3.2.29 \(\int \frac {1}{\sqrt [3]{-2-3 x^2} (6 d+d x^2)} \, dx\)

Optimal. Leaf size=119 \[ -\frac {\tan ^{-1}\left (\frac {\left (\sqrt [3]{-3 x^2-2}+\sqrt [3]{2}\right )^2}{3 \sqrt [6]{2} \sqrt {3} x}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{2} \left (\sqrt [3]{-3 x^2-2}+\sqrt [3]{2}\right )}{x}\right )}{4\ 2^{5/6} d}-\frac {\tan ^{-1}\left (\frac {x}{\sqrt {6}}\right )}{4\ 2^{5/6} \sqrt {3} d} \]

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Rubi [A]  time = 0.02, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {394} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\left (\sqrt [3]{-3 x^2-2}+\sqrt [3]{2}\right )^2}{3 \sqrt [6]{2} \sqrt {3} x}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{2} \left (\sqrt [3]{-3 x^2-2}+\sqrt [3]{2}\right )}{x}\right )}{4\ 2^{5/6} d}-\frac {\tan ^{-1}\left (\frac {x}{\sqrt {6}}\right )}{4\ 2^{5/6} \sqrt {3} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-2 - 3*x^2)^(1/3)*(6*d + d*x^2)),x]

[Out]

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

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Mathematica [C]  time = 0.11, size = 136, normalized size = 1.14 \begin {gather*} -\frac {9 x F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};-\frac {3 x^2}{2},-\frac {x^2}{6}\right )}{d \sqrt [3]{-3 x^2-2} \left (x^2+6\right ) \left (x^2 \left (F_1\left (\frac {3}{2};\frac {1}{3},2;\frac {5}{2};-\frac {3 x^2}{2},-\frac {x^2}{6}\right )+3 F_1\left (\frac {3}{2};\frac {4}{3},1;\frac {5}{2};-\frac {3 x^2}{2},-\frac {x^2}{6}\right )\right )-9 F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};-\frac {3 x^2}{2},-\frac {x^2}{6}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((-2 - 3*x^2)^(1/3)*(6*d + d*x^2)),x]

[Out]

(-9*x*AppellF1[1/2, 1/3, 1, 3/2, (-3*x^2)/2, -1/6*x^2])/(d*(-2 - 3*x^2)^(1/3)*(6 + x^2)*(-9*AppellF1[1/2, 1/3,
1, 3/2, (-3*x^2)/2, -1/6*x^2] + x^2*(AppellF1[3/2, 1/3, 2, 5/2, (-3*x^2)/2, -1/6*x^2] + 3*AppellF1[3/2, 4/3,
1, 5/2, (-3*x^2)/2, -1/6*x^2])))

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IntegrateAlgebraic [F]  time = 4.63, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{-2-3 x^2} \left (6 d+d x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((-2 - 3*x^2)^(1/3)*(6*d + d*x^2)),x]

[Out]

Defer[IntegrateAlgebraic][1/((-2 - 3*x^2)^(1/3)*(6*d + 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/(-3*x^2-2)^(1/3)/(d*x^2+6*d),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 (d x^{2} + 6 \, d\right )} {\left (-3 \, x^{2} - 2\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-3*x^2-2)^(1/3)/(d*x^2+6*d),x, algorithm="giac")

[Out]

integrate(1/((d*x^2 + 6*d)*(-3*x^2 - 2)^(1/3)), x)

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maple [C]  time = 77.11, size = 725, normalized size = 6.09

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-3*x^2-2)^(1/3)/(d*x^2+6*d),x)

[Out]

-1/24*(RootOf(_Z^6+54)*ln(-(4*RootOf(_Z^6+54)^7*x-288*RootOf(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*_Z^2)
*RootOf(_Z^6+54)^6*x+4608*RootOf(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*_Z^2)^2*RootOf(_Z^6+54)^5*x+144*R
ootOf(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*_Z^2)*RootOf(_Z^6+54)^4*(-3*x^2-2)^(1/3)*x-6912*RootOf(RootO
f(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*_Z^2)^2*RootOf(_Z^6+54)^3*(-3*x^2-2)^(1/3)*x-9*x^2*RootOf(_Z^6+54)^4+21
6*RootOf(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*_Z^2)*RootOf(_Z^6+54)^3*x^2+18*RootOf(_Z^6+54)^4-432*Root
Of(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*_Z^2)*RootOf(_Z^6+54)^3-2592*RootOf(RootOf(_Z^6+54)^2-24*_Z*Roo
tOf(_Z^6+54)+576*_Z^2)*RootOf(_Z^6+54)*(-3*x^2-2)^(1/3)+324*(-3*x^2-2)^(2/3))/(x^2+6))+ln(-(-4*RootOf(_Z^6+54)
^7*x+192*RootOf(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*_Z^2)*RootOf(_Z^6+54)^6*x-6*(-3*x^2-2)^(1/3)*RootO
f(_Z^6+54)^5*x+288*RootOf(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*_Z^2)*RootOf(_Z^6+54)^4*(-3*x^2-2)^(1/3)
*x+9*x^2*RootOf(_Z^6+54)^4-18*RootOf(_Z^6+54)^4+108*(-3*x^2-2)^(1/3)*RootOf(_Z^6+54)^2+324*(-3*x^2-2)^(2/3))/(
x^2+6))*RootOf(_Z^6+54)-24*ln(-(-4*RootOf(_Z^6+54)^7*x+192*RootOf(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*
_Z^2)*RootOf(_Z^6+54)^6*x-6*(-3*x^2-2)^(1/3)*RootOf(_Z^6+54)^5*x+288*RootOf(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^
6+54)+576*_Z^2)*RootOf(_Z^6+54)^4*(-3*x^2-2)^(1/3)*x+9*x^2*RootOf(_Z^6+54)^4-18*RootOf(_Z^6+54)^4+108*(-3*x^2-
2)^(1/3)*RootOf(_Z^6+54)^2+324*(-3*x^2-2)^(2/3))/(x^2+6))*RootOf(RootOf(_Z^6+54)^2-24*_Z*RootOf(_Z^6+54)+576*_
Z^2))/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (d x^{2} + 6 \, d\right )} {\left (-3 \, x^{2} - 2\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-3*x^2-2)^(1/3)/(d*x^2+6*d),x, algorithm="maxima")

[Out]

integrate(1/((d*x^2 + 6*d)*(-3*x^2 - 2)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (-3\,x^2-2\right )}^{1/3}\,\left (d\,x^2+6\,d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((- 3*x^2 - 2)^(1/3)*(6*d + d*x^2)),x)

[Out]

int(1/((- 3*x^2 - 2)^(1/3)*(6*d + d*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{x^{2} \sqrt [3]{- 3 x^{2} - 2} + 6 \sqrt [3]{- 3 x^{2} - 2}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-3*x**2-2)**(1/3)/(d*x**2+6*d),x)

[Out]

Integral(1/(x**2*(-3*x**2 - 2)**(1/3) + 6*(-3*x**2 - 2)**(1/3)), x)/d

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