Antiderivative of $\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}$

Without loss of generality let $a>b>c>0$. Substitute $u=x^2$ to get $$I=\int_p^q\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}\,dx=\frac12\int_{p^2}^{q^2}\sqrt{\frac{(u-a^2)(u-b^2)(u-c^2)}u}\,du$$ This is an elliptic integral and its solution depends on where the bounds are. As an example, let $c=p\le q\le b$ and as a notational convenience let an uppercase variable denote the corresponding lowercase variable squared. Then Byrd and Friedman 254.37 gives $\newcommand{sn}{\operatorname{sn}}\newcommand{cn}{\operatorname{cn}}\newcommand{dn}{\operatorname{dn}}$ $$I=\frac12\int_P^Q\sqrt{\frac{(u-A)(u-B)(u-C)}u}\,du$$ $$=\sqrt{\frac{A-C}B}(B-C)^2\frac CB\int_0^{u_1}\frac{\sn^2u\cn^2u\dn^2u}{(1-n\sn^2u)^3}\,du$$ where $n=\frac{B-C}B$, the hidden parameter $m$ in the Jacobian elliptic functions is $\frac{(B-C)A}{(A-C)B}$ and $u_1=F(\varphi,m)$ where $\sin^2\varphi=\sn^2u_1=\frac{(Q-C)B}{(B-C)Q}$. In turn, the last integral is given by B&F 362.24 as $$\frac1{n^3}\left(-mF(\varphi,m)+(3m-nm-n)\Pi(n,\varphi,m)+(2nm+2n-3m-n^2)V_2+(n-1)(n-m)V_3\right)$$ where $V_2$ and $V_3$ are in turn given by B&F 336.00 to .03 as $$V_0=F(\varphi,m)\qquad V_1=\Pi(n,\varphi,m)$$ $$V_2=\frac1{2(n-1)(m-n)}\left(nE(\varphi,m)+(m-n)F(\varphi,m)+(2nm+2n-n^2-3m)\Pi(n,\varphi,m)-\frac{n^2\sn u_1\cn u_1\dn u_1}{1-n\sn^2u_1}\right)$$ $$V_{j+3}=\frac1{2(j+2)(1-n)(m-n)}\left((2j+1)mV_j + 2(j+1)(nm+n-3m)V_{j+1} + (2j+3)(n^2-2nm-2n+3m)V_{j+2} + \frac{n^2\sn u_1\cn u_1\dn u_1}{(1-n\sn^2u_1)^{j+2}}\right)$$ As you can see, the general solution can get extremely complicated very quickly.