3.13.61 \(\int \frac {\sqrt [3]{-1+x^3}}{x^4} \, dx\)

Optimal. Leaf size=92 \[ -\frac {\sqrt [3]{x^3-1}}{3 x^3}+\frac {1}{9} \log \left (\sqrt [3]{x^3-1}+1\right )-\frac {1}{18} \log \left (\left (x^3-1\right )^{2/3}-\sqrt [3]{x^3-1}+1\right )-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x^3-1}}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 0.74, 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, 47, 58, 618, 204, 31} \begin {gather*} -\frac {\sqrt [3]{x^3-1}}{3 x^3}+\frac {1}{6} \log \left (\sqrt [3]{x^3-1}+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 \sqrt [3]{x^3-1}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\log (x)}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^3)^(1/3)/x^4,x]

[Out]

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

Rule 31

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

Rule 47

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

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[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
] && NegQ[(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 {\sqrt [3]{-1+x^3}}{x^4} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt [3]{-1+x}}{x^2} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [3]{-1+x^3}}{3 x^3}+\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{(-1+x)^{2/3} x} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [3]{-1+x^3}}{3 x^3}-\frac {\log (x)}{6}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\sqrt [3]{-1+x^3}\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\sqrt [3]{-1+x^3}\right )\\ &=-\frac {\sqrt [3]{-1+x^3}}{3 x^3}-\frac {\log (x)}{6}+\frac {1}{6} \log \left (1+\sqrt [3]{-1+x^3}\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \sqrt [3]{-1+x^3}\right )\\ &=-\frac {\sqrt [3]{-1+x^3}}{3 x^3}-\frac {\tan ^{-1}\left (\frac {1-2 \sqrt [3]{-1+x^3}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\log (x)}{6}+\frac {1}{6} \log \left (1+\sqrt [3]{-1+x^3}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 28, normalized size = 0.30 \begin {gather*} \frac {1}{4} \left (x^3-1\right )^{4/3} \, _2F_1\left (\frac {4}{3},2;\frac {7}{3};1-x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^3)^(1/3)/x^4,x]

[Out]

((-1 + x^3)^(4/3)*Hypergeometric2F1[4/3, 2, 7/3, 1 - x^3])/4

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IntegrateAlgebraic [A]  time = 0.14, size = 92, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [3]{-1+x^3}}{3 x^3}-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-1+x^3}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{9} \log \left (1+\sqrt [3]{-1+x^3}\right )-\frac {1}{18} \log \left (1-\sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^3)^(1/3)/x^4,x]

[Out]

-1/3*(-1 + x^3)^(1/3)/x^3 - ArcTan[1/Sqrt[3] - (2*(-1 + x^3)^(1/3))/Sqrt[3]]/(3*Sqrt[3]) + Log[1 + (-1 + x^3)^
(1/3)]/9 - Log[1 - (-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]/18

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fricas [A]  time = 0.45, size = 81, normalized size = 0.88 \begin {gather*} \frac {2 \, \sqrt {3} x^{3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) - x^{3} \log \left ({\left (x^{3} - 1\right )}^{\frac {2}{3}} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) + 2 \, x^{3} \log \left ({\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) - 6 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{18 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)^(1/3)/x^4,x, algorithm="fricas")

[Out]

1/18*(2*sqrt(3)*x^3*arctan(2/3*sqrt(3)*(x^3 - 1)^(1/3) - 1/3*sqrt(3)) - x^3*log((x^3 - 1)^(2/3) - (x^3 - 1)^(1
/3) + 1) + 2*x^3*log((x^3 - 1)^(1/3) + 1) - 6*(x^3 - 1)^(1/3))/x^3

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giac [A]  time = 0.35, size = 69, normalized size = 0.75 \begin {gather*} \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) - \frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{3 \, x^{3}} - \frac {1}{18} \, \log \left ({\left (x^{3} - 1\right )}^{\frac {2}{3}} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{9} \, \log \left ({\left | {\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)^(1/3)/x^4,x, algorithm="giac")

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 - 1)^(1/3) - 1)) - 1/3*(x^3 - 1)^(1/3)/x^3 - 1/18*log((x^3 - 1)^(2/3) -
(x^3 - 1)^(1/3) + 1) + 1/9*log(abs((x^3 - 1)^(1/3) + 1))

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maple [C]  time = 3.21, size = 75, normalized size = 0.82

method	result	size
meijerg	\(\frac {\mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{3}} \left (-\frac {\Gamma \left (\frac {2}{3}\right ) x^{3} \hypergeom \left (\left [1, 1, \frac {5}{3}\right ], \left [2, 3\right ], x^{3}\right )}{3}-\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}-1+3 \ln \relax (x )+i \pi \right ) \Gamma \left (\frac {2}{3}\right )-\frac {3 \Gamma \left (\frac {2}{3}\right )}{x^{3}}\right )}{9 \Gamma \left (\frac {2}{3}\right ) \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{3}}}\)	\(75\)
risch	\(-\frac {\left (x^{3}-1\right )^{\frac {1}{3}}}{3 x^{3}}+\frac {\left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {2}{3}} \left (\frac {2 \Gamma \left (\frac {2}{3}\right ) x^{3} \hypergeom \left (\left [1, 1, \frac {5}{3}\right ], \left [2, 2\right ], x^{3}\right )}{3}+\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+3 \ln \relax (x )+i \pi \right ) \Gamma \left (\frac {2}{3}\right )\right )}{9 \Gamma \left (\frac {2}{3}\right ) \mathrm {signum}\left (x^{3}-1\right )^{\frac {2}{3}}}\)	\(79\)
trager	\(-\frac {\left (x^{3}-1\right )^{\frac {1}{3}}}{3 x^{3}}+\frac {\ln \left (\frac {-13504 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )^{2} x^{3}-25008 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) x^{3}+44016 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}}-11543 x^{3}-14247 \left (x^{3}-1\right )^{\frac {2}{3}}+157992 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}+108032 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )^{2}+5502 \left (x^{3}-1\right )^{\frac {1}{3}}+127480 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )+21437}{x^{3}}\right )}{9}-\frac {\ln \left (-\frac {-105536 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )^{2} x^{3}+77464 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) x^{3}+44016 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}}+1266 x^{3}+19749 \left (x^{3}-1\right )^{\frac {2}{3}}-113976 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}+844288 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )^{2}+5502 \left (x^{3}-1\right )^{\frac {1}{3}}-52456 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )-1055}{x^{3}}\right )}{9}-\frac {8 \ln \left (-\frac {-105536 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )^{2} x^{3}+77464 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) x^{3}+44016 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}}+1266 x^{3}+19749 \left (x^{3}-1\right )^{\frac {2}{3}}-113976 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}+844288 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )^{2}+5502 \left (x^{3}-1\right )^{\frac {1}{3}}-52456 \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )-1055}{x^{3}}\right ) \RootOf \left (64 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )}{9}\)	\(429\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-1)^(1/3)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.62, size = 68, normalized size = 0.74 \begin {gather*} \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) - \frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{3 \, x^{3}} - \frac {1}{18} \, \log \left ({\left (x^{3} - 1\right )}^{\frac {2}{3}} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)^(1/3)/x^4,x, algorithm="maxima")

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 - 1)^(1/3) - 1)) - 1/3*(x^3 - 1)^(1/3)/x^3 - 1/18*log((x^3 - 1)^(2/3) -
(x^3 - 1)^(1/3) + 1) + 1/9*log((x^3 - 1)^(1/3) + 1)

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mupad [B]  time = 0.89, size = 80, normalized size = 0.87 \begin {gather*} \frac {\ln \left (\frac {{\left (x^3-1\right )}^{1/3}}{9}+\frac {1}{9}\right )}{9}+\ln \left ({\left (x^3-1\right )}^{1/3}-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )-\frac {{\left (x^3-1\right )}^{1/3}}{3\,x^3}-\ln \left (\frac {1}{2}-{\left (x^3-1\right )}^{1/3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 - 1)^(1/3)/x^4,x)

[Out]

log((x^3 - 1)^(1/3)/9 + 1/9)/9 + log((3^(1/2)*1i)/2 + (x^3 - 1)^(1/3) - 1/2)*((3^(1/2)*1i)/18 - 1/18) - (x^3 -
1)^(1/3)/(3*x^3) - log((3^(1/2)*1i)/2 - (x^3 - 1)^(1/3) + 1/2)*((3^(1/2)*1i)/18 + 1/18)

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sympy [C]  time = 0.95, size = 34, normalized size = 0.37 \begin {gather*} - \frac {\Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{3}}} \right )}}{3 x^{2} \Gamma \left (\frac {5}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-1)**(1/3)/x**4,x)

[Out]

-gamma(2/3)*hyper((-1/3, 2/3), (5/3,), exp_polar(2*I*pi)/x**3)/(3*x**2*gamma(5/3))

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