3.1246 \(\int \frac{(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx\)

Optimal. Leaf size=34 \[ \frac{10 x^2}{9}-\frac{104 x}{27}+\frac{49}{81 (3 x+2)}+\frac{91}{27} \log (3 x+2) \]

[Out]

(-104*x)/27 + (10*x^2)/9 + 49/(81*(2 + 3*x)) + (91*Log[2 + 3*x])/27

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Rubi [A]  time = 0.0142177, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {77} \[ \frac{10 x^2}{9}-\frac{104 x}{27}+\frac{49}{81 (3 x+2)}+\frac{91}{27} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-104*x)/27 + (10*x^2)/9 + 49/(81*(2 + 3*x)) + (91*Log[2 + 3*x])/27

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx &=\int \left (-\frac{104}{27}+\frac{20 x}{9}-\frac{49}{27 (2+3 x)^2}+\frac{91}{9 (2+3 x)}\right ) \, dx\\ &=-\frac{104 x}{27}+\frac{10 x^2}{9}+\frac{49}{81 (2+3 x)}+\frac{91}{27} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0126663, size = 39, normalized size = 1.15 \[ \frac{540 x^3-1512 x^2-447 x+546 (3 x+2) \log (6 x+4)+632}{162 (3 x+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(632 - 447*x - 1512*x^2 + 540*x^3 + 546*(2 + 3*x)*Log[4 + 6*x])/(162*(2 + 3*x))

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Maple [A]  time = 0.005, size = 27, normalized size = 0.8 \begin{align*} -{\frac{104\,x}{27}}+{\frac{10\,{x}^{2}}{9}}+{\frac{49}{162+243\,x}}+{\frac{91\,\ln \left ( 2+3\,x \right ) }{27}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-104/27*x+10/9*x^2+49/81/(2+3*x)+91/27*ln(2+3*x)

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Maxima [A]  time = 1.02621, size = 35, normalized size = 1.03 \begin{align*} \frac{10}{9} \, x^{2} - \frac{104}{27} \, x + \frac{49}{81 \,{\left (3 \, x + 2\right )}} + \frac{91}{27} \, \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+3*x)^2,x, algorithm="maxima")

[Out]

10/9*x^2 - 104/27*x + 49/81/(3*x + 2) + 91/27*log(3*x + 2)

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Fricas [A]  time = 1.4733, size = 105, normalized size = 3.09 \begin{align*} \frac{270 \, x^{3} - 756 \, x^{2} + 273 \,{\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 624 \, x + 49}{81 \,{\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+3*x)^2,x, algorithm="fricas")

[Out]

1/81*(270*x^3 - 756*x^2 + 273*(3*x + 2)*log(3*x + 2) - 624*x + 49)/(3*x + 2)

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Sympy [A]  time = 0.097296, size = 27, normalized size = 0.79 \begin{align*} \frac{10 x^{2}}{9} - \frac{104 x}{27} + \frac{91 \log{\left (3 x + 2 \right )}}{27} + \frac{49}{243 x + 162} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*x**2/9 - 104*x/27 + 91*log(3*x + 2)/27 + 49/(243*x + 162)

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Giac [A]  time = 2.9648, size = 65, normalized size = 1.91 \begin{align*} -\frac{2}{81} \,{\left (3 \, x + 2\right )}^{2}{\left (\frac{72}{3 \, x + 2} - 5\right )} + \frac{49}{81 \,{\left (3 \, x + 2\right )}} - \frac{91}{27} \, \log \left (\frac{{\left | 3 \, x + 2 \right |}}{3 \,{\left (3 \, x + 2\right )}^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/81*(3*x + 2)^2*(72/(3*x + 2) - 5) + 49/81/(3*x + 2) - 91/27*log(1/3*abs(3*x + 2)/(3*x + 2)^2)