\(\int (1-2 x) (3+5 x)^2 \, dx\) [1166]

Optimal result

Integrand size = 13, antiderivative size = 23 \[ \int (1-2 x) (3+5 x)^2 \, dx=9 x+6 x^2-\frac {35 x^3}{3}-\frac {25 x^4}{2} \]

[Out]

9*x+6*x^2-35/3*x^3-25/2*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} \[ \int (1-2 x) (3+5 x)^2 \, dx=-\frac {25 x^4}{2}-\frac {35 x^3}{3}+6 x^2+9 x \]

[In]

Int[(1 - 2*x)*(3 + 5*x)^2,x]

[Out]

9*x + 6*x^2 - (35*x^3)/3 - (25*x^4)/2

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 (9+12 x-35 x^2-50 x^3\right ) \, dx \\ & = 9 x+6 x^2-\frac {35 x^3}{3}-\frac {25 x^4}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int (1-2 x) (3+5 x)^2 \, dx=9 x+6 x^2-\frac {35 x^3}{3}-\frac {25 x^4}{2} \]

[In]

Integrate[(1 - 2*x)*(3 + 5*x)^2,x]

[Out]

9*x + 6*x^2 - (35*x^3)/3 - (25*x^4)/2

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
gosper	\(-\frac {x \left (75 x^{3}+70 x^{2}-36 x -54\right )}{6}\)	\(19\)
default	\(9 x +6 x^{2}-\frac {35}{3} x^{3}-\frac {25}{2} x^{4}\)	\(20\)
norman	\(9 x +6 x^{2}-\frac {35}{3} x^{3}-\frac {25}{2} x^{4}\)	\(20\)
risch	\(9 x +6 x^{2}-\frac {35}{3} x^{3}-\frac {25}{2} x^{4}\)	\(20\)
parallelrisch	\(9 x +6 x^{2}-\frac {35}{3} x^{3}-\frac {25}{2} x^{4}\)	\(20\)

[In]

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

[Out]

-1/6*x*(75*x^3+70*x^2-36*x-54)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int (1-2 x) (3+5 x)^2 \, dx=-\frac {25}{2} \, x^{4} - \frac {35}{3} \, x^{3} + 6 \, x^{2} + 9 \, x \]

[In]

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

[Out]

-25/2*x^4 - 35/3*x^3 + 6*x^2 + 9*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int (1-2 x) (3+5 x)^2 \, dx=- \frac {25 x^{4}}{2} - \frac {35 x^{3}}{3} + 6 x^{2} + 9 x \]

[In]

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

[Out]

-25*x**4/2 - 35*x**3/3 + 6*x**2 + 9*x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int (1-2 x) (3+5 x)^2 \, dx=-\frac {25}{2} \, x^{4} - \frac {35}{3} \, x^{3} + 6 \, x^{2} + 9 \, x \]

[In]

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

[Out]

-25/2*x^4 - 35/3*x^3 + 6*x^2 + 9*x

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int (1-2 x) (3+5 x)^2 \, dx=-\frac {25}{2} \, x^{4} - \frac {35}{3} \, x^{3} + 6 \, x^{2} + 9 \, x \]

[In]

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

[Out]

-25/2*x^4 - 35/3*x^3 + 6*x^2 + 9*x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int (1-2 x) (3+5 x)^2 \, dx=-\frac {25\,x^4}{2}-\frac {35\,x^3}{3}+6\,x^2+9\,x \]

[In]

int(-(2*x - 1)*(5*x + 3)^2,x)

[Out]

9*x + 6*x^2 - (35*x^3)/3 - (25*x^4)/2