I found myself with the following integral

$$ \int_{b_1}^{b_2} \sqrt{\frac{(b-b_1)(b_2-b)(b_3-b)}{(b_4-b)}} \ db $$

with $ b_1 < b_2 < b_3 < b_4 $. I know that

$$ \int_{b_1}^{b_2} \frac{db}{\sqrt{(b-b_1)(b_2-b)(b_3-b)(b_4-b)}} $$

is equal to

$$ \frac{2}{(b_4-b_2)(b_3-b_1)} K(k) $$

where $K(k)$ is the complete elliptic integral of first kind, so I suspect that this integral is somehow reducible to a linear combination of elliptic integrals, but I can't find the right way.