Integral of the Square of the Elliptic Integral Someone must know a good technique for
$$
\int E^{2}(x)dx
$$
Where $E$ is the complete elliptic integral of the second kind:
$$
E\left(k\right)=\int_{0}^{\frac{\pi}{2}}d\theta\sqrt{1-k^{2}\sin^{2}\left(\theta\right)}
$$
This is essentially the working for another integral I posted 
(Simple Integral Involving the Square of the Elliptic Integral) though this is valuable in its own right.
 A: We have that the function:
$$ B(\lambda)=\sum_{n=0}^{+\infty}\left(\frac{1}{(2n-1)4^n}\binom{2n}{n}\right)^2\lambda^{2n} = \frac{1}{\pi}\int_{0}^{\pi}\sqrt{1+\lambda^2+2\lambda\cos(2\theta)}\,d\theta $$
satisfies the relation:
$$B(\lambda) = 2\frac{1+\lambda}{\pi}\cdot E\left(\frac{4\lambda}{(1+\lambda)^2}\right)\tag{1}$$
and the differential equation:
$$ B = \left(\lambda+\frac{1}{\lambda}\right)\frac{dB}{d\lambda}+\left(1-\lambda^2\right)\frac{d^2 B}{d\lambda^2}. \tag{2}$$
Since all the coefficients of the Taylor series of $B(\lambda)$ are positive, $B(\lambda)$ is a convex function, as well as its derivatives. Now $(2)$ gives:
$$\frac{B'}{B}(\lambda)\leq \frac{\lambda}{\lambda^2+1},\tag{3}$$
so: $$ B(\lambda)\leq\sqrt{1+\lambda^2}.\tag{4}$$
Moreover,
$$\frac{B'}{B}(\lambda)= \frac{\lambda}{(\lambda^2+1)+\frac{1}{1+\lambda\cdot\frac{B''}{B'}(\lambda)}},\tag{5}$$
hence:
$$ B(\lambda)\geq\sqrt{1+\frac{\lambda^2}{2}}.\tag{6}$$
Rearranging $(4)$ and $(6)$ through $(1)$ we get:
 $$\frac{\pi}{4}\,\sqrt{\frac{3}{2}\,(2-m)+\sqrt{1-m}}\leq  E(m) \leq \frac{\pi}{2}\,\sqrt{1-\frac{m}{2}},\tag{7}$$
hence $E^2(m)$ can be very well approximated with continued-fraction-arguments that comes from $(2)$.
A: If you're looking for good numerical approximations, then for example
$$\eqalign{- 0.000038839155&+ (  2.472698342+ ( - 0.1185303776+ (\cr 
& 0.599575302+ ( - 4.237027543+ (  9.643963778+ (\cr -
& 12.22813840+ (  8.085033824- 2.189508102\,x ) x ) x
 ) x ) x ) x ) x ) x}
$$
is an optimal degree $8$ polynomial approximation on $[0,1]$, with
maximum error approximately $0.000038949$.
