3.1409 \(\int \frac{(1-2 x)^3}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=34 \[ -\frac{4 x^2}{25}+\frac{108 x}{125}-\frac{1331}{625 (5 x+3)}-\frac{726}{625} \log (5 x+3) \]

[Out]

(108*x)/125 - (4*x^2)/25 - 1331/(625*(3 + 5*x)) - (726*Log[3 + 5*x])/625

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Rubi [A]  time = 0.01423, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {43} \[ -\frac{4 x^2}{25}+\frac{108 x}{125}-\frac{1331}{625 (5 x+3)}-\frac{726}{625} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^3/(3 + 5*x)^2,x]

[Out]

(108*x)/125 - (4*x^2)/25 - 1331/(625*(3 + 5*x)) - (726*Log[3 + 5*x])/625

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(1-2 x)^3}{(3+5 x)^2} \, dx &=\int \left (\frac{108}{125}-\frac{8 x}{25}+\frac{1331}{125 (3+5 x)^2}-\frac{726}{125 (3+5 x)}\right ) \, dx\\ &=\frac{108 x}{125}-\frac{4 x^2}{25}-\frac{1331}{625 (3+5 x)}-\frac{726}{625} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0104083, size = 39, normalized size = 1.15 \[ \frac{-500 x^3+2400 x^2+395 x-726 (5 x+3) \log (10 x+6)-2066}{625 (5 x+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x)^3/(3 + 5*x)^2,x]

[Out]

(-2066 + 395*x + 2400*x^2 - 500*x^3 - 726*(3 + 5*x)*Log[6 + 10*x])/(625*(3 + 5*x))

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Maple [A]  time = 0.004, size = 27, normalized size = 0.8 \begin{align*}{\frac{108\,x}{125}}-{\frac{4\,{x}^{2}}{25}}-{\frac{1331}{1875+3125\,x}}-{\frac{726\,\ln \left ( 3+5\,x \right ) }{625}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

108/125*x-4/25*x^2-1331/625/(3+5*x)-726/625*ln(3+5*x)

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Maxima [A]  time = 1.43875, size = 35, normalized size = 1.03 \begin{align*} -\frac{4}{25} \, x^{2} + \frac{108}{125} \, x - \frac{1331}{625 \,{\left (5 \, x + 3\right )}} - \frac{726}{625} \, \log \left (5 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/25*x^2 + 108/125*x - 1331/625/(5*x + 3) - 726/625*log(5*x + 3)

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Fricas [A]  time = 1.33547, size = 113, normalized size = 3.32 \begin{align*} -\frac{500 \, x^{3} - 2400 \, x^{2} + 726 \,{\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1620 \, x + 1331}{625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/625*(500*x^3 - 2400*x^2 + 726*(5*x + 3)*log(5*x + 3) - 1620*x + 1331)/(5*x + 3)

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Sympy [A]  time = 0.095176, size = 27, normalized size = 0.79 \begin{align*} - \frac{4 x^{2}}{25} + \frac{108 x}{125} - \frac{726 \log{\left (5 x + 3 \right )}}{625} - \frac{1331}{3125 x + 1875} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x**2/25 + 108*x/125 - 726*log(5*x + 3)/625 - 1331/(3125*x + 1875)

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Giac [A]  time = 2.33065, size = 65, normalized size = 1.91 \begin{align*} \frac{4}{625} \,{\left (5 \, x + 3\right )}^{2}{\left (\frac{33}{5 \, x + 3} - 1\right )} - \frac{1331}{625 \,{\left (5 \, x + 3\right )}} + \frac{726}{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)^3/(3+5*x)^2,x, algorithm="giac")

[Out]

4/625*(5*x + 3)^2*(33/(5*x + 3) - 1) - 1331/625/(5*x + 3) + 726/625*log(1/5*abs(5*x + 3)/(5*x + 3)^2)