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.



Definite integral of the square root of a polynomial ratio
The expression obtained by @Robert can be simplified with the Imaginary-Argument Transformation from [DLMF 19.7.7](https://dlmf.nist.gov/19.7). Then, the limit $t\to 1^-$ can be performed. The result can be further simplified and written in many ways due to the large number of elliptic functions relations. The simplest expression I found reads
\begin{align}
&\int\_0^1 \mathrm db \,\sqrt{\frac{b(1-b)(b\_3-b)}{b\_4-b}}\\
&=\frac{1}{4 \sqrt{b\_4}}\left[
 b\_4 (3 b\_4-b\_3-1) E(k)
- b\_3 (b\_4+b\_3-1) K(k)\\
\quad+ \frac{b\_3}{\sqrt{b\_3-1}}
\left[(b\_3-1)^2 + 2(b\_3+1)b\_4 - 3 b\_4^2\right] \Pi
 \left(\tfrac{1}{1-b\_3},k\right)
\right]\tag{1}
\end{align}
with parameter and elliptic modulus
$$
m=k^2=\frac{b\_4-b\_3}{\left(1-b\_3\right) b\_4}.\tag{2}
$$
One can rewrite this expression with the Imaginary-Modulus Transformation [DLMF 19.7.5](https://dlmf.nist.gov/19.7) to get $0<\tilde m<1$ and real $0<\tilde k<1$, i.e., ${\tilde k}{}^2=\frac{b\_4-b\_3}{b\_3(b\_4-1)}$, however the expressions become more complicated. For $b\_3=2, b\_4=3$ this last formulation gives the representation obtained in the comment by @Robert.


