\(\int (1-2 x)^3 (3+5 x)^2 \, dx\) [1363]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int (1-2 x)^3 (3+5 x)^2 \, dx=-\frac {121}{32} (1-2 x)^4+\frac {11}{4} (1-2 x)^5-\frac {25}{48} (1-2 x)^6 \]

[Out]

-121/32*(1-2*x)^4+11/4*(1-2*x)^5-25/48*(1-2*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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int (1-2 x)^3 (3+5 x)^2 \, dx=-\frac {25}{48} (1-2 x)^6+\frac {11}{4} (1-2 x)^5-\frac {121}{32} (1-2 x)^4 \]

[In]

Int[(1 - 2*x)^3*(3 + 5*x)^2,x]

[Out]

(-121*(1 - 2*x)^4)/32 + (11*(1 - 2*x)^5)/4 - (25*(1 - 2*x)^6)/48

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 {121}{4} (1-2 x)^3-\frac {55}{2} (1-2 x)^4+\frac {25}{4} (1-2 x)^5\right ) \, dx \\ & = -\frac {121}{32} (1-2 x)^4+\frac {11}{4} (1-2 x)^5-\frac {25}{48} (1-2 x)^6 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int (1-2 x)^3 (3+5 x)^2 \, dx=9 x-12 x^2-\frac {47 x^3}{3}+\frac {69 x^4}{2}+12 x^5-\frac {100 x^6}{3} \]

[In]

Integrate[(1 - 2*x)^3*(3 + 5*x)^2,x]

[Out]

9*x - 12*x^2 - (47*x^3)/3 + (69*x^4)/2 + 12*x^5 - (100*x^6)/3

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85

method	result	size
gosper	\(-\frac {x \left (200 x^{5}-72 x^{4}-207 x^{3}+94 x^{2}+72 x -54\right )}{6}\)	\(29\)
default	\(-\frac {100}{3} x^{6}+12 x^{5}+\frac {69}{2} x^{4}-\frac {47}{3} x^{3}-12 x^{2}+9 x\)	\(30\)
norman	\(-\frac {100}{3} x^{6}+12 x^{5}+\frac {69}{2} x^{4}-\frac {47}{3} x^{3}-12 x^{2}+9 x\)	\(30\)
risch	\(-\frac {100}{3} x^{6}+12 x^{5}+\frac {69}{2} x^{4}-\frac {47}{3} x^{3}-12 x^{2}+9 x\)	\(30\)
parallelrisch	\(-\frac {100}{3} x^{6}+12 x^{5}+\frac {69}{2} x^{4}-\frac {47}{3} x^{3}-12 x^{2}+9 x\)	\(30\)

[In]

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

[Out]

-1/6*x*(200*x^5-72*x^4-207*x^3+94*x^2+72*x-54)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^3 (3+5 x)^2 \, dx=-\frac {100}{3} \, x^{6} + 12 \, x^{5} + \frac {69}{2} \, x^{4} - \frac {47}{3} \, x^{3} - 12 \, x^{2} + 9 \, x \]

[In]

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

[Out]

-100/3*x^6 + 12*x^5 + 69/2*x^4 - 47/3*x^3 - 12*x^2 + 9*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int (1-2 x)^3 (3+5 x)^2 \, dx=- \frac {100 x^{6}}{3} + 12 x^{5} + \frac {69 x^{4}}{2} - \frac {47 x^{3}}{3} - 12 x^{2} + 9 x \]

[In]

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

[Out]

-100*x**6/3 + 12*x**5 + 69*x**4/2 - 47*x**3/3 - 12*x**2 + 9*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^3 (3+5 x)^2 \, dx=-\frac {100}{3} \, x^{6} + 12 \, x^{5} + \frac {69}{2} \, x^{4} - \frac {47}{3} \, x^{3} - 12 \, x^{2} + 9 \, x \]

[In]

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

[Out]

-100/3*x^6 + 12*x^5 + 69/2*x^4 - 47/3*x^3 - 12*x^2 + 9*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^3 (3+5 x)^2 \, dx=-\frac {100}{3} \, x^{6} + 12 \, x^{5} + \frac {69}{2} \, x^{4} - \frac {47}{3} \, x^{3} - 12 \, x^{2} + 9 \, x \]

[In]

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

[Out]

-100/3*x^6 + 12*x^5 + 69/2*x^4 - 47/3*x^3 - 12*x^2 + 9*x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^3 (3+5 x)^2 \, dx=-\frac {100\,x^6}{3}+12\,x^5+\frac {69\,x^4}{2}-\frac {47\,x^3}{3}-12\,x^2+9\,x \]

[In]

int(-(2*x - 1)^3*(5*x + 3)^2,x)

[Out]

9*x - 12*x^2 - (47*x^3)/3 + (69*x^4)/2 + 12*x^5 - (100*x^6)/3