3.1294 \(\int \frac{(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx\)

Optimal. Leaf size=37 \[ \frac{9 x^4}{5}-\frac{16 x^3}{25}-\frac{431 x^2}{250}+\frac{793 x}{625}+\frac{121 \log (5 x+3)}{3125} \]

[Out]

(793*x)/625 - (431*x^2)/250 - (16*x^3)/25 + (9*x^4)/5 + (121*Log[3 + 5*x])/3125

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Rubi [A]  time = 0.0168758, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {88} \[ \frac{9 x^4}{5}-\frac{16 x^3}{25}-\frac{431 x^2}{250}+\frac{793 x}{625}+\frac{121 \log (5 x+3)}{3125} \]

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x)^2*(2 + 3*x)^2)/(3 + 5*x),x]

[Out]

(793*x)/625 - (431*x^2)/250 - (16*x^3)/25 + (9*x^4)/5 + (121*Log[3 + 5*x])/3125

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx &=\int \left (\frac{793}{625}-\frac{431 x}{125}-\frac{48 x^2}{25}+\frac{36 x^3}{5}+\frac{121}{625 (3+5 x)}\right ) \, dx\\ &=\frac{793 x}{625}-\frac{431 x^2}{250}-\frac{16 x^3}{25}+\frac{9 x^4}{5}+\frac{121 \log (3+5 x)}{3125}\\ \end{align*}

Mathematica [A]  time = 0.0100934, size = 35, normalized size = 0.95 \[ \frac{5 \left (2250 x^4-800 x^3-2155 x^2+1586 x+1263\right )+242 \log (5 x+3)}{6250} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x)^2*(2 + 3*x)^2)/(3 + 5*x),x]

[Out]

(5*(1263 + 1586*x - 2155*x^2 - 800*x^3 + 2250*x^4) + 242*Log[3 + 5*x])/6250

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Maple [A]  time = 0.002, size = 28, normalized size = 0.8 \begin{align*}{\frac{793\,x}{625}}-{\frac{431\,{x}^{2}}{250}}-{\frac{16\,{x}^{3}}{25}}+{\frac{9\,{x}^{4}}{5}}+{\frac{121\,\ln \left ( 3+5\,x \right ) }{3125}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

793/625*x-431/250*x^2-16/25*x^3+9/5*x^4+121/3125*ln(3+5*x)

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Maxima [A]  time = 0.98556, size = 36, normalized size = 0.97 \begin{align*} \frac{9}{5} \, x^{4} - \frac{16}{25} \, x^{3} - \frac{431}{250} \, x^{2} + \frac{793}{625} \, x + \frac{121}{3125} \, \log \left (5 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/5*x^4 - 16/25*x^3 - 431/250*x^2 + 793/625*x + 121/3125*log(5*x + 3)

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Fricas [A]  time = 1.48669, size = 96, normalized size = 2.59 \begin{align*} \frac{9}{5} \, x^{4} - \frac{16}{25} \, x^{3} - \frac{431}{250} \, x^{2} + \frac{793}{625} \, x + \frac{121}{3125} \, \log \left (5 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/5*x^4 - 16/25*x^3 - 431/250*x^2 + 793/625*x + 121/3125*log(5*x + 3)

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Sympy [A]  time = 0.085488, size = 34, normalized size = 0.92 \begin{align*} \frac{9 x^{4}}{5} - \frac{16 x^{3}}{25} - \frac{431 x^{2}}{250} + \frac{793 x}{625} + \frac{121 \log{\left (5 x + 3 \right )}}{3125} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9*x**4/5 - 16*x**3/25 - 431*x**2/250 + 793*x/625 + 121*log(5*x + 3)/3125

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Giac [A]  time = 2.6876, size = 38, normalized size = 1.03 \begin{align*} \frac{9}{5} \, x^{4} - \frac{16}{25} \, x^{3} - \frac{431}{250} \, x^{2} + \frac{793}{625} \, x + \frac{121}{3125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/5*x^4 - 16/25*x^3 - 431/250*x^2 + 793/625*x + 121/3125*log(abs(5*x + 3))