3.1245 \(\int \frac{(1-2 x)^2 (3+5 x)}{2+3 x} \, dx\)

Optimal. Leaf size=30 \[ \frac{20 x^3}{9}-\frac{32 x^2}{9}+\frac{65 x}{27}-\frac{49}{81} \log (3 x+2) \]

[Out]

(65*x)/27 - (32*x^2)/9 + (20*x^3)/9 - (49*Log[2 + 3*x])/81

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Rubi [A]  time = 0.0114052, 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} \[ \frac{20 x^3}{9}-\frac{32 x^2}{9}+\frac{65 x}{27}-\frac{49}{81} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(65*x)/27 - (32*x^2)/9 + (20*x^3)/9 - (49*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)^2 (3+5 x)}{2+3 x} \, dx &=\int \left (\frac{65}{27}-\frac{64 x}{9}+\frac{20 x^2}{3}-\frac{49}{27 (2+3 x)}\right ) \, dx\\ &=\frac{65 x}{27}-\frac{32 x^2}{9}+\frac{20 x^3}{9}-\frac{49}{81} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0101094, size = 27, normalized size = 0.9 \[ \frac{1}{243} \left (540 x^3-864 x^2+585 x-147 \log (3 x+2)+934\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(934 + 585*x - 864*x^2 + 540*x^3 - 147*Log[2 + 3*x])/243

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Maple [A]  time = 0.003, size = 23, normalized size = 0.8 \begin{align*}{\frac{65\,x}{27}}-{\frac{32\,{x}^{2}}{9}}+{\frac{20\,{x}^{3}}{9}}-{\frac{49\,\ln \left ( 2+3\,x \right ) }{81}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

65/27*x-32/9*x^2+20/9*x^3-49/81*ln(2+3*x)

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Maxima [A]  time = 1.02672, size = 30, normalized size = 1. \begin{align*} \frac{20}{9} \, x^{3} - \frac{32}{9} \, x^{2} + \frac{65}{27} \, x - \frac{49}{81} \, \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),x, algorithm="maxima")

[Out]

20/9*x^3 - 32/9*x^2 + 65/27*x - 49/81*log(3*x + 2)

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Fricas [A]  time = 1.45707, size = 70, normalized size = 2.33 \begin{align*} \frac{20}{9} \, x^{3} - \frac{32}{9} \, x^{2} + \frac{65}{27} \, x - \frac{49}{81} \, \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),x, algorithm="fricas")

[Out]

20/9*x^3 - 32/9*x^2 + 65/27*x - 49/81*log(3*x + 2)

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Sympy [A]  time = 0.084244, size = 27, normalized size = 0.9 \begin{align*} \frac{20 x^{3}}{9} - \frac{32 x^{2}}{9} + \frac{65 x}{27} - \frac{49 \log{\left (3 x + 2 \right )}}{81} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

20*x**3/9 - 32*x**2/9 + 65*x/27 - 49*log(3*x + 2)/81

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Giac [A]  time = 1.93782, size = 31, normalized size = 1.03 \begin{align*} \frac{20}{9} \, x^{3} - \frac{32}{9} \, x^{2} + \frac{65}{27} \, x - \frac{49}{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)^2*(3+5*x)/(2+3*x),x, algorithm="giac")

[Out]

20/9*x^3 - 32/9*x^2 + 65/27*x - 49/81*log(abs(3*x + 2))