3.1311 \(\int \frac{(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=34 \[ \frac{6 x^2}{25}-\frac{92 x}{125}-\frac{121}{625 (5 x+3)}+\frac{319}{625} \log (5 x+3) \]

[Out]

(-92*x)/125 + (6*x^2)/25 - 121/(625*(3 + 5*x)) + (319*Log[3 + 5*x])/625

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Rubi [A]  time = 0.0142219, 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} \[ \frac{6 x^2}{25}-\frac{92 x}{125}-\frac{121}{625 (5 x+3)}+\frac{319}{625} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-92*x)/125 + (6*x^2)/25 - 121/(625*(3 + 5*x)) + (319*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 (2+3 x)}{(3+5 x)^2} \, dx &=\int \left (-\frac{92}{125}+\frac{12 x}{25}+\frac{121}{125 (3+5 x)^2}+\frac{319}{125 (3+5 x)}\right ) \, dx\\ &=-\frac{92 x}{125}+\frac{6 x^2}{25}-\frac{121}{625 (3+5 x)}+\frac{319}{625} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0125446, size = 39, normalized size = 1.15 \[ \frac{1500 x^3-3700 x^2-835 x+638 (5 x+3) \log (10 x+6)+913}{1250 (5 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(913 - 835*x - 3700*x^2 + 1500*x^3 + 638*(3 + 5*x)*Log[6 + 10*x])/(1250*(3 + 5*x))

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Maple [A]  time = 0.005, size = 27, normalized size = 0.8 \begin{align*} -{\frac{92\,x}{125}}+{\frac{6\,{x}^{2}}{25}}-{\frac{121}{1875+3125\,x}}+{\frac{319\,\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*(2+3*x)/(3+5*x)^2,x)

[Out]

-92/125*x+6/25*x^2-121/625/(3+5*x)+319/625*ln(3+5*x)

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Maxima [A]  time = 2.67715, size = 35, normalized size = 1.03 \begin{align*} \frac{6}{25} \, x^{2} - \frac{92}{125} \, x - \frac{121}{625 \,{\left (5 \, x + 3\right )}} + \frac{319}{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*(2+3*x)/(3+5*x)^2,x, algorithm="maxima")

[Out]

6/25*x^2 - 92/125*x - 121/625/(5*x + 3) + 319/625*log(5*x + 3)

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Fricas [A]  time = 1.45722, size = 111, normalized size = 3.26 \begin{align*} \frac{750 \, x^{3} - 1850 \, x^{2} + 319 \,{\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1380 \, x - 121}{625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/625*(750*x^3 - 1850*x^2 + 319*(5*x + 3)*log(5*x + 3) - 1380*x - 121)/(5*x + 3)

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Sympy [A]  time = 0.09666, size = 27, normalized size = 0.79 \begin{align*} \frac{6 x^{2}}{25} - \frac{92 x}{125} + \frac{319 \log{\left (5 x + 3 \right )}}{625} - \frac{121}{3125 x + 1875} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*x**2/25 - 92*x/125 + 319*log(5*x + 3)/625 - 121/(3125*x + 1875)

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Giac [A]  time = 1.6426, size = 65, normalized size = 1.91 \begin{align*} -\frac{2}{625} \,{\left (5 \, x + 3\right )}^{2}{\left (\frac{64}{5 \, x + 3} - 3\right )} - \frac{121}{625 \,{\left (5 \, x + 3\right )}} - \frac{319}{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*(2+3*x)/(3+5*x)^2,x, algorithm="giac")

[Out]

-2/625*(5*x + 3)^2*(64/(5*x + 3) - 3) - 121/625/(5*x + 3) - 319/625*log(1/5*abs(5*x + 3)/(5*x + 3)^2)