3.1247 \(\int \frac{(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx\)

Optimal. Leaf size=38 \[ \frac{20 x}{27}-\frac{91}{27 (3 x+2)}+\frac{49}{162 (3 x+2)^2}-\frac{16}{9} \log (3 x+2) \]

[Out]

(20*x)/27 + 49/(162*(2 + 3*x)^2) - 91/(27*(2 + 3*x)) - (16*Log[2 + 3*x])/9

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Rubi [A]  time = 0.0151843, antiderivative size = 38, 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} \[ \frac{20 x}{27}-\frac{91}{27 (3 x+2)}+\frac{49}{162 (3 x+2)^2}-\frac{16}{9} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(20*x)/27 + 49/(162*(2 + 3*x)^2) - 91/(27*(2 + 3*x)) - (16*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)^2 (3+5 x)}{(2+3 x)^3} \, dx &=\int \left (\frac{20}{27}-\frac{49}{27 (2+3 x)^3}+\frac{91}{9 (2+3 x)^2}-\frac{16}{3 (2+3 x)}\right ) \, dx\\ &=\frac{20 x}{27}+\frac{49}{162 (2+3 x)^2}-\frac{91}{27 (2+3 x)}-\frac{16}{9} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0139426, size = 41, normalized size = 1.08 \[ \frac{1080 x^3+900 x^2-1878 x-288 (3 x+2)^2 \log (6 x+4)-1283}{162 (3 x+2)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1283 - 1878*x + 900*x^2 + 1080*x^3 - 288*(2 + 3*x)^2*Log[4 + 6*x])/(162*(2 + 3*x)^2)

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Maple [A]  time = 0.006, size = 31, normalized size = 0.8 \begin{align*}{\frac{20\,x}{27}}+{\frac{49}{162\, \left ( 2+3\,x \right ) ^{2}}}-{\frac{91}{54+81\,x}}-{\frac{16\,\ln \left ( 2+3\,x \right ) }{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

20/27*x+49/162/(2+3*x)^2-91/27/(2+3*x)-16/9*ln(2+3*x)

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Maxima [A]  time = 1.076, size = 42, normalized size = 1.11 \begin{align*} \frac{20}{27} \, x - \frac{7 \,{\left (234 \, x + 149\right )}}{162 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac{16}{9} \, \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)^3,x, algorithm="maxima")

[Out]

20/27*x - 7/162*(234*x + 149)/(9*x^2 + 12*x + 4) - 16/9*log(3*x + 2)

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Fricas [A]  time = 1.45406, size = 138, normalized size = 3.63 \begin{align*} \frac{1080 \, x^{3} + 1440 \, x^{2} - 288 \,{\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 1158 \, x - 1043}{162 \,{\left (9 \, x^{2} + 12 \, x + 4\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)^3,x, algorithm="fricas")

[Out]

1/162*(1080*x^3 + 1440*x^2 - 288*(9*x^2 + 12*x + 4)*log(3*x + 2) - 1158*x - 1043)/(9*x^2 + 12*x + 4)

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Sympy [A]  time = 0.112872, size = 29, normalized size = 0.76 \begin{align*} \frac{20 x}{27} - \frac{1638 x + 1043}{1458 x^{2} + 1944 x + 648} - \frac{16 \log{\left (3 x + 2 \right )}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

20*x/27 - (1638*x + 1043)/(1458*x**2 + 1944*x + 648) - 16*log(3*x + 2)/9

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Giac [A]  time = 1.7328, size = 36, normalized size = 0.95 \begin{align*} \frac{20}{27} \, x - \frac{7 \,{\left (234 \, x + 149\right )}}{162 \,{\left (3 \, x + 2\right )}^{2}} - \frac{16}{9} \, \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)^3,x, algorithm="giac")

[Out]

20/27*x - 7/162*(234*x + 149)/(3*x + 2)^2 - 16/9*log(abs(3*x + 2))