\(\int \frac {\sqrt [3]{-x^2+x^3}}{x} \, dx\) [1529]

Optimal result

Integrand size = 17, antiderivative size = 106 \[ \int \frac {\sqrt [3]{-x^2+x^3}}{x} \, dx=\sqrt [3]{-x^2+x^3}+\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x^2+x^3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (-x+\sqrt [3]{-x^2+x^3}\right )-\frac {1}{6} \log \left (x^2+x \sqrt [3]{-x^2+x^3}+\left (-x^2+x^3\right )^{2/3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.41, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2046, 2057, 61} \[ \int \frac {\sqrt [3]{-x^2+x^3}}{x} \, dx=\frac {(x-1)^{2/3} x^{4/3} \arctan \left (\frac {2 \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{x-1}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} \left (x^3-x^2\right )^{2/3}}+\sqrt [3]{x^3-x^2}+\frac {(x-1)^{2/3} x^{4/3} \log \left (\frac {\sqrt [3]{x}}{\sqrt [3]{x-1}}-1\right )}{2 \left (x^3-x^2\right )^{2/3}}+\frac {(x-1)^{2/3} x^{4/3} \log (x-1)}{6 \left (x^3-x^2\right )^{2/3}} \]

[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 61

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

Rule 2046

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 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[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*} \text {integral}& = \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} \arctan \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*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt [3]{-x^2+x^3}}{x} \, dx=\frac {(-1+x)^{2/3} x^{4/3} \left (6 \sqrt [3]{-1+x} x^{2/3}+2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{x}}\right )+2 \log \left (\sqrt [3]{-1+x}-\sqrt [3]{x}\right )-\log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{x}+x^{2/3}\right )\right )}{6 \left ((-1+x) x^2\right )^{2/3}} \]

[In]

Integrate[(-x^2 + x^3)^(1/3)/x,x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.74 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.25

method	result	size
meijerg	\(\frac {3 \operatorname {signum}\left (-1+x \right )^{\frac {1}{3}} x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x\right )}{2 \left (-\operatorname {signum}\left (-1+x \right )\right )^{\frac {1}{3}}}\)	\(27\)
pseudoelliptic	\(\frac {x^{2} \left (2 \sqrt {3}\, \arctan \left (\frac {\left (2 \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )-6 \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+\ln \left (\frac {\left (\left (-1+x \right ) x^{2}\right )^{\frac {2}{3}}+\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )-2 \ln \left (\frac {\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}-x}{x}\right )\right )}{6 \left (\left (\left (-1+x \right ) x^{2}\right )^{\frac {2}{3}}+\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}} x +x^{2}\right ) \left (\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}-x \right )}\)	\(135\)
trager	\(\left (x^{3}-x^{2}\right )^{\frac {1}{3}}+\frac {\ln \left (-\frac {36 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}-72 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x -144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+90 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}} x +60 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-78 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x -15 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}-9 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}+25 x^{2}-15 x}{x}\right )}{3}+2 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \ln \left (-\frac {144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}-288 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x +144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}-54 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}} x -66 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-36 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x +9 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+15 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}-20 x^{2}+5 x}{x}\right )\)	\(338\)
risch	\(\text {Expression too large to display}\)	\(636\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt [3]{-x^2+x^3}}{x} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} + \frac {1}{3} \, \log \left (-\frac {x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{6} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

[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)

Sympy [F]

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

[In]

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

[Out]

Integral((x**2*(x - 1))**(1/3)/x, x)

Maxima [F]

\[ \int \frac {\sqrt [3]{-x^2+x^3}}{x} \, dx=\int { \frac {{\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x} \,d x } \]

[In]

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

[Out]

integrate((x^3 - x^2)^(1/3)/x, x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt [3]{-x^2+x^3}}{x} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + x {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - \frac {1}{6} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{-x^2+x^3}}{x} \, dx=\int \frac {{\left (x^3-x^2\right )}^{1/3}}{x} \,d x \]

[In]

int((x^3 - x^2)^(1/3)/x,x)

[Out]

int((x^3 - x^2)^(1/3)/x, x)