3.12.77 \(\int \frac {1}{x^3 (1+x^2)^{2/3}} \, dx\)

Optimal. Leaf size=87 \[ -\frac {\sqrt [3]{x^2+1}}{2 x^2}-\frac {1}{3} \log \left (\sqrt [3]{x^2+1}-1\right )+\frac {1}{6} \log \left (\left (x^2+1\right )^{2/3}+\sqrt [3]{x^2+1}+1\right )+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+1}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {266, 51, 57, 618, 204, 31} \begin {gather*} -\frac {\sqrt [3]{x^2+1}}{2 x^2}-\frac {1}{2} \log \left (1-\sqrt [3]{x^2+1}\right )+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+1}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\log (x)}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 + x^2)^(2/3)),x]

[Out]

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

Rule 31

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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^3 \left (1+x^2\right )^{2/3}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 (1+x)^{2/3}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [3]{1+x^2}}{2 x^2}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x (1+x)^{2/3}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [3]{1+x^2}}{2 x^2}+\frac {\log (x)}{3}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1+x^2}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1+x^2}\right )\\ &=-\frac {\sqrt [3]{1+x^2}}{2 x^2}+\frac {\log (x)}{3}-\frac {1}{2} \log \left (1-\sqrt [3]{1+x^2}\right )-\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1+x^2}\right )\\ &=-\frac {\sqrt [3]{1+x^2}}{2 x^2}+\frac {\tan ^{-1}\left (\frac {1+2 \sqrt [3]{1+x^2}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\log (x)}{3}-\frac {1}{2} \log \left (1-\sqrt [3]{1+x^2}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 26, normalized size = 0.30 \begin {gather*} \frac {3}{2} \sqrt [3]{x^2+1} \, _2F_1\left (\frac {1}{3},2;\frac {4}{3};x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(1 + x^2)^(2/3)),x]

[Out]

(3*(1 + x^2)^(1/3)*Hypergeometric2F1[1/3, 2, 4/3, 1 + x^2])/2

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IntegrateAlgebraic [A]  time = 0.08, size = 87, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [3]{1+x^2}}{2 x^2}+\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^2}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (-1+\sqrt [3]{1+x^2}\right )+\frac {1}{6} \log \left (1+\sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(1 + x^2)^(1/3)/x^2 + ArcTan[1/Sqrt[3] + (2*(1 + x^2)^(1/3))/Sqrt[3]]/Sqrt[3] - Log[-1 + (1 + x^2)^(1/3)]
/3 + Log[1 + (1 + x^2)^(1/3) + (1 + x^2)^(2/3)]/6

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fricas [A]  time = 0.55, size = 78, normalized size = 0.90 \begin {gather*} \frac {2 \, \sqrt {3} x^{2} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{2} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + x^{2} \log \left ({\left (x^{2} + 1\right )}^{\frac {2}{3}} + {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) - 2 \, x^{2} \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) - 3 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}}}{6 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*sqrt(3)*x^2*arctan(2/3*sqrt(3)*(x^2 + 1)^(1/3) + 1/3*sqrt(3)) + x^2*log((x^2 + 1)^(2/3) + (x^2 + 1)^(1/
3) + 1) - 2*x^2*log((x^2 + 1)^(1/3) - 1) - 3*(x^2 + 1)^(1/3))/x^2

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giac [A]  time = 0.29, size = 66, normalized size = 0.76 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{2} + 1\right )}^{\frac {1}{3}}}{2 \, x^{2}} + \frac {1}{6} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {2}{3}} + {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{3} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^2 + 1)^(1/3) + 1)) - 1/2*(x^2 + 1)^(1/3)/x^2 + 1/6*log((x^2 + 1)^(2/3) +
(x^2 + 1)^(1/3) + 1) - 1/3*log((x^2 + 1)^(1/3) - 1)

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maple [C]  time = 1.96, size = 55, normalized size = 0.63

method	result	size
meijerg	\(\frac {\frac {5 \Gamma \left (\frac {2}{3}\right ) x^{2} \hypergeom \left (\left [1, 1, \frac {8}{3}\right ], \left [2, 3\right ], -x^{2}\right )}{9}-\frac {2 \left (\frac {1}{2}+\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+2 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )}{3}-\frac {\Gamma \left (\frac {2}{3}\right )}{x^{2}}}{2 \Gamma \left (\frac {2}{3}\right )}\)	\(55\)
risch	\(-\frac {\left (x^{2}+1\right )^{\frac {1}{3}}}{2 x^{2}}-\frac {-\frac {2 \Gamma \left (\frac {2}{3}\right ) x^{2} \hypergeom \left (\left [1, 1, \frac {5}{3}\right ], \left [2, 2\right ], -x^{2}\right )}{3}+\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+2 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )}{3 \Gamma \left (\frac {2}{3}\right )}\)	\(59\)
trager	\(-\frac {\left (x^{2}+1\right )^{\frac {1}{3}}}{2 x^{2}}-\frac {\ln \left (-\frac {36 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}-18 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}+9 \left (x^{2}+1\right )^{\frac {2}{3}}-144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}}-36 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}-10 x^{2}+15 \left (x^{2}+1\right )^{\frac {1}{3}}+60 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )-25}{x^{2}}\right )}{3}+\frac {\ln \left (\frac {180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}-96 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}-24 \left (x^{2}+1\right )^{\frac {2}{3}}+54 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}}-180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}+3 x^{2}+15 \left (x^{2}+1\right )^{\frac {1}{3}}-114 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )+4}{x^{2}}\right )}{3}-2 \ln \left (\frac {180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}-96 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}-24 \left (x^{2}+1\right )^{\frac {2}{3}}+54 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}}-180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}+3 x^{2}+15 \left (x^{2}+1\right )^{\frac {1}{3}}-114 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )+4}{x^{2}}\right ) \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )\)	\(428\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/GAMMA(2/3)*(5/9*GAMMA(2/3)*x^2*hypergeom([1,1,8/3],[2,3],-x^2)-2/3*(1/2+1/6*Pi*3^(1/2)-3/2*ln(3)+2*ln(x))*
GAMMA(2/3)-GAMMA(2/3)/x^2)

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maxima [A]  time = 0.57, size = 66, normalized size = 0.76 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{2} + 1\right )}^{\frac {1}{3}}}{2 \, x^{2}} + \frac {1}{6} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {2}{3}} + {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{3} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^2 + 1)^(1/3) + 1)) - 1/2*(x^2 + 1)^(1/3)/x^2 + 1/6*log((x^2 + 1)^(2/3) +
(x^2 + 1)^(1/3) + 1) - 1/3*log((x^2 + 1)^(1/3) - 1)

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mupad [B]  time = 0.89, size = 80, normalized size = 0.92 \begin {gather*} -\frac {\ln \left ({\left (x^2+1\right )}^{1/3}-1\right )}{3}-\frac {{\left (x^2+1\right )}^{1/3}}{2\,x^2}-\ln \left (3\,{\left (x^2+1\right )}^{1/3}+\frac {3}{2}-\frac {\sqrt {3}\,3{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\ln \left (3\,{\left (x^2+1\right )}^{1/3}+\frac {3}{2}+\frac {\sqrt {3}\,3{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(x^2 + 1)^(2/3)),x)

[Out]

log((3^(1/2)*3i)/2 + 3*(x^2 + 1)^(1/3) + 3/2)*((3^(1/2)*1i)/6 + 1/6) - (x^2 + 1)^(1/3)/(2*x^2) - log(3*(x^2 +
1)^(1/3) - (3^(1/2)*3i)/2 + 3/2)*((3^(1/2)*1i)/6 - 1/6) - log((x^2 + 1)^(1/3) - 1)/3

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sympy [C]  time = 0.92, size = 32, normalized size = 0.37 \begin {gather*} - \frac {\Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{2 x^{\frac {10}{3}} \Gamma \left (\frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(5/3)*hyper((2/3, 5/3), (8/3,), exp_polar(I*pi)/x**2)/(2*x**(10/3)*gamma(8/3))

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