\(\int \frac {1}{x^2 (4+6 x)^2} \, dx\) [262]

Optimal result

Integrand size = 11, antiderivative size = 35 \[ \int \frac {1}{x^2 (4+6 x)^2} \, dx=-\frac {1}{16 x}-\frac {3}{16 (2+3 x)}-\frac {3 \log (x)}{16}+\frac {3}{16} \log (2+3 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \[ \int \frac {1}{x^2 (4+6 x)^2} \, dx=-\frac {1}{16 x}-\frac {3}{16 (3 x+2)}-\frac {3 \log (x)}{16}+\frac {3}{16} \log (3 x+2) \]

[In]

Int[1/(x^2*(4 + 6*x)^2),x]

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 (4+6 x)^2} \, dx=\frac {1}{16} \left (-\frac {1}{x}-\frac {3}{2+3 x}-3 \log (x)+3 \log (2+3 x)\right ) \]

[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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {1}{16 x}-\frac {3}{16 \left (2+3 x \right )}-\frac {3 \ln \left (x \right )}{16}+\frac {3 \ln \left (2+3 x \right )}{16}\)	\(28\)
risch	\(\frac {-\frac {3 x}{8}-\frac {1}{8}}{x \left (2+3 x \right )}-\frac {3 \ln \left (x \right )}{16}+\frac {3 \ln \left (2+3 x \right )}{16}\)	\(31\)
norman	\(\frac {-\frac {1}{8}+\frac {9 x^{2}}{16}}{x \left (2+3 x \right )}-\frac {3 \ln \left (x \right )}{16}+\frac {3 \ln \left (2+3 x \right )}{16}\)	\(32\)
meijerg	\(-\frac {1}{16 x}-\frac {3}{32}-\frac {3 \ln \left (x \right )}{16}-\frac {3 \ln \left (3\right )}{16}+\frac {3 \ln \left (2\right )}{16}+\frac {27 x}{64 \left (\frac {9 x}{2}+3\right )}+\frac {3 \ln \left (1+\frac {3 x}{2}\right )}{16}\)	\(38\)
parallelrisch	\(-\frac {9 \ln \left (x \right ) x^{2}-9 \ln \left (\frac {2}{3}+x \right ) x^{2}+2+6 \ln \left (x \right ) x -6 \ln \left (\frac {2}{3}+x \right ) x -9 x^{2}}{16 x \left (2+3 x \right )}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {1}{x^2 (4+6 x)^2} \, dx=\frac {3 \, {\left (3 \, x^{2} + 2 \, x\right )} \log \left (3 \, x + 2\right ) - 3 \, {\left (3 \, x^{2} + 2 \, x\right )} \log \left (x\right ) - 6 \, x - 2}{16 \, {\left (3 \, x^{2} + 2 \, x\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 (4+6 x)^2} \, dx=\frac {- 3 x - 1}{24 x^{2} + 16 x} - \frac {3 \log {\left (x \right )}}{16} + \frac {3 \log {\left (x + \frac {2}{3} \right )}}{16} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 (4+6 x)^2} \, dx=-\frac {3 \, x + 1}{8 \, {\left (3 \, x^{2} + 2 \, x\right )}} + \frac {3}{16} \, \log \left (3 \, x + 2\right ) - \frac {3}{16} \, \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^2 (4+6 x)^2} \, dx=-\frac {3}{16 \, {\left (3 \, x + 2\right )}} + \frac {3}{32 \, {\left (\frac {2}{3 \, x + 2} - 1\right )}} - \frac {3}{16} \, \log \left ({\left | -\frac {2}{3 \, x + 2} + 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^2 (4+6 x)^2} \, dx=\frac {3\,\ln \left (\frac {6\,x+4}{x}\right )}{16}-\frac {3}{4\,\left (6\,x+4\right )}-\frac {1}{4\,x\,\left (6\,x+4\right )} \]

[In]

int(1/(x^2*(6*x + 4)^2),x)

[Out]

(3*log((6*x + 4)/x))/16 - 3/(4*(6*x + 4)) - 1/(4*x*(6*x + 4))