\(\int (1-2 x)^2 (3+5 x)^3 \, dx\) [1280]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int (1-2 x)^2 (3+5 x)^3 \, dx=\frac {121}{500} (3+5 x)^4-\frac {44}{625} (3+5 x)^5+\frac {2}{375} (3+5 x)^6 \]

[Out]

121/500*(3+5*x)^4-44/625*(3+5*x)^5+2/375*(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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int (1-2 x)^2 (3+5 x)^3 \, dx=\frac {2}{375} (5 x+3)^6-\frac {44}{625} (5 x+3)^5+\frac {121}{500} (5 x+3)^4 \]

[In]

Int[(1 - 2*x)^2*(3 + 5*x)^3,x]

[Out]

(121*(3 + 5*x)^4)/500 - (44*(3 + 5*x)^5)/625 + (2*(3 + 5*x)^6)/375

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}{25} (3+5 x)^3-\frac {44}{25} (3+5 x)^4+\frac {4}{25} (3+5 x)^5\right ) \, dx \\ & = \frac {121}{500} (3+5 x)^4-\frac {44}{625} (3+5 x)^5+\frac {2}{375} (3+5 x)^6 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int (1-2 x)^2 (3+5 x)^3 \, dx=27 x+\frac {27 x^2}{2}-69 x^3-\frac {235 x^4}{4}+80 x^5+\frac {250 x^6}{3} \]

[In]

Integrate[(1 - 2*x)^2*(3 + 5*x)^3,x]

[Out]

27*x + (27*x^2)/2 - 69*x^3 - (235*x^4)/4 + 80*x^5 + (250*x^6)/3

Maple [A] (verified)

Time = 1.87 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85

method	result	size
gosper	\(\frac {x \left (1000 x^{5}+960 x^{4}-705 x^{3}-828 x^{2}+162 x +324\right )}{12}\)	\(29\)
default	\(\frac {250}{3} x^{6}+80 x^{5}-\frac {235}{4} x^{4}-69 x^{3}+\frac {27}{2} x^{2}+27 x\)	\(30\)
norman	\(\frac {250}{3} x^{6}+80 x^{5}-\frac {235}{4} x^{4}-69 x^{3}+\frac {27}{2} x^{2}+27 x\)	\(30\)
risch	\(\frac {250}{3} x^{6}+80 x^{5}-\frac {235}{4} x^{4}-69 x^{3}+\frac {27}{2} x^{2}+27 x\)	\(30\)
parallelrisch	\(\frac {250}{3} x^{6}+80 x^{5}-\frac {235}{4} x^{4}-69 x^{3}+\frac {27}{2} x^{2}+27 x\)	\(30\)

[In]

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

[Out]

1/12*x*(1000*x^5+960*x^4-705*x^3-828*x^2+162*x+324)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^2 (3+5 x)^3 \, dx=\frac {250}{3} \, x^{6} + 80 \, x^{5} - \frac {235}{4} \, x^{4} - 69 \, x^{3} + \frac {27}{2} \, x^{2} + 27 \, x \]

[In]

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

[Out]

250/3*x^6 + 80*x^5 - 235/4*x^4 - 69*x^3 + 27/2*x^2 + 27*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int (1-2 x)^2 (3+5 x)^3 \, dx=\frac {250 x^{6}}{3} + 80 x^{5} - \frac {235 x^{4}}{4} - 69 x^{3} + \frac {27 x^{2}}{2} + 27 x \]

[In]

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

[Out]

250*x**6/3 + 80*x**5 - 235*x**4/4 - 69*x**3 + 27*x**2/2 + 27*x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^2 (3+5 x)^3 \, dx=\frac {250}{3} \, x^{6} + 80 \, x^{5} - \frac {235}{4} \, x^{4} - 69 \, x^{3} + \frac {27}{2} \, x^{2} + 27 \, x \]

[In]

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

[Out]

250/3*x^6 + 80*x^5 - 235/4*x^4 - 69*x^3 + 27/2*x^2 + 27*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^2 (3+5 x)^3 \, dx=\frac {250}{3} \, x^{6} + 80 \, x^{5} - \frac {235}{4} \, x^{4} - 69 \, x^{3} + \frac {27}{2} \, x^{2} + 27 \, x \]

[In]

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

[Out]

250/3*x^6 + 80*x^5 - 235/4*x^4 - 69*x^3 + 27/2*x^2 + 27*x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^2 (3+5 x)^3 \, dx=\frac {250\,x^6}{3}+80\,x^5-\frac {235\,x^4}{4}-69\,x^3+\frac {27\,x^2}{2}+27\,x \]

[In]

int((2*x - 1)^2*(5*x + 3)^3,x)

[Out]

27*x + (27*x^2)/2 - 69*x^3 - (235*x^4)/4 + 80*x^5 + (250*x^6)/3