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

Optimal. Leaf size=151 \[ \frac {\sqrt {b} \tan ^{-1}\left (\frac {\left (\sqrt [3]{2}-\sqrt [3]{b x^2+2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{2}-\sqrt [3]{b x^2+2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d} \]

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Rubi [A]  time = 0.03, antiderivative size = 151, 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 = {394} \begin {gather*} \frac {\sqrt {b} \tan ^{-1}\left (\frac {\left (\sqrt [3]{2}-\sqrt [3]{b x^2+2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{2}-\sqrt [3]{b x^2+2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\sqrt {b} \tan ^{-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[(Sqrt[b]*x)/(3*Sqrt[2])])/(12*2^(5/6)*d) + (Sqrt[b]*ArcTan[(2^(1/3) - (2 + b*x^2)^(1/3))^2/(3*
2^(1/6)*Sqrt[b]*x)])/(12*2^(5/6)*d) - (Sqrt[b]*ArcTanh[(2^(1/6)*Sqrt[3]*(2^(1/3) - (2 + b*x^2)^(1/3)))/(Sqrt[b
]*x)])/(4*2^(5/6)*Sqrt[3]*d)

Rule 394

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[b/a, 2]}, Simp[(q*ArcTan[
(q*x)/3])/(12*Rt[a, 3]*d), x] + (Simp[(q*ArcTan[(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*ArcTanh[(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] && PosQ[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 {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d}+\frac {\sqrt {b} \tan ^{-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}-\frac {\sqrt {b} \tanh ^{-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}\\ \end {align*}

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Mathematica [C]  time = 0.15, size = 148, normalized size = 0.98 \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 \sqrt [3]{b x^2+2} \left (b x^2+18\right ) \left (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 )-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 )\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, -1/2*(b*x^2), -1/18*(b*x^2)])/(d*(2 + b*x^2)^(1/3)*(18 + b*x^2)*(-27*Appel
lF1[1/2, 1/3, 1, 3/2, -1/2*(b*x^2), -1/18*(b*x^2)] + b*x^2*(AppellF1[3/2, 1/3, 2, 5/2, -1/2*(b*x^2), -1/18*(b*
x^2)] + 3*AppellF1[3/2, 4/3, 1, 5/2, -1/2*(b*x^2), -1/18*(b*x^2)])))

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IntegrateAlgebraic [F]  time = 5.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{2+b x^2} \left (\frac {18 d}{b}+d x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[IntegrateAlgebraic][1/((2 + b*x^2)^(1/3)*((18*d)/b + d*x^2)), 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+2)^(1/3)/(18*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 \begin {gather*} \int \frac {1}{{\left (b x^{2} + 2\right )}^{\frac {1}{3}} {\left (d x^{2} + \frac {18 \, d}{b}\right )}}\,{d x} \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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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \,x^{2}+2\right )^{\frac {1}{3}} \left (d \,x^{2}+\frac {18 d}{b}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(b*x^2+2)^(1/3)/(18/b*d+d*x^2),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} + 2\right )}^{\frac {1}{3}} {\left (d x^{2} + \frac {18 \, d}{b}\right )}}\,{d x} \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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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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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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