3.1547 \(\int \frac{3+5 x}{(1-2 x)^2 (2+3 x)} \, dx\)

Optimal. Leaf size=32 \[ \frac{11}{14 (1-2 x)}+\frac{1}{49} \log (1-2 x)-\frac{1}{49} \log (3 x+2) \]

[Out]

11/(14*(1 - 2*x)) + Log[1 - 2*x]/49 - Log[2 + 3*x]/49

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Rubi [A]  time = 0.014893, 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} \[ \frac{11}{14 (1-2 x)}+\frac{1}{49} \log (1-2 x)-\frac{1}{49} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

11/(14*(1 - 2*x)) + Log[1 - 2*x]/49 - 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 (2+3 x)} \, dx &=\int \left (\frac{11}{7 (-1+2 x)^2}+\frac{2}{49 (-1+2 x)}-\frac{3}{49 (2+3 x)}\right ) \, dx\\ &=\frac{11}{14 (1-2 x)}+\frac{1}{49} \log (1-2 x)-\frac{1}{49} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0126618, size = 37, normalized size = 1.16 \[ \frac{(4 x-2) \log (1-2 x)+(2-4 x) \log (6 x+4)-77}{98 (2 x-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-77 + (-2 + 4*x)*Log[1 - 2*x] + (2 - 4*x)*Log[4 + 6*x])/(98*(-1 + 2*x))

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Maple [A]  time = 0.006, size = 27, normalized size = 0.8 \begin{align*} -{\frac{11}{28\,x-14}}+{\frac{\ln \left ( 2\,x-1 \right ) }{49}}-{\frac{\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/(2+3*x),x)

[Out]

-11/14/(2*x-1)+1/49*ln(2*x-1)-1/49*ln(2+3*x)

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Maxima [A]  time = 1.08185, size = 35, normalized size = 1.09 \begin{align*} -\frac{11}{14 \,{\left (2 \, x - 1\right )}} - \frac{1}{49} \, \log \left (3 \, x + 2\right ) + \frac{1}{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/(2+3*x),x, algorithm="maxima")

[Out]

-11/14/(2*x - 1) - 1/49*log(3*x + 2) + 1/49*log(2*x - 1)

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Fricas [A]  time = 1.37141, size = 103, normalized size = 3.22 \begin{align*} -\frac{2 \,{\left (2 \, x - 1\right )} \log \left (3 \, x + 2\right ) - 2 \,{\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) + 77}{98 \,{\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/(2+3*x),x, algorithm="fricas")

[Out]

-1/98*(2*(2*x - 1)*log(3*x + 2) - 2*(2*x - 1)*log(2*x - 1) + 77)/(2*x - 1)

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Sympy [A]  time = 0.109262, size = 22, normalized size = 0.69 \begin{align*} \frac{\log{\left (x - \frac{1}{2} \right )}}{49} - \frac{\log{\left (x + \frac{2}{3} \right )}}{49} - \frac{11}{28 x - 14} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 1/2)/49 - log(x + 2/3)/49 - 11/(28*x - 14)

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Giac [A]  time = 2.35081, size = 34, normalized size = 1.06 \begin{align*} -\frac{11}{14 \,{\left (2 \, x - 1\right )}} - \frac{1}{49} \, \log \left ({\left | -\frac{7}{2 \, x - 1} - 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/14/(2*x - 1) - 1/49*log(abs(-7/(2*x - 1) - 3))