The arc length of a graph of a function $f(x)$ is
$$\mathcal{l} = \int_a^b \sqrt {1 + f'(x)^2} dx$$
If you choose $f$ to be a parametrization of the ellipse with $a=1, b=\frac{1}{2}$, i.e. $f(x) = \frac{1}{2} \sqrt{1-x^2}$, the arc length of the graph of this function is
$$\mathcal{l} = \int_0^1 \sqrt\frac{1-\frac{3x^2}{4} }{1-x^2}dx$$
If you enter int_0^1 sqrt((1-(3x^2)/4)/(1-x^2))dx at wolframalpha.com you will get the result:
What I would have expected here as the argument of $E$ (the complete elliptic integral of the second kind) was the eccentricity $e = \sqrt{1-(b/a)^2} = \sqrt{\frac{3}{4}}$ but it's $e^2$.
Is this a Wolframalpha bug or did I make or understand something wrong?
Wolfram has an unconventional definition of the elliptic integrals. If you look at the manual you see that the argument of
EllipticEis $m$ which is related to the usual argument $k$ of $E(k)$ via $m=k^2$.