3.2.42 \(\int \frac {1}{(3 a-b x^2) \sqrt [3]{a+b x^2}} \, dx\) [142]

Optimal. Leaf size=202 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{6\ 2^{2/3} a^{5/6} \sqrt {b}}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{2} \sqrt [3]{a+b x^2}\right )}\right )}{2\ 2^{2/3} a^{5/6} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {3} \sqrt {a}}{\sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} a^{5/6} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{2} \sqrt [3]{a+b x^2}\right )}{\sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} a^{5/6} \sqrt {b}} \]

[Out]

1/4*arctan(x*b^(1/2)/a^(1/6)/(a^(1/3)+2^(1/3)*(b*x^2+a)^(1/3)))*2^(1/3)/a^(5/6)/b^(1/2)-1/12*arctan(x*b^(1/2)/
a^(1/2))*2^(1/3)/a^(5/6)/b^(1/2)-1/12*arctanh(a^(1/6)*(a^(1/3)-2^(1/3)*(b*x^2+a)^(1/3))*3^(1/2)/x/b^(1/2))*2^(
1/3)/a^(5/6)*3^(1/2)/b^(1/2)-1/12*arctanh(3^(1/2)*a^(1/2)/x/b^(1/2))*2^(1/3)/a^(5/6)*3^(1/2)/b^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {401} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt [6]{a} \left (\sqrt [3]{2} \sqrt [3]{a+b x^2}+\sqrt [3]{a}\right )}\right )}{2\ 2^{2/3} a^{5/6} \sqrt {b}}-\frac {\text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{6\ 2^{2/3} a^{5/6} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{2} \sqrt [3]{a+b x^2}\right )}{\sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} a^{5/6} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {3} \sqrt {a}}{\sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} a^{5/6} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(2^(2/3)*a^(5/6)*Sqrt[b]) + ArcTan[(Sqrt[b]*x)/(a^(1/6)*(a^(1/3) + 2^(1/3)*(a
+ b*x^2)^(1/3)))]/(2*2^(2/3)*a^(5/6)*Sqrt[b]) - ArcTanh[(Sqrt[3]*Sqrt[a])/(Sqrt[b]*x)]/(2*2^(2/3)*Sqrt[3]*a^(
5/6)*Sqrt[b]) - ArcTanh[(Sqrt[3]*a^(1/6)*(a^(1/3) - 2^(1/3)*(a + b*x^2)^(1/3)))/(Sqrt[b]*x)]/(2*2^(2/3)*Sqrt[3
]*a^(5/6)*Sqrt[b])

Rule 401

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[b/a, 2]}, Simp[q*(ArcTanh
[Sqrt[3]/(q*x)]/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d)), x] + (-Simp[q*(ArcTan[(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*(ArcTan[q*x]/(6*2^(2/3)*a^(1/3)*d)), x] + Simp[q*(ArcTanh[Sqr
t[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] && PosQ[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 {b} x}{\sqrt {a}}\right )}{6\ 2^{2/3} a^{5/6} \sqrt {b}}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{2} \sqrt [3]{a+b x^2}\right )}\right )}{2\ 2^{2/3} a^{5/6} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {3} \sqrt {a}}{\sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} a^{5/6} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{2} \sqrt [3]{a+b x^2}\right )}{\sqrt {b} x}\right )}{2\ 2^{2/3} \sqrt {3} a^{5/6} \sqrt {b}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 6.02, size = 166, normalized size = 0.82 \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 )}{\left (3 a-b x^2\right ) \sqrt [3]{a+b x^2} \left (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 )+2 b x^2 \left (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 )-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 )\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), (b*x^2)/(3*a)])/((3*a - b*x^2)*(a + b*x^2)^(1/3)*(9*a*AppellF1
[1/2, 1/3, 1, 3/2, -((b*x^2)/a), (b*x^2)/(3*a)] + 2*b*x^2*(AppellF1[3/2, 1/3, 2, 5/2, -((b*x^2)/a), (b*x^2)/(3
*a)] - AppellF1[3/2, 4/3, 1, 5/2, -((b*x^2)/a), (b*x^2)/(3*a)])))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (-b \,x^{2}+3 a \right ) \left (b \,x^{2}+a \right )^{\frac {1}{3}}}\, dx\]

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*} \text {Failed to integrate} \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 + a)^(1/3)*(b*x^2 - 3*a)), x)

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Fricas [F(-1)] Timed out
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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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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 + a)^(1/3)*(b*x^2 - 3*a)), 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 (3\,a-b\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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