In Gradshteyn and Ryzhik, (specifically starting with the section 3.13) there are several results involving integrals of polynomials inside square root. These are given in terms of combinations of elliptic integrals. See for instance:
where $F[\alpha, p]$ is the elliptic integral of first kind. I tried to reproduce the first result above in Mathematica (version 12) but failed. I would appreciate if anyone could point out what I am doing wrong. My first attempt is
Integrate[1/Sqrt[(a - x) (b - x) (c - x)], {x, -Infinity, u}, Assumptions -> { a > b > c >= u}]
which returned no result:
Then I tried without integration limits, and take the limits after
Integrate[1/Sqrt[(a - x) (b - x) (c - x)], x, Assumptions -> { a > b > c}]
giving:
taking now the upper limit and simplifying
Simplify[Limit[(2 (a - x)^(3/2) Sqrt[(b - x)/(a - x)] Sqrt[(c - x)/(a - x)]
EllipticF[ArcSin[Sqrt[a - b]/Sqrt[a - x]], (a - c)/(a - b)])/(Sqrt[a - b] Sqrt[(a - x) (-b + x) (-c + x)]), x -> u], Assumptions -> { a > b > c >= u}]
giving:
whereas the integral vanishes when $x\rightarrow -\infty$. Clearly, the above result given by Mathematica differs from the Gradshteyn and Ryzhik's. Two results match if the substitution: $b \rightarrow c$, $c \rightarrow b$ is made but this would then be at odds with the condition: $a > b > c$.
Assumptions -> {a > b > c > u > 0}
already the definite integral works with the result(2 EllipticF[ArcSin[Sqrt[(a - b)/(a - u)]], (a - c)/(a - b)])/Sqrt[a - b]
. However, this result is, indeed, not the one you from the Handbook. I did not manage to convert one result into another. $\endgroup$Assumptions -> {a > b > c > u}
is sufficient. $\endgroup$