3.1169 \(\int \frac{(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx\)

Optimal. Leaf size=38 \[ -\frac{50 x}{27}+\frac{8}{9 (3 x+2)}-\frac{7}{162 (3 x+2)^2}+\frac{65}{27} \log (3 x+2) \]

[Out]

(-50*x)/27 - 7/(162*(2 + 3*x)^2) + 8/(9*(2 + 3*x)) + (65*Log[2 + 3*x])/27

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Rubi [A]  time = 0.0162983, 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} \[ -\frac{50 x}{27}+\frac{8}{9 (3 x+2)}-\frac{7}{162 (3 x+2)^2}+\frac{65}{27} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-50*x)/27 - 7/(162*(2 + 3*x)^2) + 8/(9*(2 + 3*x)) + (65*Log[2 + 3*x])/27

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)^3} \, dx &=\int \left (-\frac{50}{27}+\frac{7}{27 (2+3 x)^3}-\frac{8}{3 (2+3 x)^2}+\frac{65}{9 (2+3 x)}\right ) \, dx\\ &=-\frac{50 x}{27}-\frac{7}{162 (2+3 x)^2}+\frac{8}{9 (2+3 x)}+\frac{65}{27} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.022794, size = 35, normalized size = 0.92 \[ \frac{1}{162} \left (-\frac{3 \left (600 x^2+656 x+173\right )}{(3 x+2)^2}-300 x+390 \log (3 x+2)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-300*x - (3*(173 + 656*x + 600*x^2))/(2 + 3*x)^2 + 390*Log[2 + 3*x])/162

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Maple [A]  time = 0.004, size = 31, normalized size = 0.8 \begin{align*} -{\frac{50\,x}{27}}-{\frac{7}{162\, \left ( 2+3\,x \right ) ^{2}}}+{\frac{8}{18+27\,x}}+{\frac{65\,\ln \left ( 2+3\,x \right ) }{27}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-50/27*x-7/162/(2+3*x)^2+8/9/(2+3*x)+65/27*ln(2+3*x)

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Maxima [A]  time = 1.17056, size = 42, normalized size = 1.11 \begin{align*} -\frac{50}{27} \, x + \frac{432 \, x + 281}{162 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac{65}{27} \, \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)^3,x, algorithm="maxima")

[Out]

-50/27*x + 1/162*(432*x + 281)/(9*x^2 + 12*x + 4) + 65/27*log(3*x + 2)

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Fricas [A]  time = 1.92503, size = 136, normalized size = 3.58 \begin{align*} -\frac{2700 \, x^{3} + 3600 \, x^{2} - 390 \,{\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 768 \, x - 281}{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)*(3+5*x)^2/(2+3*x)^3,x, algorithm="fricas")

[Out]

-1/162*(2700*x^3 + 3600*x^2 - 390*(9*x^2 + 12*x + 4)*log(3*x + 2) + 768*x - 281)/(9*x^2 + 12*x + 4)

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Sympy [A]  time = 0.114663, size = 29, normalized size = 0.76 \begin{align*} - \frac{50 x}{27} + \frac{432 x + 281}{1458 x^{2} + 1944 x + 648} + \frac{65 \log{\left (3 x + 2 \right )}}{27} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-50*x/27 + (432*x + 281)/(1458*x**2 + 1944*x + 648) + 65*log(3*x + 2)/27

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Giac [A]  time = 2.83135, size = 36, normalized size = 0.95 \begin{align*} -\frac{50}{27} \, x + \frac{432 \, x + 281}{162 \,{\left (3 \, x + 2\right )}^{2}} + \frac{65}{27} \, \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)*(3+5*x)^2/(2+3*x)^3,x, algorithm="giac")

[Out]

-50/27*x + 1/162*(432*x + 281)/(3*x + 2)^2 + 65/27*log(abs(3*x + 2))