3.1212 \(\int \frac{(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=34 \[ -\frac{9 x^2}{25}+\frac{33 x}{125}-\frac{11}{625 (5 x+3)}+\frac{64}{625} \log (5 x+3) \]

[Out]

(33*x)/125 - (9*x^2)/25 - 11/(625*(3 + 5*x)) + (64*Log[3 + 5*x])/625

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Rubi [A]  time = 0.0148097, 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} \[ -\frac{9 x^2}{25}+\frac{33 x}{125}-\frac{11}{625 (5 x+3)}+\frac{64}{625} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(33*x)/125 - (9*x^2)/25 - 11/(625*(3 + 5*x)) + (64*Log[3 + 5*x])/625

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 x)^2}{(3+5 x)^2} \, dx &=\int \left (\frac{33}{125}-\frac{18 x}{25}+\frac{11}{125 (3+5 x)^2}+\frac{64}{125 (3+5 x)}\right ) \, dx\\ &=\frac{33 x}{125}-\frac{9 x^2}{25}-\frac{11}{625 (3+5 x)}+\frac{64}{625} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0136871, size = 41, normalized size = 1.21 \[ \frac{-1125 x^3+150 x^2+1545 x+64 (5 x+3) \log (-3 (5 x+3))+619}{625 (5 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(619 + 1545*x + 150*x^2 - 1125*x^3 + 64*(3 + 5*x)*Log[-3*(3 + 5*x)])/(625*(3 + 5*x))

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Maple [A]  time = 0.005, size = 27, normalized size = 0.8 \begin{align*}{\frac{33\,x}{125}}-{\frac{9\,{x}^{2}}{25}}-{\frac{11}{1875+3125\,x}}+{\frac{64\,\ln \left ( 3+5\,x \right ) }{625}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

33/125*x-9/25*x^2-11/625/(3+5*x)+64/625*ln(3+5*x)

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Maxima [A]  time = 0.994804, size = 35, normalized size = 1.03 \begin{align*} -\frac{9}{25} \, x^{2} + \frac{33}{125} \, x - \frac{11}{625 \,{\left (5 \, x + 3\right )}} + \frac{64}{625} \, \log \left (5 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/25*x^2 + 33/125*x - 11/625/(5*x + 3) + 64/625*log(5*x + 3)

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Fricas [A]  time = 1.46311, size = 108, normalized size = 3.18 \begin{align*} -\frac{1125 \, x^{3} - 150 \, x^{2} - 64 \,{\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 495 \, x + 11}{625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/625*(1125*x^3 - 150*x^2 - 64*(5*x + 3)*log(5*x + 3) - 495*x + 11)/(5*x + 3)

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Sympy [A]  time = 0.096639, size = 27, normalized size = 0.79 \begin{align*} - \frac{9 x^{2}}{25} + \frac{33 x}{125} + \frac{64 \log{\left (5 x + 3 \right )}}{625} - \frac{11}{3125 x + 1875} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*x**2/25 + 33*x/125 + 64*log(5*x + 3)/625 - 11/(3125*x + 1875)

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Giac [A]  time = 2.24429, size = 65, normalized size = 1.91 \begin{align*} \frac{3}{625} \,{\left (5 \, x + 3\right )}^{2}{\left (\frac{29}{5 \, x + 3} - 3\right )} - \frac{11}{625 \,{\left (5 \, x + 3\right )}} - \frac{64}{625} \, \log \left (\frac{{\left | 5 \, x + 3 \right |}}{5 \,{\left (5 \, x + 3\right )}^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/625*(5*x + 3)^2*(29/(5*x + 3) - 3) - 11/625/(5*x + 3) - 64/625*log(1/5*abs(5*x + 3)/(5*x + 3)^2)