Notes
This is a slowly-developing answer, still incomplete.
For this answer I will use the "wikipedia" argument convention $\operatorname{K}(k)$ rather than the "Wolfram Alpha" convention $\operatorname{K}(m=k^2).$
Approach
The expressions $f(x)$, $g(x)$ and $h(x)$ all output an infinite series of additive terms in powers of $x$.
For each expression I will attempt to define a formula for the numeric coefficient $C_n$ of the term containing $x^n$. Then if all formulae give the same value for $C_n$ the identity: $f(x)=g(x)=h(x)$ will be proven.
For expression $g(x)$ all the arguments are in terms of $x$, whereas $f(x)$ and $h(x)$ both have argument $\gamma= \sqrt\frac{4x}{((1+x)^2)}$ .
Coefficients of Expression $g(x)$
For $g(x)$, in order to obtain a formula for the numerical coefficient of each $x$-power term, it is relatively straightforward. We use the power series formulae for $\operatorname{K}$ and $\operatorname{E}$.
Consider the following expression of $g(x)$ :-
$$
g(x) =
-\operatorname{K}(x)
~+ x^2* \operatorname{K}(x)
~+2\operatorname{E}(x)
$$
and defining $P_n$ as the Legendre Polynomial $P_{2n}(0)$ expressed as ...
$$
P_n =\left(
\frac{(2n)!}
{4^n ~n!~n!}
\right).
$$
We then have
$$
\frac{2}{\pi}g(x) =
-\sum_{n=0}^{\infty} P^2_n x^{2n}
+x^2\sum_{n=0}^{\infty} P^2_n x^{2n}
+2\sum_{n=0}^{\infty} P^2_n \frac{1}{1-2n}x^{2n}
$$
$$
\frac{2}{\pi}g(x) =
-\sum_{n=0}^{\infty} P^2_n x^{2n}
+\sum_{n=0}^{\infty} P^2_n x^{2n+2}
+\sum_{n=0}^{\infty} 2 P^2_n \frac{1}{1-2n}x^{2n}
$$
It can be seen that the power series will only contain even powers of $x$:$(0,2,4,6...)$.
We can deduce that for a given term $C_r.x^r$ with $r$ being the particular power of $x$, the coefficient $C_r$ from $g(x)$, symbolized as $C_{r(g)}$ will be given by the sum of three contributions:-
$$C_{r(g)} = \frac{\pi}{2} \left[ - P^2_{(n=r/2)} + P^2_{(n=(r-2)/2)} + 2\frac{P^2_{(n=r/2)}}{1-2{(n=r/2)}}\right]$$
$$C_{r(g)} = \frac{\pi}{2} \left[ - P^2_{(r/2)} + P^2_{(r/2 -1)} + 2\frac{P^2_{(r/2)}}{1-r}\right]$$
This applies for $r=2,4,6,8,...$. For $r=0$ there is no contribution from the middle term:- $x^2K(x)$.
Initial terms calculated from this formula are:-
$$
\pi \left( 1+\frac{1x^2}{8} +\frac{1x^4}{128} +\frac{1x^6}{512} +\frac{25x^8}{32768} +\frac{49x^{10}}{131072} \text{...}\right)
$$
This agrees with the initial terms calculated (by Wolfram Alpha) for $h(x)$ and $g(x)$.
User's Hint
$$g(x)_{user} = \frac{\pi}{2} \sum^{\infty}_{n=0}
\left(
\binom{2n}{n} * \frac{1}{4^n} * \frac{1}{(2n-1)}
\right)^2 x^{2n}
$$
For a required power $R$ of $x$ the coefficient $C_{r(u)}$ can be calculated quite simply. First we note that there are no odd powers of $x$ since the power term is $2n$ and $n$ is an integer $(0,1,2,3...)$.
The value of $C_{r(u)}$ is given by:-
$$
C_{r(u)} = \frac{\pi}{2} \left( \binom{r}{r/2} * \frac{1}{4^{r/2}} * \frac{1}{r-1} \right)^2
$$
The initial terms of $g(x)$ calculated (by excel) with this formula are
$$
\pi \left( 1+\frac{1x^2}{8} +\frac{1x^4}{128} +\frac{1x^6}{512} +\frac{25x^8}{32768} +\frac{49x^{10}}{131072} \text{...}\right)
$$
This agrees with the initial terms calculated (by Wolfram Alpha) for $h(x)$ and $g(x)$.
Note: I cannot apply the User hint formula in a proof because (so far) I have not been able to derive it from the initial formulae for $f(x)$, $g(x)$ or $h(x)$.
Remaining tasks
(1) obtain a similar formula for the coefficients of $h(x)$ (or $f(x)$);
(2) show that for any value of $n>0$ (because these formulae dont work for $n=0$) the coefficient formulae give equal values of $C_n$;
(3) show that the coefficients for $n=0$ for expressions $g()$ and $h()$ (or $f()$) have the same values .
NOTE Need to insert factor $\pi /2$ as appropriate below
Coefficients of Expression $h(x)$
This will be more difficult than for $g(x)$ because the argument ($\gamma=\sqrt{\frac{4x}{(1+x)^2}}$) of the complete elliptic integral ($\operatorname{E}$) expands to a more complicated expression...
$$
h(x)
= (1+x) \operatorname{E}(\gamma)
= (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} \gamma^{2n}
= (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} \left( \frac{4x}{(1+x)^2}\right)^n
$$
$$
h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4x)^n(1+x)^{-2n}
$$
$$
h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4x)^n(1-2x+3x^2-4x^3 ...)^{n}
$$
$$
h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4)^n(x-2x^2+3x^3-4x^4 ...)^{n}
$$
$$
h(x) =
\sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4)^n(x-2x^2+3x^3-4x^4 ...)^{n}
+ \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} (4)^n(x-2x^2+3x^3-4x^4 ...)^{n} (x)
$$
Actually this is not the best way to go about things.
A formula is available - the Binomial Series Expansion which applies for all fixed real $a$ and for all $x$ in the interval $-1<x<+1$ (UEOM: Universal Encyclopedia of Mathematics, page 574) for obtaining the coefficients of the different $x$-powers for expressions of the form $(1+x)^{n/m}$ namely
$$(1+x)^{n/m}= 1 + \frac{n}{m}x^1 - \frac{n(m-n)}{2!m^2} x^2+ \frac{n(m-n)(2m-n)}{3!m^3} x^3 - ... + (-1)^{k+1} \frac{n(m-n)(2m-n)...[(k-1)m - n]}{k!m^k}x^k + ...$$
In the present problem the fractional form can be replaced by the simpler integer form (UEOM, page 69):
$$
(1+x)^{\alpha} = 1 + \binom{\alpha}{1}x+ \binom{\alpha}{2}x^2+ \binom{\alpha}{3}x^3 + ... = \sum_{k=0}^{\infty} \binom{\alpha}{k}x^k
$$
Where the coefficients are calculated from this formula:
$$
\binom{\alpha}{k} = \frac{\alpha(\alpha-1)(\alpha-2)...(\alpha-k+1)}{k!}
$$
Note that in the present problem the Binomial Theorem (UEOM, page 69) does NOT apply in general because the exponent $\alpha = -2n$ is usually $<-1$
For any chosen power $i$ of $x$ we can apply the Binomial Series formula to our denominator factor $(1+x)^{-2}$ to obtain a coefficient which contributes to the coefficient $C_i$ in the term $C_ix^i$ in the expansion of $h(x)$. But we must also consider the other factors, namely $(4x)^n$, $(1+x)$ and $P_n$, which contribute to the value of $C_i$.
For example
$$
\text{factor }(1+x)^{\alpha=-2} \text{ contributes the term } \binom{\alpha=-2}{k}x^{k}
$$
Let us re-express the formula for $h(x)$:-
$$
h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2}{1-2n} \left( \frac{4x}{(1+x)^2}\right)^n
$$
$$
h(x) = (1+x) \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~(1+x)^{-2n}
$$
Let us split the RHS into two "streams", the "1-stream" and the "$X$-stream."
$$
h(x) = (1) \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~(1+x)^{-2n}
+(x) \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~(1+x)^{-2n}
$$
$$
h(x) = \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~(1+x)^{-2n}
+ \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^{n+1}~(1+x)^{-2n}
$$
$$
h(x) = \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^n~\sum_{k=0}^{\infty} \binom{-2n}{k}x^k
+ \sum_{n=0}^\infty \frac{{P_n}^2~4^n}{1-2n} x^{n+1}~\sum_{k=0}^{\infty} \binom{-2n}{k}x^k
$$
For clarity let us replace
$$
\frac{{P_n}^2~4^n}{1-2n} ~\text{ by }~Q_n
$$
so now the RHS is expressed as the sum of two streams, with each stream now comprising a series in $n$ containing a "nested" series in $k$.
$$
h(x) = \sum_{n=0}^\infty Q_n x^n~\sum_{k=0}^{\infty} \binom{-2n}{k}x^k
+ \sum_{n=0}^\infty Q_n x^{n+1}~\sum_{k=0}^{\infty} \binom{-2n}{k}x^k
$$
It can be seen that a particular pair of $n,k$ values will output a value of
$$
Q_n x^n.\binom{-2n}{k}x^{k}+ Q_n x^{n+1}.\binom{-2n}{k}x^{k}
$$
$$
= Q_n x^{k+n}.\binom{-2n}{k} + Q_n x^{k+n+1}.\binom{-2n}{k}
$$
Nothing forces us to pair with the same $n$ across the two streams. We could arrange it so that like powers of $x$ are collected by pairing the $(n+1,k)$ and $(n,k)$ terms from respective streams, thus...
$$
= Q_{n+1} x^{k+n+1}.\binom{-2(n+1)}{k} + Q_n x^{k+n+1}.\binom{-2n}{k}.
$$
I will call such an arrangement a "staggered" arrangement.
Now let us consider what it takes to collect, in either stream, all the contributions $x^r$ in a particular "required" power $r:r>0$, of $x$.
Firstly note that, for any $n>=1$, expansion of the factor $(1+x)^{-2n}$ generates an infinite series of ascending positive $x$-power terms $x^i$: with indices $i = 0, 1,2,3,4,5,6...\infty$ for example:-
$$\begin{align}
{n=0}; (1+x)^{-2n} & = 1 \\
{n=1}; (1+x)^{-2n} & = 1 - 2x^1 + 3x^2 - 4x^3 + 5x^4 - 6x^5 ...\\
{n=2}; (1+x)^{-2n} & = 1 - 4x^1 + 10x^2 - 20x^3 + 35x^4 - 56x^5 ...\\
{n=3}; (1+x)^{-2n} & = 1 - 6x^1 + 21x^2 - 56x^3 + 126x^4 -252x^5 ...\\
{n=4}; (1+x)^{-2n} & = 1 - 8x^1 + 36x^2 -120x^3 + 330x^4 -792x^5 ...\\
\end{align}
$$
Therefore, in either stream, for $(n>0)$, every value of $n$ will have an associated expansion of the $(1+x)^{-2n}$ factor which contains an $x$-power term that will contribute to $x^r$.
For example lets say we wish to determine the ultimate coefficient of the term $x^4$. In each stream (without staggering), the contributions will come from two sources, the primary source (located just after $Q_n$) of either $x^{n}$ (from 1-stream) or $x^{n+1}$ (from X-stream) and the secondary source (located within the $(1+x)^{-2n}$ expansion series) of $x^{i}$.
For the 1-stream we require the condition must be satisfied that $n+i=r$, so in this example $n+i=4$ and we can see that contributions to $x^4$ will come from the following $(n,i)$ pairs:- (1,3) (2,2) (3,1) and (4,0) only. Obviously as $r$ increases so the number of contributing $(n,i)$ pairs will increase also.
This allows us to write a deterministic formula for the pre-ultimate coefficient $C_{r(h)_1}$ of the required term $C_r.x^r$ coming from the 1-stream (primary source: $x^n$):-
$$
C_{r(h)_1} = \sum_{n=1}^{r} Q_n * \binom{-2n}{r-n}
$$
A similar formula can be written for the pre-ultimate coefficient $C_{r(h)_X}$ of the required term $C_r.x^r$ coming from the $X$-stream (primary source: $x^{n+1}$):-
$$
C_{r(h)_X} = \sum_{n=1}^{r-1} Q_n * \binom{-2n}{r-1-n}
$$
Therefore the coefficient $C_r$ of the required $x$-power term $C_r.x^r$ is given by
$$
C_{r(h)}=C_{r(h)_1}+C_{r(h)_X} = \sum_{n=1}^{r} Q_n * \binom{-2n}{r-n} + \sum_{n=1}^{r-1} Q_n * \binom{-2n}{r-1-n}
$$
$$
C_{r(h)} = Q_r * \binom{-2r}{r-r=0}
+
\sum_{n=1}^{r-1} Q_n * \left( \binom{-2n}{r-n} + \binom{-2n}{r-n-1} \right ).
$$
Now $\binom{A}{0}= 1$, for any $A$ so...
$$
C_{r(h)} = Q_r * 1
+
\sum_{n=1}^{r-1} Q_n * \left( \binom{-2n}{r-n} + \binom{-2n}{r-n-1} \right )
$$
Now we can apply the Addition(Induction) Rule of binomial coefficients, namely $\binom{N}{K} = \binom{N-1}{K} +\binom{N-1}{K-1}$ from which:- $\binom{N+1}{K} = \binom{N}{K} +\binom{N}{K-1}$, so...
$$
C_{r(h)} = Q_r
+
\sum_{n=1}^{r-1} Q_n * \binom{-2n+1}{r-n}
= Q_r
+
\sum_{n=1}^{r-1} Q_n * \binom{-(2n-1)}{r-n}
$$
Upper Negation
To remove the negative $n$ terms we apply the Upper Negation Identity for binomial coefficients, namely $\binom{-N}{K} = (-1)^K * \binom{N+K-1}{K}$
$$
C_{r(h)} = Q_r
+
\sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-1 + r-n -1}{r-n}
$$
$$
C_{r(h)} = Q_r
+
\sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{r-n}
$$
Applying the Symmetry Rule:- $\binom{N}{K} = \binom{N}{N-K}$ gives...
$$
C_{r(h)} = Q_r
+
\sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{(2n-2 + r-n )-(r-n)}
$$
$$
C_{r(h)} = Q_r
+
\sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2}
$$
(This expression tested OK for $r>=2$).
Unfortunately this expression for $C_{r(h)}$ is still not very similar in pattern to the previously-derived expressions of
(i) $C_{r(g)}$ from $g(x)$:-
$$C_{r(g)} = \frac{\pi}{2} \left[ - P^2_{(r/2)} + P^2_{(r/2 -1)} + 2\frac{P^2_{(r/2)}}{1-r}\right]$$
or (ii) $C_{r(u)}$ from $g(x)_{user}$:-
$$
C_{r(u)} = \frac{\pi}{2} \left( \binom{r}{r/2} * \frac{1}{4^{r/2}} * \frac{1}{r-1} \right)^2.
$$
Proof by Induction - Pending
One possible line of investigation is to consider obtaining formulae for the change in coefficient value from $C_r$ to $C_{r+1}$ for any $r$. If (i) the formula is the same for $g(x)$ and $h(x)$; and (ii) the value of $C_r$ for some initial value of $r$ is the same for $g(x)$ and $h(x)$; then this would constitute a proof that $g(x)$ = $h(x)$.
Induction for $h(x)$
We have
$$
C_{r(h)} = Q_r
+
\sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2}
$$
and so
$$
C_{r+1(h)} = Q_{r+1}
+
\sum_{n=1}^{r+1-1} Q_n * (-1)^{r-n+1} \binom{2n-2 + r-n +1}{2n-2}
$$
and so the change in $C_{r(h)}$ from $r$ to $r+1$ is given by:-
$$\text{delta}~C_{r(h)} =
Q_{r+1} -Q_{r} $$
$$+ \sum_{n=1}^{r+1-1} Q_n * (-1)^{r-n+1} \binom{2n-2 + r-n +1}{2n-2} $$
$$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2}
$$
$$=Q_{r+1} -Q_{r} + Q_r * (-1)^{r-r+1} \binom{2r-2 + r-r +1}{2r-2} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} \binom{2n-2 + r-n +1}{2n-2} $$
$$-- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} \binom{2n-2 + r-n }{2n-2} $$
$$=Q_{r+1} -Q_{r} + Q_r * (-1) \binom{2r-2 +1}{2r-2} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1} \left( \binom{2n-2 + r-n +1}{2n-2}
+ \binom{2n-2 + r-n }{2n-2} \right) .$$
Using the Addition/Induction Rule and the Absorption/Extraction Rule $\left(\binom{N}{K}=\frac{N}{K}\binom{N-1}{K-1}\right)$ we can show that $\binom{N}{K}+\binom{N+1}{K} = \left(2-\frac{K}{N+1}\right)\binom{N+1}{K}$, hence
$$\text{delta}~C_{r(h)} = Q_{r+1} -Q_{r} - Q_r * \binom{2r-2 +1}{2r-2} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1}
*\left( 2 - \frac{2n-2}{2n-2 + r-n +1}\right)
*\binom{2n-2 + r-n +1}{2n-2} $$
$$\text{delta}~C_{r(h)} = Q_{r+1} -Q_{r} - Q_r * \binom{2r-2 +1}{2r-2} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1}
*\left( 2 - \frac{2n-2}{n-1 + r }\right)
*\binom{n-1 + r}{2n-2} $$
$$\text{delta}~C_{r(h)} = Q_{r+1} -Q_{r} - Q_r * \binom{2r-2 +1}{2r-2} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1}
*\left( \frac{ 2n-2 + 2r + 2n-2}{n-1 + r }\right)
*\binom{n-1 + r}{2n-2}$$
$$\text{delta}~C_{r(h)} = Q_{r+1} -Q_{r} - Q_r * \binom{2r-2 +1}{2r-2} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n+1}
*\left( \frac{2r}{n-1 + r }\right)
*\binom{n-1 + r}{2n-2} $$
Unfortunately this does not look very tractable :-(.
We might obtain more clarity if we treated $g(x)$ as two streams - even and odd powers of $x$. We would expect that $dC_{r(h,odd)}$ will equal $0$ which will allow us to ignore odd powers of $x$ (given the calculated seed that $C_{r(h)}x^1=0$). And we might get a simpler expression of $dC_{r(h,even)}$ which is easier to compare with that for $dC_{r(g)}$.
We have
$$
C_{r(h)} = Q_r
+
\sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2}
$$
and so
$$
C_{r+2(h)} = Q_{r+2}
+
\sum_{n=1}^{r+2-1} Q_n * (-1)^{r-n+2} \binom{2n-2 + r-n +2}{2n-2}
$$
and so the change in $C_{r(h)}$ from $r$ to $r+1$ is given by:-
$$\text{d2_}~C_{r(h)} =
Q_{r+2} -Q_{r} $$
$$+ \sum_{n=1}^{r+2-1} Q_n * (-1)^{r-n+2} \binom{2n-2 + r-n +2}{2n-2} $$
$$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2}
$$
Now we use the fact that $(-1)^{r-n+2}=(-1)^{r-n}$ to replace the former by the latter...
$$\text{d2_}~C_{r(h)} =
Q_{r+2} -Q_{r} $$
$$+ \sum_{n=1}^{r+2-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n +2}{2n-2} $$
$$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2}
$$
Now we peel off the last sum-term of the first sum so that the two sums both have final index $n=r-1$...
I need to think about the difference in the top index value between odd and even streams.
$$\text{d2_}~C_{r(h)} =
Q_{r+2} -Q_{r} $$
$$+ Q_{r+2-1} * (-1)^{r-({r+2-1})} \binom{2({r+2-1})-2 + r-({r+2-1}) +2}{2({r+2-1})-2} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n +2}{2n-2} $$
$$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2}
$$
Rationalize the first binomial term...
$$\text{d2_}~C_{r(h)} =
Q_{r+2} -Q_{r} $$
$$+ Q_{r+ 1} * (-1)^{-1} \binom{ 2r+1}{2r} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n +2}{2n-2} $$
$$- \sum_{n=1}^{r-1} Q_n * (-1)^{r-n} \binom{2n-2 + r-n }{2n-2}$$
Now merge two sums into one...
$$\text{d2_}~C_{r(h)} =
Q_{r+2} -Q_{r} $$
$$+ Q_{r+ 1} * (-1) \binom{ 2r+1}{2r} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n}
\left[
\binom{2n-2 + r-n +2}{2n-2}
- \binom{2n-2 + r-n }{2n-2}
\right]$$
It is demonstrable that $\binom{N+2}{K}-\binom{N}{K}=((N+2)(N+1)-1)\binom{N}{K}$ so...
$$\left[
\binom{2n-2 + r-n +2}{2n-2}
- \binom{2n-2 + r-n }{2n-2}
\right]$$
$$=\left[
\left((2n-2 + r-n +2)(2n-2 + r-n +1)-1
\right)
* \binom{2n-2 + r-n }{2n-2}
\right]$$
then
$$=\left[
\left((n + r )(n + r- 1)-1
\right)
* \binom{2n-2 + r-n }{2n-2}
\right]$$
then again
$$=\left[
\left(n^2 + r^2 +2nr -n - r-1
\right)
* \binom{2n-2 + r-n }{2n-2}
\right]$$
So...
$$\text{d2_}~C_{r(h)} =
Q_{r+2} -Q_{r} + Q_{r+ 1} * (-1) \binom{ 2r+1}{2r} $$
$$+ \sum_{n=1}^{r-1} Q_n * (-1)^{r-n}
*
\left(n^2 + r^2 +2nr -n - r-1
\right)
* \binom{2n-2 + r-n }{2n-2}
$$
Unfortunately this does not look very tractable either :-(.
Using the Double Factorial formulae for $K()$ and $E()$
Given the failure so far of the previous methods it might be worthwhile exploring the use of the alternative formula for $K()$ and $E()$, i.e. those which use Double Factorial terms (see wikipedia page for Elliptic Integrals).
( need to check ALL the above !!! )
see Pascal's Extended Triangle by Ken Ward
to be continued...
