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

Optimal. Leaf size=114 \[ -\frac {3 \left (x^3-x^2\right )^{2/3}}{(x-1) x}-\log \left (\sqrt [3]{x^3-x^2}-x\right )+\frac {1}{2} \log \left (x^2+\sqrt [3]{x^3-x^2} x+\left (x^3-x^2\right )^{2/3}\right )+\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-x^2}+x}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 151, normalized size of antiderivative = 1.32, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2056, 47, 59} \begin {gather*} -\frac {3 x}{\sqrt [3]{x^3-x^2}}-\frac {3 \sqrt [3]{x-1} x^{2/3} \log \left (\frac {\sqrt [3]{x-1}}{\sqrt [3]{x}}-1\right )}{2 \sqrt [3]{x^3-x^2}}-\frac {\sqrt [3]{x-1} x^{2/3} \log (x)}{2 \sqrt [3]{x^3-x^2}}-\frac {\sqrt {3} \sqrt [3]{x-1} x^{2/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x-1}}{\sqrt {3} \sqrt [3]{x}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{x^3-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 59

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, -Simp[(Sqrt
[3]*q*ArcTan[(2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3)) + 1/Sqrt[3]])/d, x] + (-Simp[(3*q*Log[(q*(a + b*x
)^(1/3))/(c + d*x)^(1/3) - 1])/(2*d), x] - Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[
b*c - a*d, 0] && PosQ[d/b]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 33, normalized size = 0.29 \begin {gather*} -\frac {3 x^{2/3} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};1-x\right )}{\sqrt [3]{(x-1) x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.23, size = 114, normalized size = 1.00 \begin {gather*} -\frac {3 \left (-x^2+x^3\right )^{2/3}}{(-1+x) x}+\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x^2+x^3}}\right )-\log \left (-x+\sqrt [3]{-x^2+x^3}\right )+\frac {1}{2} \log \left (x^2+x \sqrt [3]{-x^2+x^3}+\left (-x^2+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 3.12, size = 137, normalized size = 1.20 \begin {gather*} -\frac {2 \, \sqrt {3} {\left (x^{2} - x\right )} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) + 2 \, {\left (x^{2} - x\right )} \log \left (-\frac {x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - {\left (x^{2} - x\right )} \log \left (\frac {x^{2} + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 6 \, {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{2 \, {\left (x^{2} - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 74, normalized size = 0.65 \begin {gather*} -\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {3}{{\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}}} + \frac {1}{2} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.61, size = 27, normalized size = 0.24

method	result	size
meijerg	\(-\frac {3 \left (-\mathrm {signum}\left (-1+x \right )\right )^{\frac {1}{3}} x^{\frac {4}{3}} \hypergeom \left (\left [\frac {4}{3}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], x\right )}{4 \mathrm {signum}\left (-1+x \right )^{\frac {1}{3}}}\)	\(27\)
risch	\(-\frac {3 x}{\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}}+\frac {3 \left (-\mathrm {signum}\left (-1+x \right )\right )^{\frac {1}{3}} x^{\frac {1}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x\right )}{\mathrm {signum}\left (-1+x \right )^{\frac {1}{3}}}\)	\(40\)
trager	\(-\frac {3 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}}{\left (-1+x \right ) x}-\ln \left (-\frac {9 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right )^{2} x^{2}-18 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right )^{2} x +144 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}-90 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}} x -60 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) x^{2}+78 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) x -60 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}-36 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}+100 x^{2}-60 x}{x}\right )+\ln \left (-\frac {18 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right )^{2} x^{2}-36 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right )^{2} x -72 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+27 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}} x +33 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) x^{2}+18 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) x +18 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+30 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}-40 x^{2}+10 x}{x}\right )-\frac {3 \ln \left (-\frac {18 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right )^{2} x^{2}-36 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right )^{2} x -72 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+27 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}} x +33 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) x^{2}+18 \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right ) x +18 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+30 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}-40 x^{2}+10 x}{x}\right ) \RootOf \left (9 \textit {\_Z}^{2}-6 \textit {\_Z} +4\right )}{2}\)	\(503\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4/signum(-1+x)^(1/3)*(-signum(-1+x))^(1/3)*x^(4/3)*hypergeom([4/3,4/3],[7/3],x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (x^3-x^2\right )}^{1/3}\,\left (x-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((x^3 - x^2)^(1/3)*(x - 1)),x)

[Out]

int(x/((x^3 - x^2)^(1/3)*(x - 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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