3.1262 \(\int \frac{(1-2 x)^2 (3+5 x)^2}{2+3 x} \, dx\)

Optimal. Leaf size=37 \[ \frac{25 x^4}{3}-\frac{140 x^3}{27}-\frac{251 x^2}{54}+\frac{340 x}{81}+\frac{49}{243} \log (3 x+2) \]

[Out]

(340*x)/81 - (251*x^2)/54 - (140*x^3)/27 + (25*x^4)/3 + (49*Log[2 + 3*x])/243

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Rubi [A]  time = 0.0157756, 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} \[ \frac{25 x^4}{3}-\frac{140 x^3}{27}-\frac{251 x^2}{54}+\frac{340 x}{81}+\frac{49}{243} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(340*x)/81 - (251*x^2)/54 - (140*x^3)/27 + (25*x^4)/3 + (49*Log[2 + 3*x])/243

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 (3+5 x)^2}{2+3 x} \, dx &=\int \left (\frac{340}{81}-\frac{251 x}{27}-\frac{140 x^2}{9}+\frac{100 x^3}{3}+\frac{49}{81 (2+3 x)}\right ) \, dx\\ &=\frac{340 x}{81}-\frac{251 x^2}{54}-\frac{140 x^3}{27}+\frac{25 x^4}{3}+\frac{49}{243} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0086873, size = 32, normalized size = 0.86 \[ \frac{12150 x^4-7560 x^3-6777 x^2+6120 x+294 \log (3 x+2)+2452}{1458} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2452 + 6120*x - 6777*x^2 - 7560*x^3 + 12150*x^4 + 294*Log[2 + 3*x])/1458

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Maple [A]  time = 0.002, size = 28, normalized size = 0.8 \begin{align*}{\frac{340\,x}{81}}-{\frac{251\,{x}^{2}}{54}}-{\frac{140\,{x}^{3}}{27}}+{\frac{25\,{x}^{4}}{3}}+{\frac{49\,\ln \left ( 2+3\,x \right ) }{243}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

340/81*x-251/54*x^2-140/27*x^3+25/3*x^4+49/243*ln(2+3*x)

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Maxima [A]  time = 1.00507, size = 36, normalized size = 0.97 \begin{align*} \frac{25}{3} \, x^{4} - \frac{140}{27} \, x^{3} - \frac{251}{54} \, x^{2} + \frac{340}{81} \, x + \frac{49}{243} \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/3*x^4 - 140/27*x^3 - 251/54*x^2 + 340/81*x + 49/243*log(3*x + 2)

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Fricas [A]  time = 1.57954, size = 93, normalized size = 2.51 \begin{align*} \frac{25}{3} \, x^{4} - \frac{140}{27} \, x^{3} - \frac{251}{54} \, x^{2} + \frac{340}{81} \, x + \frac{49}{243} \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/3*x^4 - 140/27*x^3 - 251/54*x^2 + 340/81*x + 49/243*log(3*x + 2)

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Sympy [A]  time = 0.086662, size = 34, normalized size = 0.92 \begin{align*} \frac{25 x^{4}}{3} - \frac{140 x^{3}}{27} - \frac{251 x^{2}}{54} + \frac{340 x}{81} + \frac{49 \log{\left (3 x + 2 \right )}}{243} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*x**4/3 - 140*x**3/27 - 251*x**2/54 + 340*x/81 + 49*log(3*x + 2)/243

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Giac [A]  time = 1.69126, size = 38, normalized size = 1.03 \begin{align*} \frac{25}{3} \, x^{4} - \frac{140}{27} \, x^{3} - \frac{251}{54} \, x^{2} + \frac{340}{81} \, x + \frac{49}{243} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/3*x^4 - 140/27*x^3 - 251/54*x^2 + 340/81*x + 49/243*log(abs(3*x + 2))