3.1444 \(\int \frac{3+5 x}{(1-2 x) (2+3 x)^2} \, dx\)

Optimal. Leaf size=32 \[ \frac{1}{21 (3 x+2)}-\frac{11}{49} \log (1-2 x)+\frac{11}{49} \log (3 x+2) \]

[Out]

1/(21*(2 + 3*x)) - (11*Log[1 - 2*x])/49 + (11*Log[2 + 3*x])/49

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Rubi [A]  time = 0.0152345, antiderivative size = 32, 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} \[ \frac{1}{21 (3 x+2)}-\frac{11}{49} \log (1-2 x)+\frac{11}{49} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(21*(2 + 3*x)) - (11*Log[1 - 2*x])/49 + (11*Log[2 + 3*x])/49

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{3+5 x}{(1-2 x) (2+3 x)^2} \, dx &=\int \left (-\frac{22}{49 (-1+2 x)}-\frac{1}{7 (2+3 x)^2}+\frac{33}{49 (2+3 x)}\right ) \, dx\\ &=\frac{1}{21 (2+3 x)}-\frac{11}{49} \log (1-2 x)+\frac{11}{49} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0144604, size = 30, normalized size = 0.94 \[ \frac{1}{147} \left (\frac{7}{3 x+2}-33 \log (3-6 x)+33 \log (3 x+2)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7/(2 + 3*x) - 33*Log[3 - 6*x] + 33*Log[2 + 3*x])/147

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Maple [A]  time = 0.006, size = 27, normalized size = 0.8 \begin{align*} -{\frac{11\,\ln \left ( 2\,x-1 \right ) }{49}}+{\frac{1}{42+63\,x}}+{\frac{11\,\ln \left ( 2+3\,x \right ) }{49}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/49*ln(2*x-1)+1/21/(2+3*x)+11/49*ln(2+3*x)

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Maxima [A]  time = 1.03227, size = 35, normalized size = 1.09 \begin{align*} \frac{1}{21 \,{\left (3 \, x + 2\right )}} + \frac{11}{49} \, \log \left (3 \, x + 2\right ) - \frac{11}{49} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21/(3*x + 2) + 11/49*log(3*x + 2) - 11/49*log(2*x - 1)

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Fricas [A]  time = 1.46824, size = 104, normalized size = 3.25 \begin{align*} \frac{33 \,{\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 33 \,{\left (3 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 7}{147 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/147*(33*(3*x + 2)*log(3*x + 2) - 33*(3*x + 2)*log(2*x - 1) + 7)/(3*x + 2)

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Sympy [A]  time = 0.117083, size = 26, normalized size = 0.81 \begin{align*} - \frac{11 \log{\left (x - \frac{1}{2} \right )}}{49} + \frac{11 \log{\left (x + \frac{2}{3} \right )}}{49} + \frac{1}{63 x + 42} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11*log(x - 1/2)/49 + 11*log(x + 2/3)/49 + 1/(63*x + 42)

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Giac [A]  time = 3.09648, size = 34, normalized size = 1.06 \begin{align*} \frac{1}{21 \,{\left (3 \, x + 2\right )}} - \frac{11}{49} \, \log \left ({\left | -\frac{7}{3 \, x + 2} + 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21/(3*x + 2) - 11/49*log(abs(-7/(3*x + 2) + 2))