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

Optimal. Leaf size=123 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [6]{2} \left (\sqrt [3]{2}-\sqrt [3]{2-3 x^2}\right )}{x}\right )}{4\ 2^{5/6} d}+\frac {\tanh ^{-1}\left (\frac {\left (\sqrt [3]{2}-\sqrt [3]{2-3 x^2}\right )^2}{3 \sqrt [6]{2} \sqrt {3} x}\right )}{4\ 2^{5/6} \sqrt {3} d}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {6}}\right )}{4\ 2^{5/6} \sqrt {3} d} \]

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Rubi [A]  time = 0.02, antiderivative size = 123, 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 = {395} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [6]{2} \left (\sqrt [3]{2}-\sqrt [3]{2-3 x^2}\right )}{x}\right )}{4\ 2^{5/6} d}+\frac {\tanh ^{-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 {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[(2^(1/6)*(2^(1/3) - (2 - 3*x^2)^(1/3)))/x]/(4*2^(5/6)*d) - ArcTanh[x/Sqrt[6]]/(4*2^(5/6)*Sqrt[3]*d) +
ArcTanh[(2^(1/3) - (2 - 3*x^2)^(1/3))^2/(3*2^(1/6)*Sqrt[3]*x)]/(4*2^(5/6)*Sqrt[3]*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]{2-3 x^2} \left (-6 d+d x^2\right )} \, dx &=-\frac {\tan ^{-1}\left (\frac {\sqrt [6]{2} \left (\sqrt [3]{2}-\sqrt [3]{2-3 x^2}\right )}{x}\right )}{4\ 2^{5/6} d}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {6}}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\tanh ^{-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}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 136, normalized size = 1.11 \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]{2-3 x^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, x^2/6])/(d*(2 - 3*x^2)^(1/3)*(-6 + x^2)*(9*AppellF1[1/2, 1/3, 1, 3/
2, (3*x^2)/2, x^2/6] + x^2*(AppellF1[3/2, 1/3, 2, 5/2, (3*x^2)/2, x^2/6] + 3*AppellF1[3/2, 4/3, 1, 5/2, (3*x^2
)/2, x^2/6])))

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IntegrateAlgebraic [F]  time = 4.62, 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 = 59.07, size = 1061, normalized size = 8.63

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*(24*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*ln((-4608*RootOf(RootOf(_Z^6-54)^2-24*_Z*Roo
tOf(_Z^6-54)+576*_Z^2)^2*RootOf(_Z^6-54)^5*x+288*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*Root
Of(_Z^6-54)^6*x-4*RootOf(_Z^6-54)^7*x+6912*RootOf(_Z^6-54)^3*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+57
6*_Z^2)^2*(-3*x^2+2)^(1/3)*x-144*RootOf(_Z^6-54)^4*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*(-
3*x^2+2)^(1/3)*x-216*RootOf(_Z^6-54)^3*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*x^2+9*x^2*Root
Of(_Z^6-54)^4-432*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*RootOf(_Z^6-54)^3+18*RootOf(_Z^6-54
)^4-2592*RootOf(_Z^6-54)*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*(-3*x^2+2)^(1/3)+324*(-3*x^2
+2)^(2/3))/(x^2-6))-24*ln((-768*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)^2*RootOf(_Z^6-54)^5*x
+16*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*RootOf(_Z^6-54)^6*x+1152*RootOf(_Z^6-54)^3*RootOf
(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)^2*(-3*x^2+2)^(1/3)*x-72*RootOf(_Z^6-54)^4*RootOf(RootOf(_Z^
6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*(-3*x^2+2)^(1/3)*x+(-3*x^2+2)^(1/3)*RootOf(_Z^6-54)^5*x+36*RootOf(_Z^6
-54)^3*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*x^2+72*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_
Z^6-54)+576*_Z^2)*RootOf(_Z^6-54)^3+432*RootOf(_Z^6-54)*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^
2)*(-3*x^2+2)^(1/3)-18*(-3*x^2+2)^(1/3)*RootOf(_Z^6-54)^2+54*(-3*x^2+2)^(2/3))/(x^2-6))*RootOf(RootOf(_Z^6-54)
^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)+ln((-768*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)^2*RootOf(
_Z^6-54)^5*x+16*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*RootOf(_Z^6-54)^6*x+1152*RootOf(_Z^6-
54)^3*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)^2*(-3*x^2+2)^(1/3)*x-72*RootOf(_Z^6-54)^4*RootO
f(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*(-3*x^2+2)^(1/3)*x+(-3*x^2+2)^(1/3)*RootOf(_Z^6-54)^5*x+36
*RootOf(_Z^6-54)^3*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6-54)+576*_Z^2)*x^2+72*RootOf(RootOf(_Z^6-54)^2-24
*_Z*RootOf(_Z^6-54)+576*_Z^2)*RootOf(_Z^6-54)^3+432*RootOf(_Z^6-54)*RootOf(RootOf(_Z^6-54)^2-24*_Z*RootOf(_Z^6
-54)+576*_Z^2)*(-3*x^2+2)^(1/3)-18*(-3*x^2+2)^(1/3)*RootOf(_Z^6-54)^2+54*(-3*x^2+2)^(2/3))/(x^2-6))*RootOf(_Z^
6-54))/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 (2-3\,x^2\right )}^{1/3}\,\left (6\,d-d\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-int(1/((2 - 3*x^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]{2 - 3 x^{2}} - 6 \sqrt [3]{2 - 3 x^{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*(2 - 3*x**2)**(1/3) - 6*(2 - 3*x**2)**(1/3)), x)/d

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