3.2.55 \(\int \frac {1}{\sqrt [3]{-2+b x^2} (-\frac {18 d}{b}+d x^2)} \, dx\) [155]

Optimal. Leaf size=147 \[ \frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d} \]

[Out]

-1/24*arctanh(1/6*(2^(1/3)+(b*x^2-2)^(1/3))^2*2^(5/6)/x/b^(1/2))*b^(1/2)*2^(1/6)/d+1/24*arctanh(1/6*x*b^(1/2)*
2^(1/2))*b^(1/2)*2^(1/6)/d+1/24*arctan(2^(1/6)*(2^(1/3)+(b*x^2-2)^(1/3))*3^(1/2)/x/b^(1/2))*b^(1/2)*2^(1/6)/d*
3^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 147, 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 = {404} \begin {gather*} \frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{b x^2-2}+\sqrt [3]{2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\left (\sqrt [3]{b x^2-2}+\sqrt [3]{2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-2 + b*x^2)^(1/3)*((-18*d)/b + d*x^2)),x]

[Out]

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

Rule 404

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)]/(12
*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]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx &=\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 5.66, size = 148, normalized size = 1.01 \begin {gather*} \frac {27 b x F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {b x^2}{2},\frac {b x^2}{18}\right )}{d \left (-18+b x^2\right ) \sqrt [3]{-2+b x^2} \left (27 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\frac {b x^2}{2},\frac {b x^2}{18}\right )+b x^2 \left (F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};\frac {b x^2}{2},\frac {b x^2}{18}\right )+3 F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};\frac {b x^2}{2},\frac {b x^2}{18}\right )\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((-2 + b*x^2)^(1/3)*((-18*d)/b + d*x^2)),x]

[Out]

(27*b*x*AppellF1[1/2, 1/3, 1, 3/2, (b*x^2)/2, (b*x^2)/18])/(d*(-18 + b*x^2)*(-2 + b*x^2)^(1/3)*(27*AppellF1[1/
2, 1/3, 1, 3/2, (b*x^2)/2, (b*x^2)/18] + b*x^2*(AppellF1[3/2, 1/3, 2, 5/2, (b*x^2)/2, (b*x^2)/18] + 3*AppellF1
[3/2, 4/3, 1, 5/2, (b*x^2)/2, (b*x^2)/18])))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{2}-2\right )^{\frac {1}{3}} \left (-\frac {18 d}{b}+d \,x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2-2)^(1/3)/(-18*d/b+d*x^2),x)

[Out]

int(1/(b*x^2-2)^(1/3)/(-18*d/b+d*x^2),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^2-2)^(1/3)/(-18*d/b+d*x^2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 - 2)^(1/3)*(d*x^2 - 18*d/b)), 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^2-2)^(1/3)/(-18*d/b+d*x^2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {b \int \frac {1}{b x^{2} \sqrt [3]{b x^{2} - 2} - 18 \sqrt [3]{b x^{2} - 2}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2-2)**(1/3)/(-18*d/b+d*x**2),x)

[Out]

b*Integral(1/(b*x**2*(b*x**2 - 2)**(1/3) - 18*(b*x**2 - 2)**(1/3)), x)/d

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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^2-2)^(1/3)/(-18*d/b+d*x^2),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 - 2)^(1/3)*(d*x^2 - 18*d/b)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {1}{\left (\frac {18\,d}{b}-d\,x^2\right )\,{\left (b\,x^2-2\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(((18*d)/b - d*x^2)*(b*x^2 - 2)^(1/3)),x)

[Out]

int(-1/(((18*d)/b - d*x^2)*(b*x^2 - 2)^(1/3)), x)

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