Function $h$
The function $h(x)$ can be expressed in several ways. Firstly it is expressed as:- $$ h_1(x) = (1+x) \operatorname{E}\left( \frac{4x}{(1+x)^2} \right) $$ (where $\operatorname{E}$ is a Complete Ellitpic Integral function of the 2nd Kind)
Secondly, by unwrapping the formula for $E()$ we obtain a formula containing an infinite sum for $h(x)$ given by:- $$ h_2(x) = (1+x) \frac{\pi}{2} \sum_{n=0}^{\infty}\left[ \frac{(2n)!}{4^n(n!)^2} \right]^2 \frac{\left( \frac{4x}{(1+x)^2} \right)^{2n}} {1-2n}. $$
Thirdly, by expanding $h_2(x)$, $h(x)$ can be expressed as an infinite Taylor Series:- $$ h_3(x) = \pi \left( \frac{1x^0}{2} +\frac{1x^2}{8} +\frac{1x^4}{128} +\frac{1x^6}{512} +\frac{25x^6}{32768} + ... \right) $$
Also, assuming the truth of a conjectured identity here and following comment by /u/user at that link, I infer that:- $$ h_4(x) = \frac{\pi}{2} \sum_{n=0}^{\infty}\left[ \frac{(2n)!}{4^n(2n-1)(n!)^2} \right]^2~ x^{2n}. $$
Function $e4$
I am not presently clear how $h_4(x)$ is obtained from $h_2(x)$. In trying to understand how, I am looking at the simpler function, labelled $e4(x)$, defined by:-
$$e4(x) = \operatorname{E}\left( \frac{4x}{(1+x)^2} \right) = \frac{\pi}{2} \sum_{n=0}^{\infty}\left[ \frac{(2n)!}{4^n(n!)^2} \right]^2~ \frac{\left( \frac{4x}{(1+x)^2} \right)^{2n}} {1-2n}. $$
A Wolfram Alpha expansion of $e4(x)$ suggests that the output series can be neatly split into two component sub-series. I hypothesize that these two sub-series can be generated by two separate functions: $e4A(x)$ and $e4B(x)$... $$ e4(x) = e4A(x) + e4B(x) $$
Question - functions $e4A$ and $e4B$
I wish to know (or learn of a method for determining) the two specific infinite sum formulae expressions (containing $A(n,x)$ and $B(n,x)$ respectively) which correspond to the functions $e4A(x)$ and $e4B(x)$:-
$$ e4A(x)= \pi \left( \frac{1x^0}{2} +\frac{5x^2}{8} +\frac{81x^4}{128} +\frac{325x^6}{512} +\frac{20825x^8}{32768} + ... \right) = \frac{\pi}{2} \sum_{n=0}^{\infty} A(n,x) $$
$$ e4B(x) = - \pi \left( \frac{1x^1}{2} +\frac{5x^3}{8} +\frac{81x^5}{128} +\frac{325x^7}{512} +\frac{20825x^9}{32768} + ... \right) = \frac{\pi}{2} \sum_{n=0}^{\infty} B(n,x). $$