3.266 \(\int \frac{1}{x (4+6 x)^3} \, dx\)

Optimal. Leaf size=39 \[ \frac{1}{32 (3 x+2)}+\frac{1}{32 (3 x+2)^2}+\frac{\log (x)}{64}-\frac{1}{64} \log (3 x+2) \]

[Out]

1/(32*(2 + 3*x)^2) + 1/(32*(2 + 3*x)) + Log[x]/64 - Log[2 + 3*x]/64

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Rubi [A]  time = 0.0119353, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {44} \[ \frac{1}{32 (3 x+2)}+\frac{1}{32 (3 x+2)^2}+\frac{\log (x)}{64}-\frac{1}{64} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(4 + 6*x)^3),x]

[Out]

1/(32*(2 + 3*x)^2) + 1/(32*(2 + 3*x)) + Log[x]/64 - Log[2 + 3*x]/64

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{x (4+6 x)^3} \, dx &=\int \left (\frac{1}{64 x}-\frac{3}{16 (2+3 x)^3}-\frac{3}{32 (2+3 x)^2}-\frac{3}{64 (2+3 x)}\right ) \, dx\\ &=\frac{1}{32 (2+3 x)^2}+\frac{1}{32 (2+3 x)}+\frac{\log (x)}{64}-\frac{1}{64} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0346237, size = 29, normalized size = 0.74 \[ \frac{1}{64} \left (\frac{6 (x+1)}{(3 x+2)^2}+\log (-6 x)-\log (6 x+4)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(4 + 6*x)^3),x]

[Out]

((6*(1 + x))/(2 + 3*x)^2 + Log[-6*x] - Log[4 + 6*x])/64

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Maple [A]  time = 0.007, size = 32, normalized size = 0.8 \begin{align*}{\frac{1}{32\, \left ( 2+3\,x \right ) ^{2}}}+{\frac{1}{64+96\,x}}+{\frac{\ln \left ( x \right ) }{64}}-{\frac{\ln \left ( 2+3\,x \right ) }{64}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(4+6*x)^3,x)

[Out]

1/32/(2+3*x)^2+1/32/(2+3*x)+1/64*ln(x)-1/64*ln(2+3*x)

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Maxima [A]  time = 1.05525, size = 41, normalized size = 1.05 \begin{align*} \frac{3 \,{\left (x + 1\right )}}{32 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac{1}{64} \, \log \left (3 \, x + 2\right ) + \frac{1}{64} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)^3,x, algorithm="maxima")

[Out]

3/32*(x + 1)/(9*x^2 + 12*x + 4) - 1/64*log(3*x + 2) + 1/64*log(x)

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Fricas [A]  time = 1.66655, size = 132, normalized size = 3.38 \begin{align*} -\frac{{\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) -{\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (x\right ) - 6 \, x - 6}{64 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)^3,x, algorithm="fricas")

[Out]

-1/64*((9*x^2 + 12*x + 4)*log(3*x + 2) - (9*x^2 + 12*x + 4)*log(x) - 6*x - 6)/(9*x^2 + 12*x + 4)

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Sympy [A]  time = 0.212665, size = 27, normalized size = 0.69 \begin{align*} \frac{3 x + 3}{288 x^{2} + 384 x + 128} + \frac{\log{\left (x \right )}}{64} - \frac{\log{\left (x + \frac{2}{3} \right )}}{64} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)**3,x)

[Out]

(3*x + 3)/(288*x**2 + 384*x + 128) + log(x)/64 - log(x + 2/3)/64

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Giac [A]  time = 1.19562, size = 36, normalized size = 0.92 \begin{align*} \frac{3 \,{\left (x + 1\right )}}{32 \,{\left (3 \, x + 2\right )}^{2}} - \frac{1}{64} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac{1}{64} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)^3,x, algorithm="giac")

[Out]

3/32*(x + 1)/(3*x + 2)^2 - 1/64*log(abs(3*x + 2)) + 1/64*log(abs(x))