\(\int (1-2 x)^2 (2+3 x) (3+5 x)^2 \, dx\) [1260]

Optimal result

Integrand size = 20, antiderivative size = 35 \[ \int (1-2 x)^2 (2+3 x) (3+5 x)^2 \, dx=18 x+\frac {15 x^2}{2}-\frac {136 x^3}{3}-\frac {137 x^4}{4}+52 x^5+50 x^6 \]

[Out]

18*x+15/2*x^2-136/3*x^3-137/4*x^4+52*x^5+50*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} \[ \int (1-2 x)^2 (2+3 x) (3+5 x)^2 \, dx=50 x^6+52 x^5-\frac {137 x^4}{4}-\frac {136 x^3}{3}+\frac {15 x^2}{2}+18 x \]

[In]

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

[Out]

18*x + (15*x^2)/2 - (136*x^3)/3 - (137*x^4)/4 + 52*x^5 + 50*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 (18+15 x-136 x^2-137 x^3+260 x^4+300 x^5\right ) \, dx \\ & = 18 x+\frac {15 x^2}{2}-\frac {136 x^3}{3}-\frac {137 x^4}{4}+52 x^5+50 x^6 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int (1-2 x)^2 (2+3 x) (3+5 x)^2 \, dx=18 x+\frac {15 x^2}{2}-\frac {136 x^3}{3}-\frac {137 x^4}{4}+52 x^5+50 x^6 \]

[In]

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

[Out]

18*x + (15*x^2)/2 - (136*x^3)/3 - (137*x^4)/4 + 52*x^5 + 50*x^6

Maple [A] (verified)

Time = 1.86 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {x \left (600 x^{5}+624 x^{4}-411 x^{3}-544 x^{2}+90 x +216\right )}{12}\)	\(29\)
default	\(18 x +\frac {15}{2} x^{2}-\frac {136}{3} x^{3}-\frac {137}{4} x^{4}+52 x^{5}+50 x^{6}\)	\(30\)
norman	\(18 x +\frac {15}{2} x^{2}-\frac {136}{3} x^{3}-\frac {137}{4} x^{4}+52 x^{5}+50 x^{6}\)	\(30\)
risch	\(18 x +\frac {15}{2} x^{2}-\frac {136}{3} x^{3}-\frac {137}{4} x^{4}+52 x^{5}+50 x^{6}\)	\(30\)
parallelrisch	\(18 x +\frac {15}{2} x^{2}-\frac {136}{3} x^{3}-\frac {137}{4} x^{4}+52 x^{5}+50 x^{6}\)	\(30\)

[In]

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

[Out]

1/12*x*(600*x^5+624*x^4-411*x^3-544*x^2+90*x+216)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (2+3 x) (3+5 x)^2 \, dx=50 \, x^{6} + 52 \, x^{5} - \frac {137}{4} \, x^{4} - \frac {136}{3} \, x^{3} + \frac {15}{2} \, x^{2} + 18 \, x \]

[In]

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

[Out]

50*x^6 + 52*x^5 - 137/4*x^4 - 136/3*x^3 + 15/2*x^2 + 18*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int (1-2 x)^2 (2+3 x) (3+5 x)^2 \, dx=50 x^{6} + 52 x^{5} - \frac {137 x^{4}}{4} - \frac {136 x^{3}}{3} + \frac {15 x^{2}}{2} + 18 x \]

[In]

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

[Out]

50*x**6 + 52*x**5 - 137*x**4/4 - 136*x**3/3 + 15*x**2/2 + 18*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (2+3 x) (3+5 x)^2 \, dx=50 \, x^{6} + 52 \, x^{5} - \frac {137}{4} \, x^{4} - \frac {136}{3} \, x^{3} + \frac {15}{2} \, x^{2} + 18 \, x \]

[In]

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

[Out]

50*x^6 + 52*x^5 - 137/4*x^4 - 136/3*x^3 + 15/2*x^2 + 18*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (2+3 x) (3+5 x)^2 \, dx=50 \, x^{6} + 52 \, x^{5} - \frac {137}{4} \, x^{4} - \frac {136}{3} \, x^{3} + \frac {15}{2} \, x^{2} + 18 \, x \]

[In]

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

[Out]

50*x^6 + 52*x^5 - 137/4*x^4 - 136/3*x^3 + 15/2*x^2 + 18*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (2+3 x) (3+5 x)^2 \, dx=50\,x^6+52\,x^5-\frac {137\,x^4}{4}-\frac {136\,x^3}{3}+\frac {15\,x^2}{2}+18\,x \]

[In]

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

[Out]

18*x + (15*x^2)/2 - (136*x^3)/3 - (137*x^4)/4 + 52*x^5 + 50*x^6