3.1167 \(\int \frac{(1-2 x) (3+5 x)^2}{2+3 x} \, dx\)

Optimal. Leaf size=30 \[ -\frac{50 x^3}{9}-\frac{5 x^2}{18}+\frac{118 x}{27}+\frac{7}{81} \log (3 x+2) \]

[Out]

(118*x)/27 - (5*x^2)/18 - (50*x^3)/9 + (7*Log[2 + 3*x])/81

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Rubi [A]  time = 0.0126983, antiderivative size = 30, 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} \[ -\frac{50 x^3}{9}-\frac{5 x^2}{18}+\frac{118 x}{27}+\frac{7}{81} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(118*x)/27 - (5*x^2)/18 - (50*x^3)/9 + (7*Log[2 + 3*x])/81

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) (3+5 x)^2}{2+3 x} \, dx &=\int \left (\frac{118}{27}-\frac{5 x}{9}-\frac{50 x^2}{3}+\frac{7}{27 (2+3 x)}\right ) \, dx\\ &=\frac{118 x}{27}-\frac{5 x^2}{18}-\frac{50 x^3}{9}+\frac{7}{81} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0094917, size = 27, normalized size = 0.9 \[ \frac{1}{486} \left (-2700 x^3-135 x^2+2124 x+42 \log (3 x+2)+676\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(676 + 2124*x - 135*x^2 - 2700*x^3 + 42*Log[2 + 3*x])/486

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Maple [A]  time = 0.002, size = 23, normalized size = 0.8 \begin{align*}{\frac{118\,x}{27}}-{\frac{5\,{x}^{2}}{18}}-{\frac{50\,{x}^{3}}{9}}+{\frac{7\,\ln \left ( 2+3\,x \right ) }{81}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

118/27*x-5/18*x^2-50/9*x^3+7/81*ln(2+3*x)

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Maxima [A]  time = 1.10636, size = 30, normalized size = 1. \begin{align*} -\frac{50}{9} \, x^{3} - \frac{5}{18} \, x^{2} + \frac{118}{27} \, x + \frac{7}{81} \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-50/9*x^3 - 5/18*x^2 + 118/27*x + 7/81*log(3*x + 2)

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Fricas [A]  time = 1.74567, size = 72, normalized size = 2.4 \begin{align*} -\frac{50}{9} \, x^{3} - \frac{5}{18} \, x^{2} + \frac{118}{27} \, x + \frac{7}{81} \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-50/9*x^3 - 5/18*x^2 + 118/27*x + 7/81*log(3*x + 2)

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Sympy [A]  time = 0.08637, size = 27, normalized size = 0.9 \begin{align*} - \frac{50 x^{3}}{9} - \frac{5 x^{2}}{18} + \frac{118 x}{27} + \frac{7 \log{\left (3 x + 2 \right )}}{81} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-50*x**3/9 - 5*x**2/18 + 118*x/27 + 7*log(3*x + 2)/81

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Giac [A]  time = 1.86205, size = 31, normalized size = 1.03 \begin{align*} -\frac{50}{9} \, x^{3} - \frac{5}{18} \, x^{2} + \frac{118}{27} \, x + \frac{7}{81} \, \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)*(3+5*x)^2/(2+3*x),x, algorithm="giac")

[Out]

-50/9*x^3 - 5/18*x^2 + 118/27*x + 7/81*log(abs(3*x + 2))