The problem of finding an arc length of an ellipse is the origin of the name of the elliptic integrals. The equation of an ellipse with semi-major axis, $a$, and semi-minor axis, $b$, may be written in parametric form as \begin{align*} x &= a \sin\phi\\ y &= b \cos\phi \end{align*} The element of length along the ellipse's perimeter, \begin{align*} \mathrm{d}s &= \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2} = \sqrt{a^2\cos^2\phi + b^2\sin^2\phi}\;\mathrm{d}\phi\\ & = a \sqrt{1 - e^2\sin^2\phi}\;\mathrm{d}\phi, \end{align*} where $e = \sqrt{1-b^2/a^2}$ is the eccentricity. The arc length may therefore be written in terms of incomplete elliptic integrals of the second kind: $$ \int \;\mathrm{d}s = a\int_{\phi_1}^{\phi_2} \sqrt{1 - e^2\sin^2\phi}\;\mathrm{d}\phi = a[E(e; \phi_2) - E(e; \phi_1)]. $$
Earth's orbit is an ellipse with semi-major axis 149,598,261 km and eccentricity 0.01671123. We will find the distance travelled by the Earth in one orbit, and compare it with that obtained assuming a circular orbit of radius $1\;\mathrm{AU} \equiv 149597870.7\;\mathrm{km}$.
The perimeter of an ellipse may be written using the above expression with $\phi_1 = 0, \phi_2 = 2\pi$:
$$
P = a[E(e, 2\pi) - E(e, 0)] = 4aE(e),
$$
since the entire perimeter is four times the quarter-perimeters which may be written in terms of the complete elliptic integral of the second kind. Noting that SciPy's ellipe function actually takes $m=e^2$ as its argument, we have:
In [x]: import numpy as np
In [x]: from scipy.special import ellipe
In [x]: a, e = 149598261, 0.01671123 # semi-major axis (km), eccentricity
In [x]: pe = 4 * a * ellipe(e*e)
In [x]: print(pe)
939887967.974 # "exact" answer
In [x]: AU = 149597870.7 # mean orbit radius, km
In [x]: pc = 2 * np.pi * AU
In [x]: print(pc)
939951143.1675915 # assuming circular orbit
In [x]: (pc - pe) / pe * 100
0.0067215663638305143
That is, the percentage error in the perimeter in treating the orbit as circular is about 0.0067%.