# Approximations of the incomplete elliptic integral of the second kind

For a calculation I am working on I need to determine the arc length $l$ of a part of an ellipse in terms of the major axis $2a$, the minor axis $2b$ and the angle $\phi$. I know that this is a classical problem which results in an incomplete elliptic integral of the second kind: $$\tag{1} l=a E\left(\phi \left|\sqrt{1-\frac{b^2}{a^2}}\right.\right)$$

What I would like to know is whether an algebraic approximation to this equation is known which is applicable with good accuracy in the range $1 < \frac{a}{b} < 3$? I would want to use this approximation to determine $l=f(\phi)$.

Of course, I could do a Taylor series around $a=b$ but that is fairly inaccurate unless you use a significant number of terms. What I would love to have is an equation akin to the beautiful approximation due to Ramanujan (1914) for the complete elliptic integral of the second kind: $$\tag{2} \frac{1}{4} \pi (a+b) \left(\frac{3 (a-b)^2}{(a+b)^2 \left(\sqrt{4-\frac{3 (a-b)^2}{(a+b)^2}}+10\right)}+1\right)$$

-
Have you already seen this? –  Ｊ. Ｍ. Apr 3 '13 at 15:39
I hadn't seen it yet, but after taking a look it doesn't get much clearer for me. I can see that Eq 21 in the link is an approximation, but I am not sure how I would go about applying it? –  Michiel Apr 3 '13 at 15:53
@J.M. the use of pade approximants seems to me to be a fairly involved approach in which I have to determine all the prefactors, which I would probably need a computer for so then I don't see the point in using the approximation over the 'real' incomplete elliptic integral –  Michiel Apr 4 '13 at 5:23
Well, what did you need the "approximation" for anyway? If it's for manual calculations, I do believe that in this case, the simple ones would not be very accurate, and the accurate ones would not be very simple. The situation for the other two sorts of elliptic integrals is actually less hairy. –  Ｊ. Ｍ. Apr 4 '13 at 5:25
@J.M. I need the approximation to determine $l=f(\phi)$ such that I can rewrite it in a form $\phi=f(l)$. With the exact elliptic integral that doesn't seem possible. –  Michiel Apr 4 '13 at 5:31
show 3 more comments