3.34 \(\int \frac {1}{x (1-x^2)^{2/3}} \, dx\)

Optimal. Leaf size=58 \[ \frac {3}{4} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )-\frac {\log (x)}{2} \]

[Out]

-1/2*ln(x)+3/4*ln(1-(-x^2+1)^(1/3))-1/2*arctan(1/3*(1+2*(-x^2+1)^(1/3))*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {266, 57, 618, 204, 31} \[ \frac {3}{4} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )-\frac {\log (x)}{2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 - x^2)^(2/3)),x]

[Out]

-(Sqrt[3]*ArcTan[(1 + 2*(1 - x^2)^(1/3))/Sqrt[3]])/2 - Log[x]/2 + (3*Log[1 - (1 - x^2)^(1/3)])/4

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {1}{x \left (1-x^2\right )^{2/3}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1-x)^{2/3} x} \, dx,x,x^2\right )\\ &=-\frac {\log (x)}{2}-\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )\\ &=-\frac {\log (x)}{2}+\frac {3}{4} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1-x^2}\right )\\ &=-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {1+2 \sqrt [3]{1-x^2}}{\sqrt {3}}\right )-\frac {\log (x)}{2}+\frac {3}{4} \log \left (1-\sqrt [3]{1-x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 81, normalized size = 1.40 \[ \frac {1}{2} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac {1}{4} \log \left (\left (1-x^2\right )^{2/3}+\sqrt [3]{1-x^2}+1\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 - x^2)^(2/3)),x]

[Out]

-1/2*(Sqrt[3]*ArcTan[(1 + 2*(1 - x^2)^(1/3))/Sqrt[3]]) + Log[1 - (1 - x^2)^(1/3)]/2 - Log[1 + (1 - x^2)^(1/3)
+ (1 - x^2)^(2/3)]/4

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fricas [A]  time = 0.71, size = 64, normalized size = 1.10 \[ -\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{4} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*arctan(2/3*sqrt(3)*(-x^2 + 1)^(1/3) + 1/3*sqrt(3)) - 1/4*log((-x^2 + 1)^(2/3) + (-x^2 + 1)^(1/3)
+ 1) + 1/2*log((-x^2 + 1)^(1/3) - 1)

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giac [A]  time = 1.09, size = 64, normalized size = 1.10 \[ -\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{4} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left (-{\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-x^2 + 1)^(1/3) + 1)) - 1/4*log((-x^2 + 1)^(2/3) + (-x^2 + 1)^(1/3) + 1) +
1/2*log(-(-x^2 + 1)^(1/3) + 1)

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maple [C]  time = 0.10, size = 48, normalized size = 0.83 \[ \frac {\frac {2 \Gamma \left (\frac {2}{3}\right ) x^{2} \hypergeom \left (\left [1, 1, \frac {5}{3}\right ], \left [2, 2\right ], x^{2}\right )}{3}+\left (2 \ln \relax (x )+\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+i \pi \right ) \Gamma \left (\frac {2}{3}\right )}{2 \Gamma \left (\frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/GAMMA(2/3)*(2/3*GAMMA(2/3)*x^2*hypergeom([1,1,5/3],[2,2],x^2)+(1/6*Pi*3^(1/2)-3/2*ln(3)+2*ln(x)+I*Pi)*GAMM
A(2/3))

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maxima [A]  time = 1.30, size = 62, normalized size = 1.07 \[ -\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{4} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-x^2 + 1)^(1/3) + 1)) - 1/4*log((-x^2 + 1)^(2/3) + (-x^2 + 1)^(1/3) + 1) +
1/2*log((-x^2 + 1)^(1/3) - 1)

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mupad [B]  time = 0.46, size = 76, normalized size = 1.31 \[ \frac {\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {9}{4}\right )}{2}+\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{2}+\frac {9}{4}-\frac {\sqrt {3}\,9{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{2}+\frac {9}{4}+\frac {\sqrt {3}\,9{}\mathrm {i}}{4}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(1 - x^2)^(2/3)),x)

[Out]

log((9*(1 - x^2)^(1/3))/4 - 9/4)/2 + log((9*(1 - x^2)^(1/3))/2 - (3^(1/2)*9i)/4 + 9/4)*((3^(1/2)*1i)/4 - 1/4)
- log((3^(1/2)*9i)/4 + (9*(1 - x^2)^(1/3))/2 + 9/4)*((3^(1/2)*1i)/4 + 1/4)

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sympy [C]  time = 0.98, size = 37, normalized size = 0.64 \[ - \frac {e^{- \frac {2 i \pi }{3}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {1}{x^{2}}} \right )}}{2 x^{\frac {4}{3}} \Gamma \left (\frac {5}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(-2*I*pi/3)*gamma(2/3)*hyper((2/3, 2/3), (5/3,), x**(-2))/(2*x**(4/3)*gamma(5/3))

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