For a numerical representation of the inverse in terms of the angle $\varphi$
where $E(\varphi \, | \,k) = \int_0^\varphi \sqrt{1-k^2\sin^2\theta} \, \mathrm{d}\theta $ is the
elliptic integral for the second kind, one could expand $E(\varphi |k)$ in a power
series around $\varphi=0$,

$ E(k |\varphi) := \varphi
-1/6 k^2\varphi^3
+1/5(1/6k^2-1/8k^2)\varphi^5
+1/7(-1/45 k^2+1/12k^4-1/16k^6)\varphi^7
+1/9(1/630k^2-1/40k^4+1/16k^6-5/128k^8)\varphi^9 \cdots$

let $k^2=m$,The expansion coefficients in front of the order $\varphi^{n+1} ,(n=2,4,6,...)$ are
$\sum_{r=2}^{n}U(r,n)F(r,m)/r!]/(n+1)!$, $r$ is even integer and
$U(r,n) = (-1)^{(r+n)/2}/2^{r-n} \sum_{l=0}^{r}(-1)^l (l-r/2)^n \binom{r}{l}]$ and $F(r,m) = -m^{n/2}[(r-1)!!]^2/(r-1)$ ,
with $(r-1)!! = 1\times 3\times 5\times 7\times \cdots\times(r-1) $

Then invert this as outlined in **chapt 3.6.25 of the book edited by M Abramowitz
and I Stegun** you get finally your inverse :

$\varphi := E(\varphi,m)
+1/6mE(\varphi,m)^3
+1/120m(13m-4)E(\varphi,m)^5
+1/5040m(493m^2-284m+16)E(phi,m)^7
+1/362880m(37369m^3-31224m^2+4944m-64)E(\varphi,m)^9 ...$

you may use Mathematica code up to $\varphi=12$ you may try the following code for numerical verification:

```
In[2]:= InverseSeries[Series[EllipticE[z, m], {z, 0, 12}]]
Out[2]= z + (m*z^3)/6 + (1/120)*(-4*m + 13*m^2)*z^5 + ((16*m - 284*m^2 +
493*m^3)*z^7)/5040 + ((-64*m + 4944*m^2 - 31224*m^3 +
37369*m^4)*z^9)/362880 + ((256*m - 81088*m^2 + 1406832*m^3 - 5165224*m^4 +
4732249*m^5)*z^11)/39916800 + O[z]^13
```