\(\int \frac {(-1+x^2)^{2/3}}{x} \, dx\) [1211]

Optimal result

Integrand size = 13, antiderivative size = 89 \[ \int \frac {\left (-1+x^2\right )^{2/3}}{x} \, dx=\frac {3}{4} \left (-1+x^2\right )^{2/3}+\frac {1}{2} \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-1+x^2}}{\sqrt {3}}\right )+\frac {1}{2} \log \left (1+\sqrt [3]{-1+x^2}\right )-\frac {1}{4} \log \left (1-\sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.73, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {272, 52, 58, 632, 210, 31} \[ \int \frac {\left (-1+x^2\right )^{2/3}}{x} \, dx=\frac {1}{2} \sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{x^2-1}}{\sqrt {3}}\right )+\frac {3}{4} \left (x^2-1\right )^{2/3}+\frac {3}{4} \log \left (\sqrt [3]{x^2-1}+1\right )-\frac {\log (x)}{2} \]

[In]

Int[(-1 + x^2)^(2/3)/x,x]

[Out]

(3*(-1 + x^2)^(2/3))/4 + (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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && NegQ
[(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+x)^{2/3}}{x} \, dx,x,x^2\right ) \\ & = \frac {3}{4} \left (-1+x^2\right )^{2/3}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt [3]{-1+x} x} \, dx,x,x^2\right ) \\ & = \frac {3}{4} \left (-1+x^2\right )^{2/3}-\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 {3}{4} \left (-1+x^2\right )^{2/3}-\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 {3}{4} \left (-1+x^2\right )^{2/3}+\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 = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-1+x^2\right )^{2/3}}{x} \, dx=\frac {1}{4} \left (3 \left (-1+x^2\right )^{2/3}+2 \sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{-1+x^2}}{\sqrt {3}}\right )+2 \log \left (1+\sqrt [3]{-1+x^2}\right )-\log \left (1-\sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}\right )\right ) \]

[In]

Integrate[(-1 + x^2)^(2/3)/x,x]

[Out]

(3*(-1 + x^2)^(2/3) + 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 [A] (verified)

Time = 2.72 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(\frac {3 \left (x^{2}-1\right )^{\frac {2}{3}}}{4}+\frac {\ln \left (1+\left (x^{2}-1\right )^{\frac {1}{3}}\right )}{2}-\frac {\ln \left (1-\left (x^{2}-1\right )^{\frac {1}{3}}+\left (x^{2}-1\right )^{\frac {2}{3}}\right )}{4}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{2}-1\right )^{\frac {1}{3}}-1\right ) \sqrt {3}}{3}\right )}{2}\)	\(66\)
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \operatorname {signum}\left (x^{2}-1\right )^{\frac {2}{3}} \left (\frac {2 \pi \sqrt {3}\, x^{2} \operatorname {hypergeom}\left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], x^{2}\right )}{3 \Gamma \left (\frac {2}{3}\right )}-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+2 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {2}{3}}}\)	\(84\)
trager	\(\frac {3 \left (x^{2}-1\right )^{\frac {2}{3}}}{4}+\frac {\ln \left (-\frac {36 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {2}{3}}+51 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-144 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-27 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {1}{3}}+24 \left (x^{2}-1\right )^{\frac {2}{3}}+15 x^{2}-120 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )+15 \left (x^{2}-1\right )^{\frac {1}{3}}-25}{x^{2}}\right )}{2}+\frac {3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (\frac {45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {2}{3}}-18 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-180 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {1}{3}}-9 \left (x^{2}-1\right )^{\frac {2}{3}}-8 x^{2}-33 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )+15 \left (x^{2}-1\right )^{\frac {1}{3}}+4}{x^{2}}\right )}{2}\)	\(291\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-1+x^2\right )^{2/3}}{x} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{2} - 1\right )}^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + \frac {3}{4} \, {\left (x^{2} - 1\right )}^{\frac {2}{3}} - \frac {1}{4} \, \log \left ({\left (x^{2} - 1\right )}^{\frac {2}{3}} - {\left (x^{2} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} - 1\right )}^{\frac {1}{3}} + 1\right ) \]

[In]

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

[Out]

-1/2*sqrt(3)*arctan(2/3*sqrt(3)*(x^2 - 1)^(1/3) - 1/3*sqrt(3)) + 3/4*(x^2 - 1)^(2/3) - 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 = 0.61 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.44 \[ \int \frac {\left (-1+x^2\right )^{2/3}}{x} \, dx=- \frac {x^{\frac {4}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{2 \Gamma \left (\frac {1}{3}\right )} \]

[In]

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

[Out]

-x**(4/3)*gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), exp_polar(2*I*pi)/x**2)/(2*gamma(1/3))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.73 \[ \int \frac {\left (-1+x^2\right )^{2/3}}{x} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{2} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) + \frac {3}{4} \, {\left (x^{2} - 1\right )}^{\frac {2}{3}} - \frac {1}{4} \, \log \left ({\left (x^{2} - 1\right )}^{\frac {2}{3}} - {\left (x^{2} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} - 1\right )}^{\frac {1}{3}} + 1\right ) \]

[In]

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

[Out]

-1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^2 - 1)^(1/3) - 1)) + 3/4*(x^2 - 1)^(2/3) - 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 = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-1+x^2\right )^{2/3}}{x} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{2} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) + \frac {3}{4} \, {\left (x^{2} - 1\right )}^{\frac {2}{3}} - \frac {1}{4} \, \log \left ({\left (x^{2} - 1\right )}^{\frac {2}{3}} - {\left (x^{2} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left | {\left (x^{2} - 1\right )}^{\frac {1}{3}} + 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.96 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^2\right )^{2/3}}{x} \, dx=\frac {\ln \left (\frac {9\,{\left (x^2-1\right )}^{1/3}}{4}+\frac {9}{4}\right )}{2}+\ln \left (9\,{\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}^2+\frac {9\,{\left (x^2-1\right )}^{1/3}}{4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-\ln \left (9\,{\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}^2+\frac {9\,{\left (x^2-1\right )}^{1/3}}{4}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )+\frac {3\,{\left (x^2-1\right )}^{2/3}}{4} \]

[In]

int((x^2 - 1)^(2/3)/x,x)

[Out]

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