\(\int \frac {1}{x (4+6 x)^2} \, dx\) [261]

Optimal result

Integrand size = 11, antiderivative size = 28 \[ \int \frac {1}{x (4+6 x)^2} \, dx=\frac {1}{8 (2+3 x)}+\frac {\log (x)}{16}-\frac {1}{16} \log (2+3 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, 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} \[ \int \frac {1}{x (4+6 x)^2} \, dx=\frac {1}{8 (3 x+2)}+\frac {\log (x)}{16}-\frac {1}{16} \log (3 x+2) \]

[In]

Int[1/(x*(4 + 6*x)^2),x]

[Out]

1/(8*(2 + 3*x)) + Log[x]/16 - 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}-\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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x (4+6 x)^2} \, dx=\frac {1}{16} \left (\frac {2}{2+3 x}+\log (-6 x)-\log (4+6 x)\right ) \]

[In]

Integrate[1/(x*(4 + 6*x)^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {1}{16+24 x}+\frac {\ln \left (x \right )}{16}-\frac {\ln \left (2+3 x \right )}{16}\)	\(21\)
default	\(\frac {1}{16+24 x}+\frac {\ln \left (x \right )}{16}-\frac {\ln \left (2+3 x \right )}{16}\)	\(23\)
norman	\(-\frac {3 x}{16 \left (2+3 x \right )}+\frac {\ln \left (x \right )}{16}-\frac {\ln \left (2+3 x \right )}{16}\)	\(24\)
meijerg	\(\frac {1}{16}+\frac {\ln \left (x \right )}{16}+\frac {\ln \left (3\right )}{16}-\frac {\ln \left (2\right )}{16}-\frac {3 x}{16 \left (2+3 x \right )}-\frac {\ln \left (1+\frac {3 x}{2}\right )}{16}\)	\(33\)
parallelrisch	\(\frac {3 \ln \left (x \right ) x -3 \ln \left (\frac {2}{3}+x \right ) x +2 \ln \left (x \right )-2 \ln \left (\frac {2}{3}+x \right )-3 x}{32+48 x}\)	\(36\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x (4+6 x)^2} \, dx=-\frac {{\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - {\left (3 \, x + 2\right )} \log \left (x\right ) - 2}{16 \, {\left (3 \, x + 2\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x (4+6 x)^2} \, dx=\frac {\log {\left (x \right )}}{16} - \frac {\log {\left (x + \frac {2}{3} \right )}}{16} + \frac {1}{24 x + 16} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x (4+6 x)^2} \, dx=\frac {1}{8 \, {\left (3 \, x + 2\right )}} - \frac {1}{16} \, \log \left (3 \, x + 2\right ) + \frac {1}{16} \, \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x (4+6 x)^2} \, dx=\frac {1}{8 \, {\left (3 \, x + 2\right )}} + \frac {1}{16} \, \log \left ({\left | -\frac {2}{3 \, x + 2} + 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x (4+6 x)^2} \, dx=\frac {1}{8\,\left (3\,x+2\right )}-\frac {\ln \left (\frac {6\,x+4}{x}\right )}{16} \]

[In]

int(1/(x*(6*x + 4)^2),x)

[Out]

1/(8*(3*x + 2)) - log((6*x + 4)/x)/16