So, the standard parameterization of the ellipse$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$is given by$$r(t) = (a \cos t , b \sin t) \quad t \in [0,2\pi)$$The arc-length parameterization is given by$$s(t) = \int_{0}^t \|r'(\tau)\| \, d\tau$$Then$$\begin{align*}r'(t) &= (-a \sin t, b \cos t) \\\|r'\| &= \sqrt{ a^2 \sin^2 t+ b^2 \cos^2 t}\end{align*}$$so$$s(t) = \int_{0}^t \sqrt{a^2 \sin^2 \tau +b^2 \cos^2 \tau}\, d\tau$$From the Pythagorean identity,$$\cos^2 \tau + \sin^2 \tau = 1\implies\cos^2 \tau = 1 - \sin^2 \tau$$so$$s(t) = \int_{0}^t \sqrt{a^2 \sin^2 \tau +b^2 - b^2 \sin^2 \tau}\, d\tau= \int_{0}^t \sqrt{ b^2 - (b^2 - a^2) \sin^2 \tau}\, d\tau$$With a factorization, and assuming $a,b \ge 0$ without loss of generality,$$s(t) = b \int_{0}^t \sqrt{ 1 - \left( 1 - \frac{a^2}{b^2} \right) \sin^2 \tau}\, d\tau = b \cdot E \left( t \,\middle| \, \sqrt{1- \frac{a^2}{b^2} } \right)$$This final function is a special, nonelementary function known as an (incomplete) integral of the second kind (Wikipedia, Mathworld). Suffice to say, then, there won't be a meaningful closed form beyond very, very special cases.
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