# Evaluating indefinite integrals from tables of definite integrals

I'll give my motivating problem, and then ask my general question.

So, I'm attempting to integrate the following indefinite integral:

$$\int\frac{\mathrm{d}u}{\sqrt{au^3+bu^2+cu+d}}$$ Now, I learned from poking around here that this is most likely an elliptic integral; sure enough, Byrd and Friedman have the integral $$\int^y_\alpha\frac{\mathrm{d}t}{\sqrt{(t-a)(t-b)(t-c)}}=gF(\phi,k)$$

where it gives values for $$g$$, $$\phi$$, and $$k$$. However, $$k$$ varies depending on the value of $$y$$ (it is different for $$y>c>b>a$$ than for $$c\geq c>b>a$$, for example). I'm not exactly sure what values $$y$$ will take nor what range it is in (this integral is from physics), so I can't use their recommended strategy of splitting up the integral given in the introduction.

So, here's my general question: in cases like these, how does one construct the general indefinite integral from tables of definite integrals?

• The indefinite integral is the function of $y$? Commented Sep 27, 2020 at 2:50

Assuming $$a\neq0$$, write$$I=\int\frac{du}{\sqrt{au^3+bu^2+cu+d}}=\frac 1 {\sqrt a}\int\frac{du}{\sqrt{(u-r_1)(u-r_2)(u-r_3)}}$$where $$r_1,r_2,r_3$$ are the roots of the cubic.
Then, without any assumptions we have $$I=-\frac 2 {\sqrt a\,\sqrt{r_2-r_1}}\color{blue}{\frac{ (u-r_1)^{3/2} \sqrt{\frac{u-r_2}{u-r_1}} \sqrt{\frac{u-r_3}{u-r_1}} }{ \sqrt{(u-r_1) (u-r_2) (u-r_3)}}}F\left(\sin ^{-1}\left(\frac{\sqrt{r_2-r_1}}{\sqrt{u-r_1}}\right)|\frac{r_1-r_3}{r_1-r_2}\right)$$ which could simplify a lot depending on the bounds of integration (depending where they locate with respect to the roots. The factor is blue is not necessay equal to $$1$$; this depends on $$\alpha$$ and $$y$$.