3.2.11 \(\int \frac {1}{(-3 a-b x^2) \sqrt [3]{-a+b x^2}} \, dx\)

Optimal. Leaf size=252 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt {a} \left (\sqrt [3]{-a}-\sqrt [3]{2} \sqrt [3]{b x^2-a}\right )}{\sqrt [3]{-a} \sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [3]{-a} \sqrt {b} x}{\sqrt {a} \left (\sqrt [3]{2} \sqrt [3]{b x^2-a}+\sqrt [3]{-a}\right )}\right )}{2\ 2^{2/3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt {a}}{\sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{6\ 2^{2/3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}} \]

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Rubi [A]  time = 0.07, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {393} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt {a} \left (\sqrt [3]{-a}-\sqrt [3]{2} \sqrt [3]{b x^2-a}\right )}{\sqrt [3]{-a} \sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [3]{-a} \sqrt {b} x}{\sqrt {a} \left (\sqrt [3]{2} \sqrt [3]{b x^2-a}+\sqrt [3]{-a}\right )}\right )}{2\ 2^{2/3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt {a}}{\sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{6\ 2^{2/3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-3*a - b*x^2)*(-a + b*x^2)^(1/3)),x]

[Out]

-ArcTan[(Sqrt[3]*Sqrt[a])/(Sqrt[b]*x)]/(2*2^(2/3)*Sqrt[3]*(-a)^(1/3)*Sqrt[a]*Sqrt[b]) - ArcTan[(Sqrt[3]*Sqrt[a
]*((-a)^(1/3) - 2^(1/3)*(-a + b*x^2)^(1/3)))/((-a)^(1/3)*Sqrt[b]*x)]/(2*2^(2/3)*Sqrt[3]*(-a)^(1/3)*Sqrt[a]*Sqr
t[b]) + ArcTanh[(Sqrt[b]*x)/Sqrt[a]]/(6*2^(2/3)*(-a)^(1/3)*Sqrt[a]*Sqrt[b]) - ArcTanh[((-a)^(1/3)*Sqrt[b]*x)/(
Sqrt[a]*((-a)^(1/3) + 2^(1/3)*(-a + b*x^2)^(1/3)))]/(2*2^(2/3)*(-a)^(1/3)*Sqrt[a]*Sqrt[b])

Rule 393

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[-(b/a), 2]}, Simp[(q*ArcT
an[Sqrt[3]/(q*x)])/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d), x] + (Simp[(q*ArcTanh[(a^(1/3)*q*x)/(a^(1/3) + 2^(1/3)*(a +
b*x^2)^(1/3))])/(2*2^(2/3)*a^(1/3)*d), x] - Simp[(q*ArcTanh[q*x])/(6*2^(2/3)*a^(1/3)*d), x] + Simp[(q*ArcTan[(
Sqrt[3]*(a^(1/3) - 2^(1/3)*(a + b*x^2)^(1/3)))/(a^(1/3)*q*x)])/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d), x])] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[b*c + 3*a*d, 0] && NegQ[b/a]

Rubi steps

\begin {align*} \int \frac {1}{\left (-3 a-b x^2\right ) \sqrt [3]{-a+b x^2}} \, dx &=-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt {a}}{\sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt {a} \left (\sqrt [3]{-a}-\sqrt [3]{2} \sqrt [3]{-a+b x^2}\right )}{\sqrt [3]{-a} \sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{6\ 2^{2/3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [3]{-a} \sqrt {b} x}{\sqrt {a} \left (\sqrt [3]{-a}+\sqrt [3]{2} \sqrt [3]{-a+b x^2}\right )}\right )}{2\ 2^{2/3} \sqrt [3]{-a} \sqrt {a} \sqrt {b}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 163, normalized size = 0.65 \begin {gather*} -\frac {9 a x F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {b x^2}{a},-\frac {b x^2}{3 a}\right )}{\sqrt [3]{b x^2-a} \left (3 a+b x^2\right ) \left (2 b x^2 \left (F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};\frac {b x^2}{a},-\frac {b x^2}{3 a}\right )-F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};\frac {b x^2}{a},-\frac {b x^2}{3 a}\right )\right )+9 a F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {b x^2}{a},-\frac {b x^2}{3 a}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((-3*a - b*x^2)*(-a + b*x^2)^(1/3)),x]

[Out]

(-9*a*x*AppellF1[1/2, 1/3, 1, 3/2, (b*x^2)/a, -1/3*(b*x^2)/a])/((-a + b*x^2)^(1/3)*(3*a + b*x^2)*(9*a*AppellF1
[1/2, 1/3, 1, 3/2, (b*x^2)/a, -1/3*(b*x^2)/a] + 2*b*x^2*(-AppellF1[3/2, 1/3, 2, 5/2, (b*x^2)/a, -1/3*(b*x^2)/a
] + AppellF1[3/2, 4/3, 1, 5/2, (b*x^2)/a, -1/3*(b*x^2)/a])))

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IntegrateAlgebraic [F]  time = 5.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (-3 a-b x^2\right ) \sqrt [3]{-a+b x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((-3*a - b*x^2)*(-a + b*x^2)^(1/3)),x]

[Out]

Defer[IntegrateAlgebraic][1/((-3*a - b*x^2)*(-a + b*x^2)^(1/3)), x]

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2-3*a)/(b*x^2-a)^(1/3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {1}{{\left (b x^{2} + 3 \, a\right )} {\left (b x^{2} - a\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2-3*a)/(b*x^2-a)^(1/3),x, algorithm="giac")

[Out]

integrate(-1/((b*x^2 + 3*a)*(b*x^2 - a)^(1/3)), x)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (-b \,x^{2}-3 a \right ) \left (b \,x^{2}-a \right )^{\frac {1}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^2-3*a)/(b*x^2-a)^(1/3),x)

[Out]

int(1/(-b*x^2-3*a)/(b*x^2-a)^(1/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {1}{{\left (b x^{2} + 3 \, a\right )} {\left (b x^{2} - a\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2-3*a)/(b*x^2-a)^(1/3),x, algorithm="maxima")

[Out]

-integrate(1/((b*x^2 + 3*a)*(b*x^2 - a)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} -\int \frac {1}{{\left (b\,x^2-a\right )}^{1/3}\,\left (b\,x^2+3\,a\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/((b*x^2 - a)^(1/3)*(3*a + b*x^2)),x)

[Out]

-int(1/((b*x^2 - a)^(1/3)*(3*a + b*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{3 a \sqrt [3]{- a + b x^{2}} + b x^{2} \sqrt [3]{- a + b x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**2-3*a)/(b*x**2-a)**(1/3),x)

[Out]

-Integral(1/(3*a*(-a + b*x**2)**(1/3) + b*x**2*(-a + b*x**2)**(1/3)), x)

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