\(\int (1-2 x)^2 (3+5 x) \, dx\) [1244]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 23 \[ \int (1-2 x)^2 (3+5 x) \, dx=3 x-\frac {7 x^2}{2}-\frac {8 x^3}{3}+5 x^4 \]

[Out]

3*x-7/2*x^2-8/3*x^3+5*x^4

Rubi [A] (verified)

Time = 0.01 (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} \[ \int (1-2 x)^2 (3+5 x) \, dx=5 x^4-\frac {8 x^3}{3}-\frac {7 x^2}{2}+3 x \]

[In]

Int[(1 - 2*x)^2*(3 + 5*x),x]

[Out]

3*x - (7*x^2)/2 - (8*x^3)/3 + 5*x^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 (3-7 x-8 x^2+20 x^3\right ) \, dx \\ & = 3 x-\frac {7 x^2}{2}-\frac {8 x^3}{3}+5 x^4 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int (1-2 x)^2 (3+5 x) \, dx=3 x-\frac {7 x^2}{2}-\frac {8 x^3}{3}+5 x^4 \]

[In]

Integrate[(1 - 2*x)^2*(3 + 5*x),x]

[Out]

3*x - (7*x^2)/2 - (8*x^3)/3 + 5*x^4

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method result size
gosper \(\frac {x \left (30 x^{3}-16 x^{2}-21 x +18\right )}{6}\) \(19\)
default \(3 x -\frac {7}{2} x^{2}-\frac {8}{3} x^{3}+5 x^{4}\) \(20\)
norman \(3 x -\frac {7}{2} x^{2}-\frac {8}{3} x^{3}+5 x^{4}\) \(20\)
risch \(3 x -\frac {7}{2} x^{2}-\frac {8}{3} x^{3}+5 x^{4}\) \(20\)
parallelrisch \(3 x -\frac {7}{2} x^{2}-\frac {8}{3} x^{3}+5 x^{4}\) \(20\)

[In]

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

[Out]

1/6*x*(30*x^3-16*x^2-21*x+18)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (3+5 x) \, dx=5 \, x^{4} - \frac {8}{3} \, x^{3} - \frac {7}{2} \, x^{2} + 3 \, x \]

[In]

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

[Out]

5*x^4 - 8/3*x^3 - 7/2*x^2 + 3*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int (1-2 x)^2 (3+5 x) \, dx=5 x^{4} - \frac {8 x^{3}}{3} - \frac {7 x^{2}}{2} + 3 x \]

[In]

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

[Out]

5*x**4 - 8*x**3/3 - 7*x**2/2 + 3*x

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (3+5 x) \, dx=5 \, x^{4} - \frac {8}{3} \, x^{3} - \frac {7}{2} \, x^{2} + 3 \, x \]

[In]

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

[Out]

5*x^4 - 8/3*x^3 - 7/2*x^2 + 3*x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (3+5 x) \, dx=5 \, x^{4} - \frac {8}{3} \, x^{3} - \frac {7}{2} \, x^{2} + 3 \, x \]

[In]

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

[Out]

5*x^4 - 8/3*x^3 - 7/2*x^2 + 3*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int (1-2 x)^2 (3+5 x) \, dx=5\,x^4-\frac {8\,x^3}{3}-\frac {7\,x^2}{2}+3\,x \]

[In]

int((2*x - 1)^2*(5*x + 3),x)

[Out]

3*x - (7*x^2)/2 - (8*x^3)/3 + 5*x^4