3.1261 \(\int (1-2 x)^2 (3+5 x)^2 \, dx\)

Optimal. Leaf size=26 \[ 20 x^5+5 x^4-\frac {59 x^3}{3}-3 x^2+9 x \]

[Out]

9*x-3*x^2-59/3*x^3+5*x^4+20*x^5

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {43} \[ 20 x^5+5 x^4-\frac {59 x^3}{3}-3 x^2+9 x \]

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^2*(3 + 5*x)^2,x]

[Out]

9*x - 3*x^2 - (59*x^3)/3 + 5*x^4 + 20*x^5

Rule 43

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*} \int (1-2 x)^2 (3+5 x)^2 \, dx &=\int \left (9-6 x-59 x^2+20 x^3+100 x^4\right ) \, dx\\ &=9 x-3 x^2-\frac {59 x^3}{3}+5 x^4+20 x^5\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 26, normalized size = 1.00 \[ 20 x^5+5 x^4-\frac {59 x^3}{3}-3 x^2+9 x \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x)^2*(3 + 5*x)^2,x]

[Out]

9*x - 3*x^2 - (59*x^3)/3 + 5*x^4 + 20*x^5

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fricas [A]  time = 0.65, size = 24, normalized size = 0.92 \[ 20 x^{5} + 5 x^{4} - \frac {59}{3} x^{3} - 3 x^{2} + 9 x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

20*x^5 + 5*x^4 - 59/3*x^3 - 3*x^2 + 9*x

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giac [A]  time = 1.42, size = 24, normalized size = 0.92 \[ 20 \, x^{5} + 5 \, x^{4} - \frac {59}{3} \, x^{3} - 3 \, x^{2} + 9 \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

20*x^5 + 5*x^4 - 59/3*x^3 - 3*x^2 + 9*x

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maple [A]  time = 0.00, size = 25, normalized size = 0.96 \[ 20 x^{5}+5 x^{4}-\frac {59}{3} x^{3}-3 x^{2}+9 x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9*x-3*x^2-59/3*x^3+5*x^4+20*x^5

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maxima [A]  time = 0.63, size = 24, normalized size = 0.92 \[ 20 \, x^{5} + 5 \, x^{4} - \frac {59}{3} \, x^{3} - 3 \, x^{2} + 9 \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

20*x^5 + 5*x^4 - 59/3*x^3 - 3*x^2 + 9*x

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mupad [B]  time = 0.02, size = 24, normalized size = 0.92 \[ 20\,x^5+5\,x^4-\frac {59\,x^3}{3}-3\,x^2+9\,x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - 1)^2*(5*x + 3)^2,x)

[Out]

9*x - 3*x^2 - (59*x^3)/3 + 5*x^4 + 20*x^5

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sympy [A]  time = 0.06, size = 24, normalized size = 0.92 \[ 20 x^{5} + 5 x^{4} - \frac {59 x^{3}}{3} - 3 x^{2} + 9 x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

20*x**5 + 5*x**4 - 59*x**3/3 - 3*x**2 + 9*x

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