\(\int (1-2 x)^3 (3+5 x) \, dx\) [1346]

Optimal result

Integrand size = 13, antiderivative size = 23 \[ \int (1-2 x)^3 (3+5 x) \, dx=-\frac {11}{16} (1-2 x)^4+\frac {1}{4} (1-2 x)^5 \]

[Out]

-11/16*(1-2*x)^4+1/4*(1-2*x)^5

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int (1-2 x)^3 (3+5 x) \, dx=\frac {1}{4} (1-2 x)^5-\frac {11}{16} (1-2 x)^4 \]

[In]

Int[(1 - 2*x)^3*(3 + 5*x),x]

[Out]

(-11*(1 - 2*x)^4)/16 + (1 - 2*x)^5/4

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {11}{2} (1-2 x)^3-\frac {5}{2} (1-2 x)^4\right ) \, dx \\ & = -\frac {11}{16} (1-2 x)^4+\frac {1}{4} (1-2 x)^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int (1-2 x)^3 (3+5 x) \, dx=3 x-\frac {13 x^2}{2}+2 x^3+9 x^4-8 x^5 \]

[In]

Integrate[(1 - 2*x)^3*(3 + 5*x),x]

[Out]

3*x - (13*x^2)/2 + 2*x^3 + 9*x^4 - 8*x^5

Maple [A] (verified)

Time = 1.93 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
gosper	\(-\frac {x \left (16 x^{4}-18 x^{3}-4 x^{2}+13 x -6\right )}{2}\)	\(24\)
default	\(-8 x^{5}+9 x^{4}+2 x^{3}-\frac {13}{2} x^{2}+3 x\)	\(25\)
norman	\(-8 x^{5}+9 x^{4}+2 x^{3}-\frac {13}{2} x^{2}+3 x\)	\(25\)
risch	\(-8 x^{5}+9 x^{4}+2 x^{3}-\frac {13}{2} x^{2}+3 x\)	\(25\)
parallelrisch	\(-8 x^{5}+9 x^{4}+2 x^{3}-\frac {13}{2} x^{2}+3 x\)	\(25\)

[In]

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

[Out]

-1/2*x*(16*x^4-18*x^3-4*x^2+13*x-6)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (1-2 x)^3 (3+5 x) \, dx=-8 \, x^{5} + 9 \, x^{4} + 2 \, x^{3} - \frac {13}{2} \, x^{2} + 3 \, x \]

[In]

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

[Out]

-8*x^5 + 9*x^4 + 2*x^3 - 13/2*x^2 + 3*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (1-2 x)^3 (3+5 x) \, dx=- 8 x^{5} + 9 x^{4} + 2 x^{3} - \frac {13 x^{2}}{2} + 3 x \]

[In]

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

[Out]

-8*x**5 + 9*x**4 + 2*x**3 - 13*x**2/2 + 3*x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (1-2 x)^3 (3+5 x) \, dx=-8 \, x^{5} + 9 \, x^{4} + 2 \, x^{3} - \frac {13}{2} \, x^{2} + 3 \, x \]

[In]

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

[Out]

-8*x^5 + 9*x^4 + 2*x^3 - 13/2*x^2 + 3*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (1-2 x)^3 (3+5 x) \, dx=-8 \, x^{5} + 9 \, x^{4} + 2 \, x^{3} - \frac {13}{2} \, x^{2} + 3 \, x \]

[In]

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

[Out]

-8*x^5 + 9*x^4 + 2*x^3 - 13/2*x^2 + 3*x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (1-2 x)^3 (3+5 x) \, dx=-8\,x^5+9\,x^4+2\,x^3-\frac {13\,x^2}{2}+3\,x \]

[In]

int(-(2*x - 1)^3*(5*x + 3),x)

[Out]

3*x - (13*x^2)/2 + 2*x^3 + 9*x^4 - 8*x^5