\(\int (-1+5 x) (-1-x+x^2)^2 \, dx\) [480]

Optimal result

Integrand size = 16, antiderivative size = 39 \[ \int (-1+5 x) \left (-1-x+x^2\right )^2 \, dx=-x+\frac {3 x^2}{2}+\frac {11 x^3}{3}-\frac {3 x^4}{4}-\frac {11 x^5}{5}+\frac {5 x^6}{6} \]

[Out]

-x+3/2*x^2+11/3*x^3-3/4*x^4-11/5*x^5+5/6*x^6

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {645} \[ \int (-1+5 x) \left (-1-x+x^2\right )^2 \, dx=\frac {5 x^6}{6}-\frac {11 x^5}{5}-\frac {3 x^4}{4}+\frac {11 x^3}{3}+\frac {3 x^2}{2}-x \]

[In]

Int[(-1 + 5*x)*(-1 - x + x^2)^2,x]

[Out]

-x + (3*x^2)/2 + (11*x^3)/3 - (3*x^4)/4 - (11*x^5)/5 + (5*x^6)/6

Rule 645

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (-1+3 x+11 x^2-3 x^3-11 x^4+5 x^5\right ) \, dx \\ & = -x+\frac {3 x^2}{2}+\frac {11 x^3}{3}-\frac {3 x^4}{4}-\frac {11 x^5}{5}+\frac {5 x^6}{6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int (-1+5 x) \left (-1-x+x^2\right )^2 \, dx=-x+\frac {3 x^2}{2}+\frac {11 x^3}{3}-\frac {3 x^4}{4}-\frac {11 x^5}{5}+\frac {5 x^6}{6} \]

[In]

Integrate[(-1 + 5*x)*(-1 - x + x^2)^2,x]

[Out]

-x + (3*x^2)/2 + (11*x^3)/3 - (3*x^4)/4 - (11*x^5)/5 + (5*x^6)/6

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.77

method	result	size
gosper	\(-x +\frac {3}{2} x^{2}+\frac {11}{3} x^{3}-\frac {3}{4} x^{4}-\frac {11}{5} x^{5}+\frac {5}{6} x^{6}\)	\(30\)
default	\(-x +\frac {3}{2} x^{2}+\frac {11}{3} x^{3}-\frac {3}{4} x^{4}-\frac {11}{5} x^{5}+\frac {5}{6} x^{6}\)	\(30\)
norman	\(-x +\frac {3}{2} x^{2}+\frac {11}{3} x^{3}-\frac {3}{4} x^{4}-\frac {11}{5} x^{5}+\frac {5}{6} x^{6}\)	\(30\)
risch	\(-x +\frac {3}{2} x^{2}+\frac {11}{3} x^{3}-\frac {3}{4} x^{4}-\frac {11}{5} x^{5}+\frac {5}{6} x^{6}\)	\(30\)
parallelrisch	\(-x +\frac {3}{2} x^{2}+\frac {11}{3} x^{3}-\frac {3}{4} x^{4}-\frac {11}{5} x^{5}+\frac {5}{6} x^{6}\)	\(30\)

[In]

int((-1+5*x)*(x^2-x-1)^2,x,method=_RETURNVERBOSE)

[Out]

-x+3/2*x^2+11/3*x^3-3/4*x^4-11/5*x^5+5/6*x^6

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int (-1+5 x) \left (-1-x+x^2\right )^2 \, dx=\frac {5}{6} \, x^{6} - \frac {11}{5} \, x^{5} - \frac {3}{4} \, x^{4} + \frac {11}{3} \, x^{3} + \frac {3}{2} \, x^{2} - x \]

[In]

integrate((-1+5*x)*(x^2-x-1)^2,x, algorithm="fricas")

[Out]

5/6*x^6 - 11/5*x^5 - 3/4*x^4 + 11/3*x^3 + 3/2*x^2 - x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87 \[ \int (-1+5 x) \left (-1-x+x^2\right )^2 \, dx=\frac {5 x^{6}}{6} - \frac {11 x^{5}}{5} - \frac {3 x^{4}}{4} + \frac {11 x^{3}}{3} + \frac {3 x^{2}}{2} - x \]

[In]

integrate((-1+5*x)*(x**2-x-1)**2,x)

[Out]

5*x**6/6 - 11*x**5/5 - 3*x**4/4 + 11*x**3/3 + 3*x**2/2 - x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int (-1+5 x) \left (-1-x+x^2\right )^2 \, dx=\frac {5}{6} \, x^{6} - \frac {11}{5} \, x^{5} - \frac {3}{4} \, x^{4} + \frac {11}{3} \, x^{3} + \frac {3}{2} \, x^{2} - x \]

[In]

integrate((-1+5*x)*(x^2-x-1)^2,x, algorithm="maxima")

[Out]

5/6*x^6 - 11/5*x^5 - 3/4*x^4 + 11/3*x^3 + 3/2*x^2 - x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int (-1+5 x) \left (-1-x+x^2\right )^2 \, dx=\frac {5}{6} \, x^{6} - \frac {11}{5} \, x^{5} - \frac {3}{4} \, x^{4} + \frac {11}{3} \, x^{3} + \frac {3}{2} \, x^{2} - x \]

[In]

integrate((-1+5*x)*(x^2-x-1)^2,x, algorithm="giac")

[Out]

5/6*x^6 - 11/5*x^5 - 3/4*x^4 + 11/3*x^3 + 3/2*x^2 - x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int (-1+5 x) \left (-1-x+x^2\right )^2 \, dx=\frac {5\,x^6}{6}-\frac {11\,x^5}{5}-\frac {3\,x^4}{4}+\frac {11\,x^3}{3}+\frac {3\,x^2}{2}-x \]

[In]

int((5*x - 1)*(x - x^2 + 1)^2,x)

[Out]

(3*x^2)/2 - x + (11*x^3)/3 - (3*x^4)/4 - (11*x^5)/5 + (5*x^6)/6