\(\int (1-2 x) (2+3 x) (3+5 x)^2 \, dx\) [1165]

Optimal result

Integrand size = 18, antiderivative size = 30 \[ \int (1-2 x) (2+3 x) (3+5 x)^2 \, dx=18 x+\frac {51 x^2}{2}-\frac {34 x^3}{3}-\frac {205 x^4}{4}-30 x^5 \]

[Out]

18*x+51/2*x^2-34/3*x^3-205/4*x^4-30*x^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, 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} \[ \int (1-2 x) (2+3 x) (3+5 x)^2 \, dx=-30 x^5-\frac {205 x^4}{4}-\frac {34 x^3}{3}+\frac {51 x^2}{2}+18 x \]

[In]

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

[Out]

18*x + (51*x^2)/2 - (34*x^3)/3 - (205*x^4)/4 - 30*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 (18+51 x-34 x^2-205 x^3-150 x^4\right ) \, dx \\ & = 18 x+\frac {51 x^2}{2}-\frac {34 x^3}{3}-\frac {205 x^4}{4}-30 x^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int (1-2 x) (2+3 x) (3+5 x)^2 \, dx=18 x+\frac {51 x^2}{2}-\frac {34 x^3}{3}-\frac {205 x^4}{4}-30 x^5 \]

[In]

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

[Out]

18*x + (51*x^2)/2 - (34*x^3)/3 - (205*x^4)/4 - 30*x^5

Maple [A] (verified)

Time = 1.75 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80

method	result	size
gosper	\(-\frac {x \left (360 x^{4}+615 x^{3}+136 x^{2}-306 x -216\right )}{12}\)	\(24\)
default	\(18 x +\frac {51}{2} x^{2}-\frac {34}{3} x^{3}-\frac {205}{4} x^{4}-30 x^{5}\)	\(25\)
norman	\(18 x +\frac {51}{2} x^{2}-\frac {34}{3} x^{3}-\frac {205}{4} x^{4}-30 x^{5}\)	\(25\)
risch	\(18 x +\frac {51}{2} x^{2}-\frac {34}{3} x^{3}-\frac {205}{4} x^{4}-30 x^{5}\)	\(25\)
parallelrisch	\(18 x +\frac {51}{2} x^{2}-\frac {34}{3} x^{3}-\frac {205}{4} x^{4}-30 x^{5}\)	\(25\)

[In]

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

[Out]

-1/12*x*(360*x^4+615*x^3+136*x^2-306*x-216)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int (1-2 x) (2+3 x) (3+5 x)^2 \, dx=-30 \, x^{5} - \frac {205}{4} \, x^{4} - \frac {34}{3} \, x^{3} + \frac {51}{2} \, x^{2} + 18 \, x \]

[In]

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

[Out]

-30*x^5 - 205/4*x^4 - 34/3*x^3 + 51/2*x^2 + 18*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int (1-2 x) (2+3 x) (3+5 x)^2 \, dx=- 30 x^{5} - \frac {205 x^{4}}{4} - \frac {34 x^{3}}{3} + \frac {51 x^{2}}{2} + 18 x \]

[In]

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

[Out]

-30*x**5 - 205*x**4/4 - 34*x**3/3 + 51*x**2/2 + 18*x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int (1-2 x) (2+3 x) (3+5 x)^2 \, dx=-30 \, x^{5} - \frac {205}{4} \, x^{4} - \frac {34}{3} \, x^{3} + \frac {51}{2} \, x^{2} + 18 \, x \]

[In]

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

[Out]

-30*x^5 - 205/4*x^4 - 34/3*x^3 + 51/2*x^2 + 18*x

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int (1-2 x) (2+3 x) (3+5 x)^2 \, dx=-30 \, x^{5} - \frac {205}{4} \, x^{4} - \frac {34}{3} \, x^{3} + \frac {51}{2} \, x^{2} + 18 \, x \]

[In]

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

[Out]

-30*x^5 - 205/4*x^4 - 34/3*x^3 + 51/2*x^2 + 18*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int (1-2 x) (2+3 x) (3+5 x)^2 \, dx=-30\,x^5-\frac {205\,x^4}{4}-\frac {34\,x^3}{3}+\frac {51\,x^2}{2}+18\,x \]

[In]

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

[Out]

18*x + (51*x^2)/2 - (34*x^3)/3 - (205*x^4)/4 - 30*x^5