3.151 \(\int \frac {1}{\sqrt [3]{a-b x^2} (-\frac {9 a d}{b}+d x^2)} \, dx\)

Optimal. Leaf size=153 \[ -\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{a-b x^2}\right )}{\sqrt {b} x}\right )}{4 \sqrt {3} a^{5/6} d}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\left (\sqrt [3]{a}-\sqrt [3]{a-b x^2}\right )^2}{3 \sqrt [6]{a} \sqrt {b} x}\right )}{12 a^{5/6} d}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {395} \[ -\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{a-b x^2}\right )}{\sqrt {b} x}\right )}{4 \sqrt {3} a^{5/6} d}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\left (\sqrt [3]{a}-\sqrt [3]{a-b x^2}\right )^2}{3 \sqrt [6]{a} \sqrt {b} x}\right )}{12 a^{5/6} d}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d} \]

Antiderivative was successfully verified.

[In]

Int[1/((a - b*x^2)^(1/3)*((-9*a*d)/b + d*x^2)),x]

[Out]

-(Sqrt[b]*ArcTan[(Sqrt[3]*a^(1/6)*(a^(1/3) - (a - b*x^2)^(1/3)))/(Sqrt[b]*x)])/(4*Sqrt[3]*a^(5/6)*d) - (Sqrt[b
]*ArcTanh[(Sqrt[b]*x)/(3*Sqrt[a])])/(12*a^(5/6)*d) + (Sqrt[b]*ArcTanh[(a^(1/3) - (a - b*x^2)^(1/3))^2/(3*a^(1/
6)*Sqrt[b]*x)])/(12*a^(5/6)*d)

Rule 395

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Simp[(q*Arc
Tanh[(q*x)/3])/(12*Rt[a, 3]*d), x] + (Simp[(q*ArcTanh[(Rt[a, 3] - (a + b*x^2)^(1/3))^2/(3*Rt[a, 3]^2*q*x)])/(1
2*Rt[a, 3]*d), x] - Simp[(q*ArcTan[(Sqrt[3]*(Rt[a, 3] - (a + b*x^2)^(1/3)))/(Rt[a, 3]*q*x)])/(4*Sqrt[3]*Rt[a,
3]*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[b*c - 9*a*d, 0] && NegQ[b/a]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [3]{a-b x^2} \left (-\frac {9 a d}{b}+d x^2\right )} \, dx &=-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{a-b x^2}\right )}{\sqrt {b} x}\right )}{4 \sqrt {3} a^{5/6} d}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {a}}\right )}{12 a^{5/6} d}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\left (\sqrt [3]{a}-\sqrt [3]{a-b x^2}\right )^2}{3 \sqrt [6]{a} \sqrt {b} x}\right )}{12 a^{5/6} d}\\ \end {align*}

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Mathematica [C]  time = 0.18, size = 167, normalized size = 1.09 \[ -\frac {27 a b x F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {b x^2}{a},\frac {b x^2}{9 a}\right )}{d \sqrt [3]{a-b x^2} \left (9 a-b x^2\right ) \left (2 b x^2 \left (F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};\frac {b x^2}{a},\frac {b x^2}{9 a}\right )+3 F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};\frac {b x^2}{a},\frac {b x^2}{9 a}\right )\right )+27 a F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {b x^2}{a},\frac {b x^2}{9 a}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a - b*x^2)^(1/3)*((-9*a*d)/b + d*x^2)),x]

[Out]

(-27*a*b*x*AppellF1[1/2, 1/3, 1, 3/2, (b*x^2)/a, (b*x^2)/(9*a)])/(d*(a - b*x^2)^(1/3)*(9*a - b*x^2)*(27*a*Appe
llF1[1/2, 1/3, 1, 3/2, (b*x^2)/a, (b*x^2)/(9*a)] + 2*b*x^2*(AppellF1[3/2, 1/3, 2, 5/2, (b*x^2)/a, (b*x^2)/(9*a
)] + 3*AppellF1[3/2, 4/3, 1, 5/2, (b*x^2)/a, (b*x^2)/(9*a)])))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(1/3)/(-9*a*d/b+d*x^2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {1}{3}} {\left (d x^{2} - \frac {9 \, a d}{b}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(1/3)/(-9*a*d/b+d*x^2),x, algorithm="giac")

[Out]

integrate(1/((-b*x^2 + a)^(1/3)*(d*x^2 - 9*a*d/b)), x)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {1}{3}} \left (d \,x^{2}-\frac {9 a d}{b}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^2+a)^(1/3)/(-9*a/b*d+d*x^2),x)

[Out]

int(1/(-b*x^2+a)^(1/3)/(-9*a/b*d+d*x^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {1}{3}} {\left (d x^{2} - \frac {9 \, a d}{b}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(1/3)/(-9*a*d/b+d*x^2),x, algorithm="maxima")

[Out]

integrate(1/((-b*x^2 + a)^(1/3)*(d*x^2 - 9*a*d/b)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a-b\,x^2\right )}^{1/3}\,\left (d\,x^2-\frac {9\,a\,d}{b}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a - b*x^2)^(1/3)*(d*x^2 - (9*a*d)/b)),x)

[Out]

int(1/((a - b*x^2)^(1/3)*(d*x^2 - (9*a*d)/b)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b \int \frac {1}{- 9 a \sqrt [3]{a - b x^{2}} + b x^{2} \sqrt [3]{a - b x^{2}}}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**2+a)**(1/3)/(-9*a*d/b+d*x**2),x)

[Out]

b*Integral(1/(-9*a*(a - b*x**2)**(1/3) + b*x**2*(a - b*x**2)**(1/3)), x)/d

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