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

Optimal. Leaf size=106 \[ \sqrt [3]{x^3-x^2}+\frac {1}{3} \log \left (\sqrt [3]{x^3-x^2}-x\right )-\frac {1}{6} \log \left (x^2+\sqrt [3]{x^3-x^2} x+\left (x^3-x^2\right )^{2/3}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-x^2}+x}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.07, antiderivative size = 149, normalized size of antiderivative = 1.41, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2021, 2032, 59} \begin {gather*} \sqrt [3]{x^3-x^2}+\frac {(x-1)^{2/3} x^{4/3} \log \left (\frac {\sqrt [3]{x}}{\sqrt [3]{x-1}}-1\right )}{2 \left (x^3-x^2\right )^{2/3}}+\frac {(x-1)^{2/3} x^{4/3} \log (x-1)}{6 \left (x^3-x^2\right )^{2/3}}+\frac {(x-1)^{2/3} x^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{x-1}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} \left (x^3-x^2\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^2 + x^3)^(1/3)/x,x]

[Out]

(-x^2 + x^3)^(1/3) + ((-1 + x)^(2/3)*x^(4/3)*ArcTan[1/Sqrt[3] + (2*x^(1/3))/(Sqrt[3]*(-1 + x)^(1/3))])/(Sqrt[3
]*(-x^2 + x^3)^(2/3)) + ((-1 + x)^(2/3)*x^(4/3)*Log[-1 + x^(1/3)/(-1 + x)^(1/3)])/(2*(-x^2 + x^3)^(2/3)) + ((-
1 + x)^(2/3)*x^(4/3)*Log[-1 + x])/(6*(-x^2 + x^3)^(2/3))

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 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{-x^2+x^3}}{x} \, dx &=\sqrt [3]{-x^2+x^3}-\frac {1}{3} \int \frac {x}{\left (-x^2+x^3\right )^{2/3}} \, dx\\ &=\sqrt [3]{-x^2+x^3}-\frac {\left ((-1+x)^{2/3} x^{4/3}\right ) \int \frac {1}{(-1+x)^{2/3} \sqrt [3]{x}} \, dx}{3 \left (-x^2+x^3\right )^{2/3}}\\ &=\sqrt [3]{-x^2+x^3}+\frac {(-1+x)^{2/3} x^{4/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{-1+x}}\right )}{\sqrt {3} \left (-x^2+x^3\right )^{2/3}}+\frac {(-1+x)^{2/3} x^{4/3} \log \left (-1+\frac {\sqrt [3]{x}}{\sqrt [3]{-1+x}}\right )}{2 \left (-x^2+x^3\right )^{2/3}}+\frac {(-1+x)^{2/3} x^{4/3} \log (-1+x)}{6 \left (-x^2+x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.33 \begin {gather*} \frac {3 \left ((x-1) x^2\right )^{4/3} \, _2F_1\left (\frac {1}{3},\frac {4}{3};\frac {7}{3};1-x\right )}{4 x^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^2 + x^3)^(1/3)/x,x]

[Out]

(3*((-1 + x)*x^2)^(4/3)*Hypergeometric2F1[1/3, 4/3, 7/3, 1 - x])/(4*x^(8/3))

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IntegrateAlgebraic [A]  time = 0.21, size = 106, normalized size = 1.00 \begin {gather*} \sqrt [3]{x^3-x^2}+\frac {1}{3} \log \left (\sqrt [3]{x^3-x^2}-x\right )-\frac {1}{6} \log \left (x^2+\sqrt [3]{x^3-x^2} x+\left (x^3-x^2\right )^{2/3}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-x^2}+x}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 103, normalized size = 0.97 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} + \frac {1}{3} \, \log \left (-\frac {x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{6} \, \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 74, normalized size = 0.70 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + x {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - \frac {1}{6} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \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^3-x^2)^(1/3)/x,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.50, size = 434, normalized size = 4.09 \begin {gather*} \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+\frac {\left (\frac {\ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-2 x^{2}+x \right )^{\frac {2}{3}}-24 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-2 x^{2}+x \right )^{\frac {1}{3}} x -3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +10 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-9 \left (x^{3}-2 x^{2}+x \right )^{\frac {2}{3}}+24 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-2 x^{2}+x \right )^{\frac {1}{3}}-15 \left (x^{3}-2 x^{2}+x \right )^{\frac {1}{3}} x +2 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-23 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x +25 x^{2}+15 \left (x^{3}-2 x^{2}+x \right )^{\frac {1}{3}}+13 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-40 x +15}{-1+x}\right )}{3}+\frac {\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-2 x^{2}+x \right )^{\frac {2}{3}}+9 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-2 x^{2}+x \right )^{\frac {1}{3}} x -15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x -19 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+24 \left (x^{3}-2 x^{2}+x \right )^{\frac {2}{3}}-9 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}-2 x^{2}+x \right )^{\frac {1}{3}}-15 \left (x^{3}-2 x^{2}+x \right )^{\frac {1}{3}} x +10 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+22 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x -4 x^{2}+15 \left (x^{3}-2 x^{2}+x \right )^{\frac {1}{3}}-3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+5 x -1}{-1+x}\right )}{3}\right ) \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}} \left (\left (-1+x \right )^{2} x \right )^{\frac {1}{3}}}{x \left (-1+x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-1+x)*x^2)^(1/3)+(1/3*ln(-(RootOf(_Z^2+_Z+1)^2*x^2+15*RootOf(_Z^2+_Z+1)*(x^3-2*x^2+x)^(2/3)-24*RootOf(_Z^2+_
Z+1)*(x^3-2*x^2+x)^(1/3)*x-3*RootOf(_Z^2+_Z+1)^2*x+10*RootOf(_Z^2+_Z+1)*x^2-9*(x^3-2*x^2+x)^(2/3)+24*RootOf(_Z
^2+_Z+1)*(x^3-2*x^2+x)^(1/3)-15*(x^3-2*x^2+x)^(1/3)*x+2*RootOf(_Z^2+_Z+1)^2-23*RootOf(_Z^2+_Z+1)*x+25*x^2+15*(
x^3-2*x^2+x)^(1/3)+13*RootOf(_Z^2+_Z+1)-40*x+15)/(-1+x))+1/3*RootOf(_Z^2+_Z+1)*ln((5*RootOf(_Z^2+_Z+1)^2*x^2+1
5*RootOf(_Z^2+_Z+1)*(x^3-2*x^2+x)^(2/3)+9*RootOf(_Z^2+_Z+1)*(x^3-2*x^2+x)^(1/3)*x-15*RootOf(_Z^2+_Z+1)^2*x-19*
RootOf(_Z^2+_Z+1)*x^2+24*(x^3-2*x^2+x)^(2/3)-9*RootOf(_Z^2+_Z+1)*(x^3-2*x^2+x)^(1/3)-15*(x^3-2*x^2+x)^(1/3)*x+
10*RootOf(_Z^2+_Z+1)^2+22*RootOf(_Z^2+_Z+1)*x-4*x^2+15*(x^3-2*x^2+x)^(1/3)-3*RootOf(_Z^2+_Z+1)+5*x-1)/(-1+x)))
*((-1+x)*x^2)^(1/3)/x*((-1+x)^2*x)^(1/3)/(-1+x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 - x^2)^(1/3)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 - x^2)^(1/3)/x,x)

[Out]

int((x^3 - x^2)^(1/3)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**2*(x - 1))**(1/3)/x, x)

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