\(\int \frac {1}{x (1-x^2)^{2/3}} \, dx\) [34]

Optimal result

Integrand size = 15, antiderivative size = 58 \[ \int \frac {1}{x \left (1-x^2\right )^{2/3}} \, dx=-\frac {1}{2} \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1-x^2}}{\sqrt {3}}\right )-\frac {\log (x)}{2}+\frac {3}{4} \log \left (1-\sqrt [3]{1-x^2}\right ) \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 59, 632, 210, 31} \[ \int \frac {1}{x \left (1-x^2\right )^{2/3}} \, dx=-\frac {1}{2} \sqrt {3} \arctan \left (\frac {2 \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )+\frac {3}{4} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac {\log (x)}{2} \]

[In]

Int[1/(x*(1 - x^2)^(2/3)),x]

[Out]

-1/2*(Sqrt[3]*ArcTan[(1 + 2*(1 - x^2)^(1/3))/Sqrt[3]]) - 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 59

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 210

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

Rule 272

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 632

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(1-x)^{2/3} x} \, dx,x,x^2\right ) \\ & = -\frac {\log (x)}{2}-\frac {3}{4} \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac {3}{4} \text {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} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1-x^2}\right ) \\ & = -\frac {1}{2} \sqrt {3} \arctan \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*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x \left (1-x^2\right )^{2/3}} \, dx=\frac {1}{4} \left (-2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1-x^2}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{1-x^2}\right )-\log \left (1+\sqrt [3]{1-x^2}+\left (1-x^2\right )^{2/3}\right )\right ) \]

[In]

Integrate[1/(x*(1 - x^2)^(2/3)),x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 1.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83

method	result	size
meijerg	\(\frac {\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+2 \ln \left (x \right )+i \pi \right ) \Gamma \left (\frac {2}{3}\right )+\frac {2 \Gamma \left (\frac {2}{3}\right ) x^{2} {}_{3}^{}{\moversetsp {}{\mundersetsp {}{F_{2}^{}}}}\left (1,1,\frac {5}{3};2,2;x^{2}\right )}{3}}{2 \Gamma \left (\frac {2}{3}\right )}\)	\(48\)
pseudoelliptic	\(-\frac {\ln \left (\left (-x^{2}+1\right )^{\frac {2}{3}}+\left (-x^{2}+1\right )^{\frac {1}{3}}+1\right )}{4}-\frac {\arctan \left (\frac {\left (1+2 \left (-x^{2}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{2}+\frac {\ln \left (\left (-x^{2}+1\right )^{\frac {1}{3}}-1\right )}{2}\)	\(63\)
trager	\(\frac {\ln \left (\frac {-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{2}+1\right )^{\frac {2}{3}}+9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-9 \left (-x^{2}+1\right )^{\frac {2}{3}}+9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{2}+1\right )^{\frac {1}{3}}+16 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-2 x^{2}+24 \left (-x^{2}+1\right )^{\frac {1}{3}}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+1}{x^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{2}+1\right )^{\frac {2}{3}}-16 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+24 \left (-x^{2}+1\right )^{\frac {2}{3}}-24 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{2}+1\right )^{\frac {1}{3}}+20 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-3 x^{2}-9 \left (-x^{2}+1\right )^{\frac {1}{3}}+29 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+5}{x^{2}}\right )}{2}\)	\(246\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x \left (1-x^2\right )^{2/3}} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{4} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x \left (1-x^2\right )^{2/3}} \, dx=- \frac {e^{- \frac {2 i \pi }{3}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {1}{x^{2}}} \right )}}{2 x^{\frac {4}{3}} \Gamma \left (\frac {5}{3}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x \left (1-x^2\right )^{2/3}} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{4} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x \left (1-x^2\right )^{2/3}} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{4} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left (-{\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.31 \[ \int \frac {1}{x \left (1-x^2\right )^{2/3}} \, dx=\frac {\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {9}{4}\right )}{2}+\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{2}+\frac {9}{4}-\frac {\sqrt {3}\,9{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{2}+\frac {9}{4}+\frac {\sqrt {3}\,9{}\mathrm {i}}{4}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right ) \]

[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)