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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.

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.

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] *)

which differs from the Gradshteyn and Ryzhik result, by the interchange of b and c and the absence of Sqrt in the second argument of EllipticF.

gr = 2 EllipticF[ArcSin[Sqrt[(a - c)/(a - u)]], Sqrt[(a - b)/(a - c)]]/Sqrt[a - c]

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} *)

gr /. vals
(* 1.48528 *)

Chop[s /. vals]
(* 1.38108 *)

and in each case they differ. To see which is correct, evaluate the integral numerically.

NIntegrate[1/Sqrt[(a - x) (b - x) (c - x)] /. vals, {x, -Infinity, u /. vals}, 
    Method -> {Automatic, "SymbolicProcessing" -> False}]
(* 1.38108 *)

It appears, therefore, that Gradshteyn and Ryzhik (7th edition, 2007) is incorrect.

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.

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] *)

which differs from the Gradshteyn and Ryzhik result, by the interchange of b and c and the absence of Sqrt in the second argument of EllipticF.

gr = 2 EllipticF[ArcSin[Sqrt[(a - c)/(a - u)]], Sqrt[(a - b)/(a - c)]]/Sqrt[a - c]

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} *)

gr /. vals
(* 1.48528 *)

Chop[s /. vals]
(* 1.38108 *)

To see which is correct, evaluate the integral numerically.

NIntegrate[1/Sqrt[(a - x) (b - x) (c - x)] /. vals, {x, -Infinity, u /. vals}, 
    Method -> {Automatic, "SymbolicProcessing" -> False}]
(* 1.38108 *)

It appears, therefore, that Gradshteyn and Ryzhik (4th edition, 1965) is incorrect.

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] *)

which differs from the Gradshteyn and Ryzhik result, by the interchange of b and c and the absence of Sqrt in the second argument of EllipticF.

gr = 2 EllipticF[ArcSin[Sqrt[(a - c)/(a - u)]], Sqrt[(a - b)/(a - c)]]/Sqrt[a - c]

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} *)

gr /. vals
(* 1.48528 *)

Chop[s /. vals]
(* 1.38108 *)

and in each case they differ. To see which is correct, evaluate the integral numerically.

NIntegrate[1/Sqrt[(a - x) (b - x) (c - x)] /. vals, {x, -Infinity, u /. vals}, 
    Method -> {Automatic, "SymbolicProcessing" -> False}]
(* 1.38108 *)

It appears, therefore, that Gradshteyn and Ryzhik (7th edition, 2007) is incorrect.