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

Optimal. Leaf size=104 \[ \frac {1}{27} \log \left (\sqrt [3]{x^3-1}-x\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-1}+x}\right )}{9 \sqrt {3}}-\frac {1}{54} \log \left (\sqrt [3]{x^3-1} x+\left (x^3-1\right )^{2/3}+x^2\right )+\frac {1}{18} \sqrt [3]{x^3-1} \left (3 x^5-x^2\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {279, 321, 331, 292, 31, 634, 618, 204, 628} \begin {gather*} \frac {1}{27} \log \left (1-\frac {x}{\sqrt [3]{x^3-1}}\right )+\frac {\tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{9 \sqrt {3}}+\frac {1}{6} \sqrt [3]{x^3-1} x^5-\frac {1}{18} \sqrt [3]{x^3-1} x^2-\frac {1}{54} \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[x^4*(-1 + x^3)^(1/3),x]

[Out]

-1/18*(x^2*(-1 + x^3)^(1/3)) + (x^5*(-1 + x^3)^(1/3))/6 + ArcTan[(1 + (2*x)/(-1 + x^3)^(1/3))/Sqrt[3]]/(9*Sqrt
[3]) + Log[1 - x/(-1 + x^3)^(1/3)]/27 - Log[1 + x^2/(-1 + x^3)^(2/3) + x/(-1 + x^3)^(1/3)]/54

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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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 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 x^4 \sqrt [3]{-1+x^3} \, dx &=\frac {1}{6} x^5 \sqrt [3]{-1+x^3}-\frac {1}{6} \int \frac {x^4}{\left (-1+x^3\right )^{2/3}} \, dx\\ &=-\frac {1}{18} x^2 \sqrt [3]{-1+x^3}+\frac {1}{6} x^5 \sqrt [3]{-1+x^3}-\frac {1}{9} \int \frac {x}{\left (-1+x^3\right )^{2/3}} \, dx\\ &=-\frac {1}{18} x^2 \sqrt [3]{-1+x^3}+\frac {1}{6} x^5 \sqrt [3]{-1+x^3}-\frac {1}{9} \operatorname {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )\\ &=-\frac {1}{18} x^2 \sqrt [3]{-1+x^3}+\frac {1}{6} x^5 \sqrt [3]{-1+x^3}-\frac {1}{27} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )+\frac {1}{27} \operatorname {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )\\ &=-\frac {1}{18} x^2 \sqrt [3]{-1+x^3}+\frac {1}{6} x^5 \sqrt [3]{-1+x^3}+\frac {1}{27} \log \left (1-\frac {x}{\sqrt [3]{-1+x^3}}\right )-\frac {1}{54} \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )+\frac {1}{18} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x}{\sqrt [3]{-1+x^3}}\right )\\ &=-\frac {1}{18} x^2 \sqrt [3]{-1+x^3}+\frac {1}{6} x^5 \sqrt [3]{-1+x^3}+\frac {1}{27} \log \left (1-\frac {x}{\sqrt [3]{-1+x^3}}\right )-\frac {1}{54} \log \left (1+\frac {x^2}{\left (-1+x^3\right )^{2/3}}+\frac {x}{\sqrt [3]{-1+x^3}}\right )-\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x}{\sqrt [3]{-1+x^3}}\right )\\ &=-\frac {1}{18} x^2 \sqrt [3]{-1+x^3}+\frac {1}{6} x^5 \sqrt [3]{-1+x^3}+\frac {\tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3}}+\frac {1}{27} \log \left (1-\frac {x}{\sqrt [3]{-1+x^3}}\right )-\frac {1}{54} \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 = 0.02, size = 54, normalized size = 0.52 \begin {gather*} \frac {x^2 \sqrt [3]{x^3-1} \left (\, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )-\left (1-x^3\right )^{4/3}\right )}{6 \sqrt [3]{1-x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(-1 + x^3)^(1/3),x]

[Out]

(x^2*(-1 + x^3)^(1/3)*(-(1 - x^3)^(4/3) + Hypergeometric2F1[-1/3, 2/3, 5/3, x^3]))/(6*(1 - x^3)^(1/3))

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IntegrateAlgebraic [A]  time = 0.18, size = 104, normalized size = 1.00 \begin {gather*} \frac {1}{18} \sqrt [3]{-1+x^3} \left (-x^2+3 x^5\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{9 \sqrt {3}}+\frac {1}{27} \log \left (-x+\sqrt [3]{-1+x^3}\right )-\frac {1}{54} \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[x^4*(-1 + x^3)^(1/3),x]

[Out]

((-1 + x^3)^(1/3)*(-x^2 + 3*x^5))/18 + ArcTan[(Sqrt[3]*x)/(x + 2*(-1 + x^3)^(1/3))]/(9*Sqrt[3]) + Log[-x + (-1
+ x^3)^(1/3)]/27 - Log[x^2 + x*(-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]/54

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fricas [A]  time = 0.99, size = 96, normalized size = 0.92 \begin {gather*} -\frac {1}{27} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {1}{18} \, {\left (3 \, x^{5} - x^{2}\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}} + \frac {1}{27} \, \log \left (-\frac {x - {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{54} \, \log \left (\frac {x^{2} + {\left (x^{3} - 1\right )}^{\frac {1}{3}} x + {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x^3 - 1)^(1/3))/x) + 1/18*(3*x^5 - x^2)*(x^3 - 1)^(1/3) + 1/2
7*log(-(x - (x^3 - 1)^(1/3))/x) - 1/54*log((x^2 + (x^3 - 1)^(1/3)*x + (x^3 - 1)^(2/3))/x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 - 1)^(1/3)*x^4, x)

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maple [C]  time = 1.50, size = 33, normalized size = 0.32

method	result	size
meijerg	\(\frac {\mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{5} \hypergeom \left (\left [-\frac {1}{3}, \frac {5}{3}\right ], \left [\frac {8}{3}\right ], x^{3}\right )}{5 \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{3}}}\)	\(33\)
risch	\(\frac {x^{2} \left (3 x^{3}-1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}}{18}-\frac {\left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {2}{3}} x^{2} \hypergeom \left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{18 \mathrm {signum}\left (x^{3}-1\right )^{\frac {2}{3}}}\)	\(53\)
trager	\(\frac {x^{2} \left (3 x^{3}-1\right ) \left (x^{3}-1\right )^{\frac {1}{3}}}{18}+\frac {\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}-1\right )^{\frac {2}{3}}+3 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}+2 x^{3}-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2\right )}{27}-\frac {\ln \left (\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x +3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}+4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}-1\right )^{\frac {2}{3}}+3 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}+4 x^{3}-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2\right ) \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{27}-\frac {\ln \left (\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {2}{3}} x +3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-1\right )^{\frac {1}{3}} x^{2}+4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}-1\right )^{\frac {2}{3}}+3 x^{2} \left (x^{3}-1\right )^{\frac {1}{3}}+4 x^{3}-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2\right )}{27}\)	\(306\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*signum(x^3-1)^(1/3)/(-signum(x^3-1))^(1/3)*x^5*hypergeom([-1/3,5/3],[8/3],x^3)

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maxima [A]  time = 0.46, size = 121, normalized size = 1.16 \begin {gather*} -\frac {1}{27} \, \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 {\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} - 1\right )}^{\frac {4}{3}}}{x^{4}}}{18 \, {\left (\frac {2 \, {\left (x^{3} - 1\right )}}{x^{3}} - \frac {{\left (x^{3} - 1\right )}^{2}}{x^{6}} - 1\right )}} - \frac {1}{54} \, \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 {1}{27} \, \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^4*(x^3-1)^(1/3),x, algorithm="maxima")

[Out]

-1/27*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 - 1)^(1/3)/x + 1)) - 1/18*(2*(x^3 - 1)^(1/3)/x + (x^3 - 1)^(4/3)/x^4)
/(2*(x^3 - 1)/x^3 - (x^3 - 1)^2/x^6 - 1) - 1/54*log((x^3 - 1)^(1/3)/x + (x^3 - 1)^(2/3)/x^2 + 1) + 1/27*log((x
^3 - 1)^(1/3)/x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,{\left (x^3-1\right )}^{1/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(x^3 - 1)^(1/3),x)

[Out]

int(x^4*(x^3 - 1)^(1/3), x)

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sympy [C]  time = 1.00, size = 32, normalized size = 0.31 \begin {gather*} \frac {x^{5} e^{\frac {i \pi }{3}} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**5*exp(I*pi/3)*gamma(5/3)*hyper((-1/3, 5/3), (8/3,), x**3)/(3*gamma(8/3))

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