3.261 \(\int \frac {1}{x (4+6 x)^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{8 (3 x+2)}+\frac {\log (x)}{16}-\frac {1}{16} \log (3 x+2) \]

[Out]

1/8/(2+3*x)+1/16*ln(x)-1/16*ln(2+3*x)

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, 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} \[ \frac {1}{8 (3 x+2)}+\frac {\log (x)}{16}-\frac {1}{16} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(4 + 6*x)^2),x]

[Out]

1/(8*(2 + 3*x)) + Log[x]/16 - Log[2 + 3*x]/16

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)^2} \, dx &=\int \left (\frac {1}{16 x}-\frac {3}{8 (2+3 x)^2}-\frac {3}{16 (2+3 x)}\right ) \, dx\\ &=\frac {1}{8 (2+3 x)}+\frac {\log (x)}{16}-\frac {1}{16} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.93 \[ \frac {1}{16} \left (\frac {2}{3 x+2}+\log (-6 x)-\log (6 x+4)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(4 + 6*x)^2),x]

[Out]

(2/(2 + 3*x) + Log[-6*x] - Log[4 + 6*x])/16

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fricas [A]  time = 0.45, size = 32, normalized size = 1.14 \[ -\frac {{\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - {\left (3 \, x + 2\right )} \log \relax (x) - 2}{16 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)^2,x, algorithm="fricas")

[Out]

-1/16*((3*x + 2)*log(3*x + 2) - (3*x + 2)*log(x) - 2)/(3*x + 2)

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giac [A]  time = 1.11, size = 25, normalized size = 0.89 \[ \frac {1}{8 \, {\left (3 \, x + 2\right )}} + \frac {1}{16} \, \log \left ({\left | -\frac {2}{3 \, x + 2} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)^2,x, algorithm="giac")

[Out]

1/8/(3*x + 2) + 1/16*log(abs(-2/(3*x + 2) + 1))

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maple [A]  time = 0.01, size = 23, normalized size = 0.82 \[ \frac {\ln \relax (x )}{16}-\frac {\ln \left (3 x +2\right )}{16}+\frac {1}{24 x +16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(4+6*x)^2,x)

[Out]

1/8/(3*x+2)+1/16*ln(x)-1/16*ln(3*x+2)

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maxima [A]  time = 1.37, size = 22, normalized size = 0.79 \[ \frac {1}{8 \, {\left (3 \, x + 2\right )}} - \frac {1}{16} \, \log \left (3 \, x + 2\right ) + \frac {1}{16} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)^2,x, algorithm="maxima")

[Out]

1/8/(3*x + 2) - 1/16*log(3*x + 2) + 1/16*log(x)

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mupad [B]  time = 0.06, size = 20, normalized size = 0.71 \[ \frac {1}{8\,\left (3\,x+2\right )}-\frac {\ln \left (\frac {6\,x+4}{x}\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(6*x + 4)^2),x)

[Out]

1/(8*(3*x + 2)) - log((6*x + 4)/x)/16

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sympy [A]  time = 0.14, size = 19, normalized size = 0.68 \[ \frac {\log {\relax (x )}}{16} - \frac {\log {\left (x + \frac {2}{3} \right )}}{16} + \frac {1}{24 x + 16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(4+6*x)**2,x)

[Out]

log(x)/16 - log(x + 2/3)/16 + 1/(24*x + 16)

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