\(\int (1-2 x) (2+3 x)^2 (3+5 x) \, dx\) [1149]

Optimal result

Integrand size = 18, antiderivative size = 28 \[ \int (1-2 x) (2+3 x)^2 (3+5 x) \, dx=12 x+16 x^2-\frac {25 x^3}{3}-\frac {129 x^4}{4}-18 x^5 \]

[Out]

12*x+16*x^2-25/3*x^3-129/4*x^4-18*x^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int (1-2 x) (2+3 x)^2 (3+5 x) \, dx=-18 x^5-\frac {129 x^4}{4}-\frac {25 x^3}{3}+16 x^2+12 x \]

[In]

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

[Out]

12*x + 16*x^2 - (25*x^3)/3 - (129*x^4)/4 - 18*x^5

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+32 x-25 x^2-129 x^3-90 x^4\right ) \, dx \\ & = 12 x+16 x^2-\frac {25 x^3}{3}-\frac {129 x^4}{4}-18 x^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int (1-2 x) (2+3 x)^2 (3+5 x) \, dx=12 x+16 x^2-\frac {25 x^3}{3}-\frac {129 x^4}{4}-18 x^5 \]

[In]

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

[Out]

12*x + 16*x^2 - (25*x^3)/3 - (129*x^4)/4 - 18*x^5

Maple [A] (verified)

Time = 1.75 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86

method	result	size
gosper	\(-\frac {x \left (216 x^{4}+387 x^{3}+100 x^{2}-192 x -144\right )}{12}\)	\(24\)
default	\(12 x +16 x^{2}-\frac {25}{3} x^{3}-\frac {129}{4} x^{4}-18 x^{5}\)	\(25\)
norman	\(12 x +16 x^{2}-\frac {25}{3} x^{3}-\frac {129}{4} x^{4}-18 x^{5}\)	\(25\)
risch	\(12 x +16 x^{2}-\frac {25}{3} x^{3}-\frac {129}{4} x^{4}-18 x^{5}\)	\(25\)
parallelrisch	\(12 x +16 x^{2}-\frac {25}{3} x^{3}-\frac {129}{4} x^{4}-18 x^{5}\)	\(25\)

[In]

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

[Out]

-1/12*x*(216*x^4+387*x^3+100*x^2-192*x-144)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int (1-2 x) (2+3 x)^2 (3+5 x) \, dx=-18 \, x^{5} - \frac {129}{4} \, x^{4} - \frac {25}{3} \, x^{3} + 16 \, x^{2} + 12 \, x \]

[In]

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

[Out]

-18*x^5 - 129/4*x^4 - 25/3*x^3 + 16*x^2 + 12*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int (1-2 x) (2+3 x)^2 (3+5 x) \, dx=- 18 x^{5} - \frac {129 x^{4}}{4} - \frac {25 x^{3}}{3} + 16 x^{2} + 12 x \]

[In]

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

[Out]

-18*x**5 - 129*x**4/4 - 25*x**3/3 + 16*x**2 + 12*x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int (1-2 x) (2+3 x)^2 (3+5 x) \, dx=-18 \, x^{5} - \frac {129}{4} \, x^{4} - \frac {25}{3} \, x^{3} + 16 \, x^{2} + 12 \, x \]

[In]

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

[Out]

-18*x^5 - 129/4*x^4 - 25/3*x^3 + 16*x^2 + 12*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int (1-2 x) (2+3 x)^2 (3+5 x) \, dx=-18 \, x^{5} - \frac {129}{4} \, x^{4} - \frac {25}{3} \, x^{3} + 16 \, x^{2} + 12 \, x \]

[In]

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

[Out]

-18*x^5 - 129/4*x^4 - 25/3*x^3 + 16*x^2 + 12*x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int (1-2 x) (2+3 x)^2 (3+5 x) \, dx=-18\,x^5-\frac {129\,x^4}{4}-\frac {25\,x^3}{3}+16\,x^2+12\,x \]

[In]

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

[Out]

12*x + 16*x^2 - (25*x^3)/3 - (129*x^4)/4 - 18*x^5