3.15.15 \(\int \frac {\sqrt [3]{-1+x^3} (-1+2 x^3)}{x^5} \, dx\)

Optimal. Leaf size=101 \[ -\frac {2}{3} \log \left (\sqrt [3]{x^3-1}-x\right )-\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-1}+x}\right )}{\sqrt {3}}+\frac {\sqrt [3]{x^3-1} \left (1-9 x^3\right )}{4 x^4}+\frac {1}{3} \log \left (\sqrt [3]{x^3-1} x+\left (x^3-1\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 109, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {451, 277, 331, 292, 31, 634, 618, 204, 628} \begin {gather*} -\frac {2 \sqrt [3]{x^3-1}}{x}-\frac {2}{3} \log \left (1-\frac {x}{\sqrt [3]{x^3-1}}\right )-\frac {2 \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\left (x^3-1\right )^{4/3}}{4 x^4}+\frac {1}{3} \log \left (\frac {x}{\sqrt [3]{x^3-1}}+\frac {x^2}{\left (x^3-1\right )^{2/3}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^3)^(1/3)*(-1 + 2*x^3))/x^5,x]

[Out]

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

Rule 31

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

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 451

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

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]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{-1+x^3} \left (-1+2 x^3\right )}{x^5} \, dx &=-\frac {\left (-1+x^3\right )^{4/3}}{4 x^4}+2 \int \frac {\sqrt [3]{-1+x^3}}{x^2} \, dx\\ &=-\frac {2 \sqrt [3]{-1+x^3}}{x}-\frac {\left (-1+x^3\right )^{4/3}}{4 x^4}+2 \int \frac {x}{\left (-1+x^3\right )^{2/3}} \, dx\\ &=-\frac {2 \sqrt [3]{-1+x^3}}{x}-\frac {\left (-1+x^3\right )^{4/3}}{4 x^4}+2 \operatorname {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )\\ &=-\frac {2 \sqrt [3]{-1+x^3}}{x}-\frac {\left (-1+x^3\right )^{4/3}}{4 x^4}+\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )\\ &=-\frac {2 \sqrt [3]{-1+x^3}}{x}-\frac {\left (-1+x^3\right )^{4/3}}{4 x^4}-\frac {2}{3} \log \left (1-\frac {x}{\sqrt [3]{-1+x^3}}\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )-\operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )\\ &=-\frac {2 \sqrt [3]{-1+x^3}}{x}-\frac {\left (-1+x^3\right )^{4/3}}{4 x^4}-\frac {2}{3} \log \left (1-\frac {x}{\sqrt [3]{-1+x^3}}\right )+\frac {1}{3} \log \left (1+\frac {x^2}{\left (-1+x^3\right )^{2/3}}+\frac {x}{\sqrt [3]{-1+x^3}}\right )+2 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x}{\sqrt [3]{-1+x^3}}\right )\\ &=-\frac {2 \sqrt [3]{-1+x^3}}{x}-\frac {\left (-1+x^3\right )^{4/3}}{4 x^4}-\frac {2 \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \log \left (1-\frac {x}{\sqrt [3]{-1+x^3}}\right )+\frac {1}{3} \log \left (1+\frac {x^2}{\left (-1+x^3\right )^{2/3}}+\frac {x}{\sqrt [3]{-1+x^3}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 52, normalized size = 0.51 \begin {gather*} \frac {\sqrt [3]{x^3-1} \left (-\frac {8 x^3 \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};x^3\right )}{\sqrt [3]{1-x^3}}-x^3+1\right )}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^3)^(1/3)*(-1 + 2*x^3))/x^5,x]

[Out]

((-1 + x^3)^(1/3)*(1 - x^3 - (8*x^3*Hypergeometric2F1[-1/3, -1/3, 2/3, x^3])/(1 - x^3)^(1/3)))/(4*x^4)

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IntegrateAlgebraic [A]  time = 0.15, size = 101, normalized size = 1.00 \begin {gather*} \frac {\left (1-9 x^3\right ) \sqrt [3]{-1+x^3}}{4 x^4}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{\sqrt {3}}-\frac {2}{3} \log \left (-x+\sqrt [3]{-1+x^3}\right )+\frac {1}{3} \log \left (x^2+x \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)*(-1 + 2*x^3))/x^5,x]

[Out]

((1 - 9*x^3)*(-1 + x^3)^(1/3))/(4*x^4) - (2*ArcTan[(Sqrt[3]*x)/(x + 2*(-1 + x^3)^(1/3))])/Sqrt[3] - (2*Log[-x
+ (-1 + x^3)^(1/3)])/3 + Log[x^2 + x*(-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]/3

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fricas [A]  time = 0.69, size = 112, normalized size = 1.11 \begin {gather*} -\frac {8 \, \sqrt {3} x^{4} \arctan \left (-\frac {25382 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{2} - 13720 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (5831 \, x^{3} - 7200\right )}}{58653 \, x^{3} - 8000}\right ) + 4 \, x^{4} \log \left (-3 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{2} + 3 \, {\left (x^{3} - 1\right )}^{\frac {2}{3}} x + 1\right ) + 3 \, {\left (9 \, x^{3} - 1\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)^(1/3)*(2*x^3-1)/x^5,x, algorithm="fricas")

[Out]

-1/12*(8*sqrt(3)*x^4*arctan(-(25382*sqrt(3)*(x^3 - 1)^(1/3)*x^2 - 13720*sqrt(3)*(x^3 - 1)^(2/3)*x + sqrt(3)*(5
831*x^3 - 7200))/(58653*x^3 - 8000)) + 4*x^4*log(-3*(x^3 - 1)^(1/3)*x^2 + 3*(x^3 - 1)^(2/3)*x + 1) + 3*(9*x^3
- 1)*(x^3 - 1)^(1/3))/x^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{3} - 1\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)^(1/3)*(2*x^3-1)/x^5,x, algorithm="giac")

[Out]

integrate((2*x^3 - 1)*(x^3 - 1)^(1/3)/x^5, x)

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maple [C]  time = 2.45, size = 57, normalized size = 0.56

method	result	size
risch	\(-\frac {9 x^{6}-10 x^{3}+1}{4 x^{4} \left (x^{3}-1\right )^{\frac {2}{3}}}+\frac {\left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {2}{3}} x^{2} \hypergeom \left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{\mathrm {signum}\left (x^{3}-1\right )^{\frac {2}{3}}}\)	\(57\)
meijerg	\(-\frac {2 \mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{3}} \hypergeom \left (\left [-\frac {1}{3}, -\frac {1}{3}\right ], \left [\frac {2}{3}\right ], x^{3}\right )}{\left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{3}} x}+\frac {\mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{3}} \left (-x^{3}+1\right )^{\frac {4}{3}}}{4 \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{3}} x^{4}}\)	\(66\)
trager	\(-\frac {\left (9 x^{3}-1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}}{4 x^{4}}+\frac {2 \ln \left (2916347648 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x^{3}[phone] \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x [phone] \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}[phone] \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{3}-1508552373 x \left (x^{3}-1\right )^{\frac {2}{3}}-1508552373 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}-1497160390 x^{3}-23330781184 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2}[phone] \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )[phone]\right )}{3}-\frac {32 \ln \left (2916347648 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x^{3}[phone] \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x [phone] \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}[phone] \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{3}-1508552373 x \left (x^{3}-1\right )^{\frac {2}{3}}-1508552373 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}-1497160390 x^{3}-23330781184 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2}[phone] \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )[phone]\right ) \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )}{3}+\frac {32 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \ln \left (2916347648 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x^{3}-3103405824 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x -3103405824 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}-3285677552 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{3}-1314589509 x \left (x^{3}-1\right )^{\frac {2}{3}}-1314589509 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}-1303197526 x^{3}-23330781184 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2}-4518501616 \RootOf \left (256 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )[phone]\right )}{3}\)	\(457\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-1)^(1/3)*(2*x^3-1)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*(9*x^6-10*x^3+1)/x^4/(x^3-1)^(2/3)+1/signum(x^3-1)^(2/3)*(-signum(x^3-1))^(2/3)*x^2*hypergeom([2/3,2/3],[
5/3],x^3)

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maxima [A]  time = 0.44, size = 93, normalized size = 0.92 \begin {gather*} \frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - \frac {{\left (x^{3} - 1\right )}^{\frac {4}{3}}}{4 \, x^{4}} + \frac {1}{3} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {2}{3} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)^(1/3)*(2*x^3-1)/x^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.12, size = 55, normalized size = 0.54 \begin {gather*} \frac {{\left (x^3-1\right )}^{1/3}-x^3\,{\left (x^3-1\right )}^{1/3}}{4\,x^4}-\frac {2\,{\left (x^3-1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},-\frac {1}{3};\ \frac {2}{3};\ x^3\right )}{x\,{\left (1-x^3\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3 - 1)^(1/3)*(2*x^3 - 1))/x^5,x)

[Out]

((x^3 - 1)^(1/3) - x^3*(x^3 - 1)^(1/3))/(4*x^4) - (2*(x^3 - 1)^(1/3)*hypergeom([-1/3, -1/3], 2/3, x^3))/(x*(1
- x^3)^(1/3))

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sympy [C]  time = 2.35, size = 167, normalized size = 1.65 \begin {gather*} - \begin {cases} \frac {\sqrt [3]{-1 + \frac {1}{x^{3}}} e^{- \frac {2 i \pi }{3}} \Gamma \left (- \frac {4}{3}\right )}{3 \Gamma \left (- \frac {1}{3}\right )} - \frac {\sqrt [3]{-1 + \frac {1}{x^{3}}} e^{- \frac {2 i \pi }{3}} \Gamma \left (- \frac {4}{3}\right )}{3 x^{3} \Gamma \left (- \frac {1}{3}\right )} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {\sqrt [3]{1 - \frac {1}{x^{3}}} \Gamma \left (- \frac {4}{3}\right )}{3 \Gamma \left (- \frac {1}{3}\right )} + \frac {\sqrt [3]{1 - \frac {1}{x^{3}}} \Gamma \left (- \frac {4}{3}\right )}{3 x^{3} \Gamma \left (- \frac {1}{3}\right )} & \text {otherwise} \end {cases} + \frac {2 e^{\frac {i \pi }{3}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-1)**(1/3)*(2*x**3-1)/x**5,x)

[Out]

-Piecewise(((-1 + x**(-3))**(1/3)*exp(-2*I*pi/3)*gamma(-4/3)/(3*gamma(-1/3)) - (-1 + x**(-3))**(1/3)*exp(-2*I*
pi/3)*gamma(-4/3)/(3*x**3*gamma(-1/3)), 1/Abs(x**3) > 1), (-(1 - 1/x**3)**(1/3)*gamma(-4/3)/(3*gamma(-1/3)) +
(1 - 1/x**3)**(1/3)*gamma(-4/3)/(3*x**3*gamma(-1/3)), True)) + 2*exp(I*pi/3)*gamma(-1/3)*hyper((-1/3, -1/3), (
2/3,), x**3)/(3*x*gamma(2/3))

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