Important Edit Made
Mathematica can perform the 3.31 integral, if Assumptions is changed from { a > b > c >= u} to {a > b > c > u}.
s = Integrate[1/Sqrt[(a - x) (b - x) (c - x)], {x, -Infinity, u},
Assumptions -> {a > b > c > u}]
(* (2 EllipticF[ArcSin[Sqrt[(a - b)/(a - u)]], (a - c)/(a - b)])/Sqrt[a - b] *)
To compare this with the expression as given in Gradshteyn and Ryzhik (7th edition, 2007), it is important to realize that this compendium defines the second argument of Elliptic Integral F differently than from that in Mathematica. Comparing the second example under "Possible Issues" in the EllipticF documentation with 11.112 #2 of Gradshteyn and Ryzhik indicates that p^2 (see question) should be used as the second argument of the Gradshteyn and Ryzhik expression when employing the Mathematica representation of EllipticF; i. e.,
gr = 2 EllipticF[ArcSin[Sqrt[(a - c)/(a - u)]], (a - b)/(a - c)]/Sqrt[a - c]
which differs from s only by the interchange of b and c. But, it is obvious from the integrand of the integral above that the relative order of {a, b, c} is irrelevant. As verification, I have evaluated s and gr numerically for several parameters, for instance,
vals = Thread[{a, b, c, u} -> Reverse@Sort@RandomReal[{-5, 5}, 4]]
(* {a -> 3.47807, b -> 2.65797, c -> -1.04855, u -> -1.17253} *)
Chop[gr /. vals]
(* 1.38108 *)
Chop[s /. vals]
(* 1.38108 *)
and in each case they are the same and agree with the original integral evaluated numerically.
NIntegrate[1/Sqrt[(a - x) (b - x) (c - x)] /. vals, {x, -Infinity, u /. vals},
Method -> {Automatic, "SymbolicProcessing" -> False}]
(* 1.38108 *)
Therefore, the apparent discrepancy between the result in Gradshteyn and Ryzhik (7th edition, 2007) and the corresponding Mathematica result is due merely to differences in notation.