How do I evaluate the integral of the square root of a quartic equation? I'm currently trying to evaluate the integral
$$\int^1_0 \text{d}t\sqrt{(1-t^2)(1-k^2t^2)}$$
where $k\in(0,1)$. Is it the case that this can be expressed in terms of elliptic integrals? I'm struggling to see how this can be done as the elliptic integral that looks closest to this expression is 
$$\int\frac{\text{d}t}{\sqrt{(1-t^2)(1-k^2t^2)}}$$
and I'm not sure how to relate the two.
 A: Rewriting the integrand and integrating by parts we find
\begin{align}f(k) &\equiv \int \limits_0^1 \sqrt{(1-x^2)(1-k^2 x^2)} \, \mathrm{d} x = \int \limits_0^1 \frac{\sqrt{1-x^2}}{x} x \sqrt{1-k^2 x^2} \, \mathrm{d} x \\
&= \left[\frac{\sqrt{1-x^2}}{x} \frac{1-(1-k^2 x^2)^{3/2}}{3 k^2}\right]_{x=0}^{x=1} - \int \limits_0^1 \left(-\frac{1}{x^2 \sqrt{1-x^2}}\right) \frac{1-(1-k^2 x^2)^{3/2}}{3 k^2} \, \mathrm{d} x \\
&= \frac{1}{3k^2} \int \limits_0^1 \frac{1-(1-k^2 x^2)^{3/2}}{x^2 \sqrt{1-x^2}} \, \mathrm{d} x \, .
\end{align}
Mathematica can do the remaining integral. In order to get the correct result by hand, we rewrite the integrand again (in a less than obvious way, admittedly):
$$ f(k) = \frac{1}{3k^2} \int \limits_0^1 \left[\frac{(1+k^2)\sqrt{1-k^2 x^2}}{\sqrt{1-x^2}} - \frac{(1-k^2)}{\sqrt{(1-x^2)(1-k^2 x^2)}} - \frac{1 - k^2 x^4 -\sqrt{1-k^2 x^2}}{x^2 \sqrt{(1-x^2)(1-k^2 x^2)}} \right]\mathrm{d} x \, . $$
The first to terms lead to elliptic integrals and the third one has a surprisingly simple antiderivative, so we obtain
\begin{align}
f(k) &= \frac{1}{3k^2}\left[(1+k^2)\operatorname{E}(k) - (1-k^2) \operatorname{K}(k) - \left[\frac{\sqrt{1-x^2}\left(1-\sqrt{1-k^2x^2}\right)}{x}\right]_{x=0}^{x=1}\right] \\
&= \frac{(1+k^2)\operatorname{E}(k) - (1-k^2) \operatorname{K}(k)}{3k^2} \, .
\end{align}
