\(\int (1-2 x) (2+3 x) (3+5 x)^3 \, dx\) [1182]

Optimal result

Integrand size = 18, antiderivative size = 34 \[ \int (1-2 x) (2+3 x) (3+5 x)^3 \, dx=\frac {11}{500} (3+5 x)^4+\frac {31}{625} (3+5 x)^5-\frac {1}{125} (3+5 x)^6 \]

[Out]

11/500*(3+5*x)^4+31/625*(3+5*x)^5-1/125*(3+5*x)^6

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, 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} \[ \int (1-2 x) (2+3 x) (3+5 x)^3 \, dx=-\frac {1}{125} (5 x+3)^6+\frac {31}{625} (5 x+3)^5+\frac {11}{500} (5 x+3)^4 \]

[In]

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

[Out]

(11*(3 + 5*x)^4)/500 + (31*(3 + 5*x)^5)/625 - (3 + 5*x)^6/125

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 (\frac {11}{25} (3+5 x)^3+\frac {31}{25} (3+5 x)^4-\frac {6}{25} (3+5 x)^5\right ) \, dx \\ & = \frac {11}{500} (3+5 x)^4+\frac {31}{625} (3+5 x)^5-\frac {1}{125} (3+5 x)^6 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int (1-2 x) (2+3 x) (3+5 x)^3 \, dx=54 x+\frac {243 x^2}{2}+51 x^3-\frac {785 x^4}{4}-295 x^5-125 x^6 \]

[In]

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

[Out]

54*x + (243*x^2)/2 + 51*x^3 - (785*x^4)/4 - 295*x^5 - 125*x^6

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85

method	result	size
gosper	\(-\frac {x \left (500 x^{5}+1180 x^{4}+785 x^{3}-204 x^{2}-486 x -216\right )}{4}\)	\(29\)
default	\(-125 x^{6}-295 x^{5}-\frac {785}{4} x^{4}+51 x^{3}+\frac {243}{2} x^{2}+54 x\)	\(30\)
norman	\(-125 x^{6}-295 x^{5}-\frac {785}{4} x^{4}+51 x^{3}+\frac {243}{2} x^{2}+54 x\)	\(30\)
risch	\(-125 x^{6}-295 x^{5}-\frac {785}{4} x^{4}+51 x^{3}+\frac {243}{2} x^{2}+54 x\)	\(30\)
parallelrisch	\(-125 x^{6}-295 x^{5}-\frac {785}{4} x^{4}+51 x^{3}+\frac {243}{2} x^{2}+54 x\)	\(30\)

[In]

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

[Out]

-1/4*x*(500*x^5+1180*x^4+785*x^3-204*x^2-486*x-216)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x) (2+3 x) (3+5 x)^3 \, dx=-125 \, x^{6} - 295 \, x^{5} - \frac {785}{4} \, x^{4} + 51 \, x^{3} + \frac {243}{2} \, x^{2} + 54 \, x \]

[In]

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

[Out]

-125*x^6 - 295*x^5 - 785/4*x^4 + 51*x^3 + 243/2*x^2 + 54*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int (1-2 x) (2+3 x) (3+5 x)^3 \, dx=- 125 x^{6} - 295 x^{5} - \frac {785 x^{4}}{4} + 51 x^{3} + \frac {243 x^{2}}{2} + 54 x \]

[In]

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

[Out]

-125*x**6 - 295*x**5 - 785*x**4/4 + 51*x**3 + 243*x**2/2 + 54*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x) (2+3 x) (3+5 x)^3 \, dx=-125 \, x^{6} - 295 \, x^{5} - \frac {785}{4} \, x^{4} + 51 \, x^{3} + \frac {243}{2} \, x^{2} + 54 \, x \]

[In]

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

[Out]

-125*x^6 - 295*x^5 - 785/4*x^4 + 51*x^3 + 243/2*x^2 + 54*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x) (2+3 x) (3+5 x)^3 \, dx=-125 \, x^{6} - 295 \, x^{5} - \frac {785}{4} \, x^{4} + 51 \, x^{3} + \frac {243}{2} \, x^{2} + 54 \, x \]

[In]

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

[Out]

-125*x^6 - 295*x^5 - 785/4*x^4 + 51*x^3 + 243/2*x^2 + 54*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x) (2+3 x) (3+5 x)^3 \, dx=-125\,x^6-295\,x^5-\frac {785\,x^4}{4}+51\,x^3+\frac {243\,x^2}{2}+54\,x \]

[In]

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

[Out]

54*x + (243*x^2)/2 + 51*x^3 - (785*x^4)/4 - 295*x^5 - 125*x^6