3.1168 \(\int \frac{(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx\)

Optimal. Leaf size=34 \[ -\frac{25 x^2}{9}+\frac{95 x}{27}-\frac{7}{81 (3 x+2)}-\frac{8}{9} \log (3 x+2) \]

[Out]

(95*x)/27 - (25*x^2)/9 - 7/(81*(2 + 3*x)) - (8*Log[2 + 3*x])/9

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Rubi [A]  time = 0.0153246, 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} \[ -\frac{25 x^2}{9}+\frac{95 x}{27}-\frac{7}{81 (3 x+2)}-\frac{8}{9} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(95*x)/27 - (25*x^2)/9 - 7/(81*(2 + 3*x)) - (8*Log[2 + 3*x])/9

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)^2} \, dx &=\int \left (\frac{95}{27}-\frac{50 x}{9}+\frac{7}{27 (2+3 x)^2}-\frac{8}{3 (2+3 x)}\right ) \, dx\\ &=\frac{95 x}{27}-\frac{25 x^2}{9}-\frac{7}{81 (2+3 x)}-\frac{8}{9} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0098726, size = 36, normalized size = 1.06 \[ \frac{-225 x^3+135 x^2+480 x-24 (3 x+2) \log (3 x+2)+191}{81 x+54} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(191 + 480*x + 135*x^2 - 225*x^3 - 24*(2 + 3*x)*Log[2 + 3*x])/(54 + 81*x)

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Maple [A]  time = 0.005, size = 27, normalized size = 0.8 \begin{align*}{\frac{95\,x}{27}}-{\frac{25\,{x}^{2}}{9}}-{\frac{7}{162+243\,x}}-{\frac{8\,\ln \left ( 2+3\,x \right ) }{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

95/27*x-25/9*x^2-7/81/(2+3*x)-8/9*ln(2+3*x)

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Maxima [A]  time = 1.08661, size = 35, normalized size = 1.03 \begin{align*} -\frac{25}{9} \, x^{2} + \frac{95}{27} \, x - \frac{7}{81 \,{\left (3 \, x + 2\right )}} - \frac{8}{9} \, \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)^2,x, algorithm="maxima")

[Out]

-25/9*x^2 + 95/27*x - 7/81/(3*x + 2) - 8/9*log(3*x + 2)

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Fricas [A]  time = 1.73467, size = 104, normalized size = 3.06 \begin{align*} -\frac{675 \, x^{3} - 405 \, x^{2} + 72 \,{\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 570 \, x + 7}{81 \,{\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)^2,x, algorithm="fricas")

[Out]

-1/81*(675*x^3 - 405*x^2 + 72*(3*x + 2)*log(3*x + 2) - 570*x + 7)/(3*x + 2)

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Sympy [A]  time = 0.096711, size = 27, normalized size = 0.79 \begin{align*} - \frac{25 x^{2}}{9} + \frac{95 x}{27} - \frac{8 \log{\left (3 x + 2 \right )}}{9} - \frac{7}{243 x + 162} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25*x**2/9 + 95*x/27 - 8*log(3*x + 2)/9 - 7/(243*x + 162)

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Giac [A]  time = 3.36877, size = 65, normalized size = 1.91 \begin{align*} \frac{5}{81} \,{\left (3 \, x + 2\right )}^{2}{\left (\frac{39}{3 \, x + 2} - 5\right )} - \frac{7}{81 \,{\left (3 \, x + 2\right )}} + \frac{8}{9} \, \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)*(3+5*x)^2/(2+3*x)^2,x, algorithm="giac")

[Out]

5/81*(3*x + 2)^2*(39/(3*x + 2) - 5) - 7/81/(3*x + 2) + 8/9*log(1/3*abs(3*x + 2)/(3*x + 2)^2)