\(\int (1-2 x)^2 (2+3 x)^2 (3+5 x) \, dx\) [1242]

Optimal result

Integrand size = 20, antiderivative size = 35 \[ \int (1-2 x)^2 (2+3 x)^2 (3+5 x) \, dx=12 x+4 x^2-\frac {89 x^3}{3}-\frac {79 x^4}{4}+\frac {168 x^5}{5}+30 x^6 \]

[Out]

12*x+4*x^2-89/3*x^3-79/4*x^4+168/5*x^5+30*x^6

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int (1-2 x)^2 (2+3 x)^2 (3+5 x) \, dx=30 x^6+\frac {168 x^5}{5}-\frac {79 x^4}{4}-\frac {89 x^3}{3}+4 x^2+12 x \]

[In]

Int[(1 - 2*x)^2*(2 + 3*x)^2*(3 + 5*x),x]

[Out]

12*x + 4*x^2 - (89*x^3)/3 - (79*x^4)/4 + (168*x^5)/5 + 30*x^6

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (12+8 x-89 x^2-79 x^3+168 x^4+180 x^5\right ) \, dx \\ & = 12 x+4 x^2-\frac {89 x^3}{3}-\frac {79 x^4}{4}+\frac {168 x^5}{5}+30 x^6 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int (1-2 x)^2 (2+3 x)^2 (3+5 x) \, dx=12 x+4 x^2-\frac {89 x^3}{3}-\frac {79 x^4}{4}+\frac {168 x^5}{5}+30 x^6 \]

[In]

Integrate[(1 - 2*x)^2*(2 + 3*x)^2*(3 + 5*x),x]

[Out]

12*x + 4*x^2 - (89*x^3)/3 - (79*x^4)/4 + (168*x^5)/5 + 30*x^6

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {x \left (1800 x^{5}+2016 x^{4}-1185 x^{3}-1780 x^{2}+240 x +720\right )}{60}\)	\(29\)
default	\(12 x +4 x^{2}-\frac {89}{3} x^{3}-\frac {79}{4} x^{4}+\frac {168}{5} x^{5}+30 x^{6}\)	\(30\)
norman	\(12 x +4 x^{2}-\frac {89}{3} x^{3}-\frac {79}{4} x^{4}+\frac {168}{5} x^{5}+30 x^{6}\)	\(30\)
risch	\(12 x +4 x^{2}-\frac {89}{3} x^{3}-\frac {79}{4} x^{4}+\frac {168}{5} x^{5}+30 x^{6}\)	\(30\)
parallelrisch	\(12 x +4 x^{2}-\frac {89}{3} x^{3}-\frac {79}{4} x^{4}+\frac {168}{5} x^{5}+30 x^{6}\)	\(30\)

[In]

int((1-2*x)^2*(2+3*x)^2*(3+5*x),x,method=_RETURNVERBOSE)

[Out]

1/60*x*(1800*x^5+2016*x^4-1185*x^3-1780*x^2+240*x+720)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (2+3 x)^2 (3+5 x) \, dx=30 \, x^{6} + \frac {168}{5} \, x^{5} - \frac {79}{4} \, x^{4} - \frac {89}{3} \, x^{3} + 4 \, x^{2} + 12 \, x \]

[In]

integrate((1-2*x)^2*(2+3*x)^2*(3+5*x),x, algorithm="fricas")

[Out]

30*x^6 + 168/5*x^5 - 79/4*x^4 - 89/3*x^3 + 4*x^2 + 12*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int (1-2 x)^2 (2+3 x)^2 (3+5 x) \, dx=30 x^{6} + \frac {168 x^{5}}{5} - \frac {79 x^{4}}{4} - \frac {89 x^{3}}{3} + 4 x^{2} + 12 x \]

[In]

integrate((1-2*x)**2*(2+3*x)**2*(3+5*x),x)

[Out]

30*x**6 + 168*x**5/5 - 79*x**4/4 - 89*x**3/3 + 4*x**2 + 12*x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (2+3 x)^2 (3+5 x) \, dx=30 \, x^{6} + \frac {168}{5} \, x^{5} - \frac {79}{4} \, x^{4} - \frac {89}{3} \, x^{3} + 4 \, x^{2} + 12 \, x \]

[In]

integrate((1-2*x)^2*(2+3*x)^2*(3+5*x),x, algorithm="maxima")

[Out]

30*x^6 + 168/5*x^5 - 79/4*x^4 - 89/3*x^3 + 4*x^2 + 12*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (2+3 x)^2 (3+5 x) \, dx=30 \, x^{6} + \frac {168}{5} \, x^{5} - \frac {79}{4} \, x^{4} - \frac {89}{3} \, x^{3} + 4 \, x^{2} + 12 \, x \]

[In]

integrate((1-2*x)^2*(2+3*x)^2*(3+5*x),x, algorithm="giac")

[Out]

30*x^6 + 168/5*x^5 - 79/4*x^4 - 89/3*x^3 + 4*x^2 + 12*x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (2+3 x)^2 (3+5 x) \, dx=30\,x^6+\frac {168\,x^5}{5}-\frac {79\,x^4}{4}-\frac {89\,x^3}{3}+4\,x^2+12\,x \]

[In]

int((2*x - 1)^2*(3*x + 2)^2*(5*x + 3),x)

[Out]

12*x + 4*x^2 - (89*x^3)/3 - (79*x^4)/4 + (168*x^5)/5 + 30*x^6