\(\int \sqrt [3]{-x^2+x^3} \, dx\) [1746]

Optimal result

Integrand size = 13, antiderivative size = 118 \[ \int \sqrt [3]{-x^2+x^3} \, dx=\frac {1}{6} (-1+3 x) \sqrt [3]{-x^2+x^3}+\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x^2+x^3}}\right )}{3 \sqrt {3}}+\frac {1}{9} \log \left (-x+\sqrt [3]{-x^2+x^3}\right )-\frac {1}{18} \log \left (x^2+x \sqrt [3]{-x^2+x^3}+\left (-x^2+x^3\right )^{2/3}\right ) \] Output:

1/6*(-1+3*x)*(x^3-x^2)^(1/3)+1/9*3^(1/2)*arctan(3^(1/2)*x/(x+2*(x^3-x^2)^( 
1/3)))+1/9*ln(-x+(x^3-x^2)^(1/3))-1/18*ln(x^2+x*(x^3-x^2)^(1/3)+(x^3-x^2)^ 
(2/3))
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.18 \[ \int \sqrt [3]{-x^2+x^3} \, dx=\frac {(-1+x)^{2/3} x^{4/3} \left (-3 \sqrt [3]{-1+x} x^{2/3}+9 \sqrt [3]{-1+x} x^{5/3}+2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{x}}\right )+2 \log \left (\sqrt [3]{-1+x}-\sqrt [3]{x}\right )-\log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{x}+x^{2/3}\right )\right )}{18 \left ((-1+x) x^2\right )^{2/3}} \] Input:

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

Output:

((-1 + x)^(2/3)*x^(4/3)*(-3*(-1 + x)^(1/3)*x^(2/3) + 9*(-1 + x)^(1/3)*x^(5 
/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)]))/(18*((-1 + x)*x^2)^(2/3))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1910, 1930, 1938, 71}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1910

\(\displaystyle \frac {1}{2} x \sqrt [3]{x^3-x^2}-\frac {1}{6} \int \frac {x^2}{\left (x^3-x^2\right )^{2/3}}dx\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{6} \left (-\frac {2}{3} \int \frac {x}{\left (x^3-x^2\right )^{2/3}}dx-\sqrt [3]{x^3-x^2}\right )+\frac {1}{2} \sqrt [3]{x^3-x^2} x\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {1}{6} \left (-\frac {2 (x-1)^{2/3} x^{4/3} \int \frac {1}{(x-1)^{2/3} \sqrt [3]{x}}dx}{3 \left (x^3-x^2\right )^{2/3}}-\sqrt [3]{x^3-x^2}\right )+\frac {1}{2} \sqrt [3]{x^3-x^2} x\)

\(\Big \downarrow \) 71

\(\displaystyle \frac {1}{6} \left (-\frac {2 (x-1)^{2/3} x^{4/3} \left (-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{x-1}}+\frac {1}{\sqrt {3}}\right )-\frac {3}{2} \log \left (\frac {\sqrt [3]{x}}{\sqrt [3]{x-1}}-1\right )-\frac {1}{2} \log (x-1)\right )}{3 \left (x^3-x^2\right )^{2/3}}-\sqrt [3]{x^3-x^2}\right )+\frac {1}{2} \sqrt [3]{x^3-x^2} x\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 71

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] && PosQ[d/b]
 

rule 1910

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

rule 1930

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

rule 1938

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[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] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 

Maple [C] (warning: unable to verify)

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

Time = 0.68 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.23

method	result	size
meijerg	\(\frac {3 \operatorname {signum}\left (-1+x \right )^{\frac {1}{3}} x^{\frac {5}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {5}{3}\right ], \left [\frac {8}{3}\right ], x\right )}{5 \left (-\operatorname {signum}\left (-1+x \right )\right )^{\frac {1}{3}}}\)	\(27\)
pseudoelliptic	\(-\frac {x^{4} \left (2 \sqrt {3}\, \arctan \left (\frac {\left (2 \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )-9 \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}} x +\ln \left (\frac {\left (\left (-1+x \right ) x^{2}\right )^{\frac {2}{3}}+\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )-2 \ln \left (\frac {\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}-x}{x}\right )+3 \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}\right )}{18 {\left (\left (\left (-1+x \right ) x^{2}\right )^{\frac {2}{3}}+\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}} x +x^{2}\right )}^{2} {\left (\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}-x \right )}^{2}}\)	\(147\)
trager	\(\left (-\frac {1}{6}+\frac {x}{2}\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}}+\frac {\operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (\frac {45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+72 \left (x^{3}-x^{2}\right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x -90 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x +87 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+15 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+15 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}-69 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +20 x^{2}-12 x}{x}\right )}{3}-\frac {\ln \left (-\frac {-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+72 \left (x^{3}-x^{2}\right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +90 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x +57 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+9 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+9 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}-9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +4 x^{2}-x}{x}\right )}{9}-\frac {\ln \left (-\frac {-45 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+72 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+72 \left (x^{3}-x^{2}\right )^{\frac {1}{3}} \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +90 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x +57 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+9 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+9 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}-9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x +4 x^{2}-x}{x}\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{3}\)	\(511\)
risch	\(\text {Expression too large to display}\)	\(643\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.93 \[ \int \sqrt [3]{-x^2+x^3} \, dx=-\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {1}{6} \, {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} {\left (3 \, x - 1\right )} + \frac {1}{9} \, \log \left (-\frac {x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{18} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) \] Input:

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

Output:

-1/9*sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x^3 - x^2)^(1/3))/x) + 1/6 
*(x^3 - x^2)^(1/3)*(3*x - 1) + 1/9*log(-(x - (x^3 - x^2)^(1/3))/x) - 1/18* 
log((x^2 + (x^3 - x^2)^(1/3)*x + (x^3 - x^2)^(2/3))/x^2)
 

Sympy [F]

\[ \int \sqrt [3]{-x^2+x^3} \, dx=\int \sqrt [3]{x^{3} - x^{2}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \sqrt [3]{-x^2+x^3} \, dx=\frac {1}{6} \, {\left ({\left (-\frac {1}{x} + 1\right )}^{\frac {4}{3}} + 2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}}\right )} x^{2} - \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{18} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{9} \, \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \] Input:

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

Output:

1/6*((-1/x + 1)^(4/3) + 2*(-1/x + 1)^(1/3))*x^2 - 1/9*sqrt(3)*arctan(1/3*s 
qrt(3)*(2*(-1/x + 1)^(1/3) + 1)) - 1/18*log((-1/x + 1)^(2/3) + (-1/x + 1)^ 
(1/3) + 1) + 1/9*log(abs((-1/x + 1)^(1/3) - 1))
 

Mupad [B] (verification not implemented)

Time = 7.78 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.23 \[ \int \sqrt [3]{-x^2+x^3} \, dx=\frac {3\,x\,{\left (x^3-x^2\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {5}{3};\ \frac {8}{3};\ x\right )}{5\,{\left (1-x\right )}^{1/3}} \] Input:

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

Output:

(3*x*(x^3 - x^2)^(1/3)*hypergeom([-1/3, 5/3], 8/3, x))/(5*(1 - x)^(1/3))
 

Reduce [F]

\[ \int \sqrt [3]{-x^2+x^3} \, dx=\frac {x^{\frac {5}{3}} \left (x -1\right )^{\frac {1}{3}}}{2}-\frac {x^{\frac {2}{3}} \left (x -1\right )^{\frac {1}{3}}}{6}-\frac {\left (\int \frac {\left (x -1\right )^{\frac {1}{3}}}{x^{\frac {4}{3}}-x^{\frac {1}{3}}}d x \right )}{9} \] Input:

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

Output:

(9*x**(2/3)*(x - 1)**(1/3)*x - 3*x**(2/3)*(x - 1)**(1/3) - 2*int((x - 1)** 
(1/3)/(x**(1/3)*x - x**(1/3)),x))/18