\(\int x^2 (a+8 x-8 x^2+4 x^3-x^4) \, dx\) [133]

Optimal result

Integrand size = 24, antiderivative size = 35 \[ \int x^2 \left (a+8 x-8 x^2+4 x^3-x^4\right ) \, dx=\frac {a x^3}{3}+2 x^4-\frac {8 x^5}{5}+\frac {2 x^6}{3}-\frac {x^7}{7} \]

[Out]

1/3*a*x^3+2*x^4-8/5*x^5+2/3*x^6-1/7*x^7

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {14} \[ \int x^2 \left (a+8 x-8 x^2+4 x^3-x^4\right ) \, dx=\frac {a x^3}{3}-\frac {x^7}{7}+\frac {2 x^6}{3}-\frac {8 x^5}{5}+2 x^4 \]

[In]

Int[x^2*(a + 8*x - 8*x^2 + 4*x^3 - x^4),x]

[Out]

(a*x^3)/3 + 2*x^4 - (8*x^5)/5 + (2*x^6)/3 - x^7/7

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (a x^2+8 x^3-8 x^4+4 x^5-x^6\right ) \, dx \\ & = \frac {a x^3}{3}+2 x^4-\frac {8 x^5}{5}+\frac {2 x^6}{3}-\frac {x^7}{7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+8 x-8 x^2+4 x^3-x^4\right ) \, dx=\frac {a x^3}{3}+2 x^4-\frac {8 x^5}{5}+\frac {2 x^6}{3}-\frac {x^7}{7} \]

[In]

Integrate[x^2*(a + 8*x - 8*x^2 + 4*x^3 - x^4),x]

[Out]

(a*x^3)/3 + 2*x^4 - (8*x^5)/5 + (2*x^6)/3 - x^7/7

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
gosper	\(\frac {1}{3} a \,x^{3}+2 x^{4}-\frac {8}{5} x^{5}+\frac {2}{3} x^{6}-\frac {1}{7} x^{7}\)	\(28\)
default	\(\frac {1}{3} a \,x^{3}+2 x^{4}-\frac {8}{5} x^{5}+\frac {2}{3} x^{6}-\frac {1}{7} x^{7}\)	\(28\)
norman	\(\frac {1}{3} a \,x^{3}+2 x^{4}-\frac {8}{5} x^{5}+\frac {2}{3} x^{6}-\frac {1}{7} x^{7}\)	\(28\)
risch	\(\frac {1}{3} a \,x^{3}+2 x^{4}-\frac {8}{5} x^{5}+\frac {2}{3} x^{6}-\frac {1}{7} x^{7}\)	\(28\)
parallelrisch	\(\frac {1}{3} a \,x^{3}+2 x^{4}-\frac {8}{5} x^{5}+\frac {2}{3} x^{6}-\frac {1}{7} x^{7}\)	\(28\)

[In]

int(x^2*(-x^4+4*x^3-8*x^2+a+8*x),x,method=_RETURNVERBOSE)

[Out]

1/3*a*x^3+2*x^4-8/5*x^5+2/3*x^6-1/7*x^7

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int x^2 \left (a+8 x-8 x^2+4 x^3-x^4\right ) \, dx=-\frac {1}{7} \, x^{7} + \frac {2}{3} \, x^{6} - \frac {8}{5} \, x^{5} + \frac {1}{3} \, a x^{3} + 2 \, x^{4} \]

[In]

integrate(x^2*(-x^4+4*x^3-8*x^2+a+8*x),x, algorithm="fricas")

[Out]

-1/7*x^7 + 2/3*x^6 - 8/5*x^5 + 1/3*a*x^3 + 2*x^4

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x^2 \left (a+8 x-8 x^2+4 x^3-x^4\right ) \, dx=\frac {a x^{3}}{3} - \frac {x^{7}}{7} + \frac {2 x^{6}}{3} - \frac {8 x^{5}}{5} + 2 x^{4} \]

[In]

integrate(x**2*(-x**4+4*x**3-8*x**2+a+8*x),x)

[Out]

a*x**3/3 - x**7/7 + 2*x**6/3 - 8*x**5/5 + 2*x**4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int x^2 \left (a+8 x-8 x^2+4 x^3-x^4\right ) \, dx=-\frac {1}{7} \, x^{7} + \frac {2}{3} \, x^{6} - \frac {8}{5} \, x^{5} + \frac {1}{3} \, a x^{3} + 2 \, x^{4} \]

[In]

integrate(x^2*(-x^4+4*x^3-8*x^2+a+8*x),x, algorithm="maxima")

[Out]

-1/7*x^7 + 2/3*x^6 - 8/5*x^5 + 1/3*a*x^3 + 2*x^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int x^2 \left (a+8 x-8 x^2+4 x^3-x^4\right ) \, dx=-\frac {1}{7} \, x^{7} + \frac {2}{3} \, x^{6} - \frac {8}{5} \, x^{5} + \frac {1}{3} \, a x^{3} + 2 \, x^{4} \]

[In]

integrate(x^2*(-x^4+4*x^3-8*x^2+a+8*x),x, algorithm="giac")

[Out]

-1/7*x^7 + 2/3*x^6 - 8/5*x^5 + 1/3*a*x^3 + 2*x^4

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int x^2 \left (a+8 x-8 x^2+4 x^3-x^4\right ) \, dx=-\frac {x^7}{7}+\frac {2\,x^6}{3}-\frac {8\,x^5}{5}+2\,x^4+\frac {a\,x^3}{3} \]

[In]

int(x^2*(a + 8*x - 8*x^2 + 4*x^3 - x^4),x)

[Out]

(a*x^3)/3 + 2*x^4 - (8*x^5)/5 + (2*x^6)/3 - x^7/7