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

Optimal. Leaf size=35 \[ -\frac{1}{16 x}-\frac{3}{16 (3 x+2)}-\frac{3 \log (x)}{16}+\frac{3}{16} \log (3 x+2) \]

[Out]

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

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Rubi [A]  time = 0.0099562, antiderivative size = 35, 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} \[ -\frac{1}{16 x}-\frac{3}{16 (3 x+2)}-\frac{3 \log (x)}{16}+\frac{3}{16} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(4 + 6*x)^2),x]

[Out]

-1/(16*x) - 3/(16*(2 + 3*x)) - (3*Log[x])/16 + (3*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^2 (4+6 x)^2} \, dx &=\int \left (\frac{1}{16 x^2}-\frac{3}{16 x}+\frac{9}{16 (2+3 x)^2}+\frac{9}{16 (2+3 x)}\right ) \, dx\\ &=-\frac{1}{16 x}-\frac{3}{16 (2+3 x)}-\frac{3 \log (x)}{16}+\frac{3}{16} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0163434, size = 31, normalized size = 0.89 \[ \frac{1}{16} \left (-\frac{1}{x}-\frac{3}{3 x+2}-3 \log (x)+3 \log (3 x+2)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(4 + 6*x)^2),x]

[Out]

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

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Maple [A]  time = 0.01, size = 28, normalized size = 0.8 \begin{align*} -{\frac{1}{16\,x}}-{\frac{3}{32+48\,x}}-{\frac{3\,\ln \left ( x \right ) }{16}}+{\frac{3\,\ln \left ( 2+3\,x \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01809, size = 42, normalized size = 1.2 \begin{align*} -\frac{3 \, x + 1}{8 \,{\left (3 \, x^{2} + 2 \, x\right )}} + \frac{3}{16} \, \log \left (3 \, x + 2\right ) - \frac{3}{16} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.73492, size = 116, normalized size = 3.31 \begin{align*} \frac{3 \,{\left (3 \, x^{2} + 2 \, x\right )} \log \left (3 \, x + 2\right ) - 3 \,{\left (3 \, x^{2} + 2 \, x\right )} \log \left (x\right ) - 6 \, x - 2}{16 \,{\left (3 \, x^{2} + 2 \, x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*(3*x^2 + 2*x)*log(3*x + 2) - 3*(3*x^2 + 2*x)*log(x) - 6*x - 2)/(3*x^2 + 2*x)

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Sympy [A]  time = 0.164169, size = 29, normalized size = 0.83 \begin{align*} - \frac{3 x + 1}{24 x^{2} + 16 x} - \frac{3 \log{\left (x \right )}}{16} + \frac{3 \log{\left (x + \frac{2}{3} \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2096, size = 54, normalized size = 1.54 \begin{align*} -\frac{3}{16 \,{\left (3 \, x + 2\right )}} + \frac{3}{32 \,{\left (\frac{2}{3 \, x + 2} - 1\right )}} - \frac{3}{16} \, \log \left ({\left | -\frac{2}{3 \, x + 2} + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16/(3*x + 2) + 3/32/(2/(3*x + 2) - 1) - 3/16*log(abs(-2/(3*x + 2) + 1))