\(\int x (1-x^3)^2 \, dx\) [462]

Optimal result

Integrand size = 11, antiderivative size = 22 \[ \int x \left (1-x^3\right )^2 \, dx=\frac {x^2}{2}-\frac {2 x^5}{5}+\frac {x^8}{8} \]

[Out]

1/2*x^2-2/5*x^5+1/8*x^8

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {276} \[ \int x \left (1-x^3\right )^2 \, dx=\frac {x^8}{8}-\frac {2 x^5}{5}+\frac {x^2}{2} \]

[In]

Int[x*(1 - x^3)^2,x]

[Out]

x^2/2 - (2*x^5)/5 + x^8/8

Rule 276

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

Rubi steps \begin{align*} \text {integral}& = \int \left (x-2 x^4+x^7\right ) \, dx \\ & = \frac {x^2}{2}-\frac {2 x^5}{5}+\frac {x^8}{8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int x \left (1-x^3\right )^2 \, dx=\frac {x^2}{2}-\frac {2 x^5}{5}+\frac {x^8}{8} \]

[In]

Integrate[x*(1 - x^3)^2,x]

[Out]

x^2/2 - (2*x^5)/5 + x^8/8

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77

method	result	size
default	\(\frac {1}{2} x^{2}-\frac {2}{5} x^{5}+\frac {1}{8} x^{8}\)	\(17\)
norman	\(\frac {1}{2} x^{2}-\frac {2}{5} x^{5}+\frac {1}{8} x^{8}\)	\(17\)
risch	\(\frac {1}{2} x^{2}-\frac {2}{5} x^{5}+\frac {1}{8} x^{8}\)	\(17\)
parallelrisch	\(\frac {1}{2} x^{2}-\frac {2}{5} x^{5}+\frac {1}{8} x^{8}\)	\(17\)
gosper	\(\frac {x^{2} \left (5 x^{6}-16 x^{3}+20\right )}{40}\)	\(18\)

[In]

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

[Out]

1/2*x^2-2/5*x^5+1/8*x^8

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int x \left (1-x^3\right )^2 \, dx=\frac {1}{8} \, x^{8} - \frac {2}{5} \, x^{5} + \frac {1}{2} \, x^{2} \]

[In]

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

[Out]

1/8*x^8 - 2/5*x^5 + 1/2*x^2

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int x \left (1-x^3\right )^2 \, dx=\frac {x^{8}}{8} - \frac {2 x^{5}}{5} + \frac {x^{2}}{2} \]

[In]

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

[Out]

x**8/8 - 2*x**5/5 + x**2/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int x \left (1-x^3\right )^2 \, dx=\frac {1}{8} \, x^{8} - \frac {2}{5} \, x^{5} + \frac {1}{2} \, x^{2} \]

[In]

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

[Out]

1/8*x^8 - 2/5*x^5 + 1/2*x^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int x \left (1-x^3\right )^2 \, dx=\frac {1}{8} \, x^{8} - \frac {2}{5} \, x^{5} + \frac {1}{2} \, x^{2} \]

[In]

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

[Out]

1/8*x^8 - 2/5*x^5 + 1/2*x^2

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int x \left (1-x^3\right )^2 \, dx=\frac {x^2\,\left (5\,x^6-16\,x^3+20\right )}{40} \]

[In]

int(x*(x^3 - 1)^2,x)

[Out]

(x^2*(5*x^6 - 16*x^3 + 20))/40