<s>[INST] What is the fastest way to estimate the Arc Length of an Ellipse? To estimate the circumference of an ellipse there are some good approximations.
$a$ is the semi-major radius and $b$ is the semi-minor radius.
$$L \approx \pi(a+b) \frac{(64-3d^4)}{(64-16d^2 )},\quad \text{where}\;d =  \frac{(a - b)}{(a+b)}$$
Are there any similar formulas to approximate the arc length of an ellipse from $\theta_1$ to $\theta_2$? 
If not what are some computationally fast ways to approximate the arc length to within about $1\%$ to $0.1\%$ of $a$?
 [/INST] I'll assume $\theta_1$ and $\theta_2$ refer to the parametrization
$$ \eqalign{x &= a \cos(\theta)\cr y &= b \sin(\theta)}$$ 
of the ellipse.
Thus the arc length in question is
$$ L = \int_{\theta_1}^{\theta_2} \sqrt{a^2 \sin^2(\theta) + b^2 \cos^2(\theta)}\; d\theta $$
We want a good approximation of the integrand that is easy to integrate.
It may be best to look at two cases, depending on which of the terms inside the square root is larger.
For $a |\sin(\theta)| \ge b |\cos(\theta)|$, we take
$$\sqrt{a^2 \sin^2(\theta) + b^2 \cos^2(\theta)} = a |\sin(\theta)| \sqrt{1 + \frac{b^2}{a^2} \cot^2(\theta)}$$
and look for a good approximation of $\sqrt{1+t^2}$ for $0 \le t \le 1$.
The best polynomial approximation of degree $3$ for this is approximately
$$ 1.000127929-0.00619431946 \;t+.5478616944\; t^2-.1274538129\; t^3$$
with maximum absolute error $\approx .0001280863448$.  Write these coefficients as $c_0, \ldots, c_3$.  Thus on the part of the interval where $a |\sin(\theta)| \ge b |\cos(\theta)|$, we can integrate
$$ \pm a  \sin(\theta) \left(c_0 + c_1 \frac{b}{a} \cot(\theta) + c_2 \frac{b^2}{a^2} \cot^2(\theta) + \frac{b^3}{a^3} \cot^3(\theta)\right)$$
($+$ on an interval where $\sin(\theta) \ge 0$, $-$ where $\sin(\theta)<0$).
An antiderivative is
$$ \pm\left( - a c_0 \cos(\theta) + b c_1 \sin(\theta) + \frac{b^2}{a} c_2 \left(\cos(\theta)+\ln(\csc(\theta)-\cot(\theta))\right) - \frac{b^3}{a^2} (\csc(\theta)+\sin(\theta))\right)$$
Similarly, for $a |\sin(\theta)| \le b |\cos(\theta)|$ take
$$\sqrt{a^2 \sin^2(\theta) + b^2 \cos^2(\theta)} = b |\cos(\theta)| \sqrt{1 + \frac{a^2}{b^2} \tan^2(\theta)}$$
and integrate
$$\pm b \cos(\theta) \left(c_0 + c_1 \frac{a}{b} \tan(\theta) + c_2 \frac{a^2}{b^2} \tan^2(\theta) + c_3 \frac{a^3}{b^3} \tan^3(\theta) \right)$$
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