.\tag{4}\\
\end{align}$$
Wolfram Alpha is in fact capable of returning an antiderivative for $\operatorname{E}{\left(x^{2}\right)}$ in the form of a generalized hypergeometric function (the antiderivative can also be found here), giving us
$$\begin{align}
\mathcal{I}{\left(z\right)}
&=\frac{\pi z}{2}\,{_3F_2}{\left(-\frac12,\frac12,\frac12;1,\frac32;z^{2}\right)}.\tag{5}\\
\end{align}$$
I couldn't find any known expressions for the function in the tables here, and my own attempts to derive one failed as well. So unfortunately, this hypergeometric function might be the simplest closed-form that we can find.

