\(\int \frac {1}{x (4+6 x)^3} \, dx\) [266]

Optimal result

Integrand size = 11, antiderivative size = 39 \[ \int \frac {1}{x (4+6 x)^3} \, dx=\frac {1}{32 (2+3 x)^2}+\frac {1}{32 (2+3 x)}+\frac {\log (x)}{64}-\frac {1}{64} \log (2+3 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, 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} \[ \int \frac {1}{x (4+6 x)^3} \, dx=\frac {1}{32 (3 x+2)}+\frac {1}{32 (3 x+2)^2}+\frac {\log (x)}{64}-\frac {1}{64} \log (3 x+2) \]

[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 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}{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] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x (4+6 x)^3} \, dx=\frac {1}{64} \left (\frac {6 (1+x)}{(2+3 x)^2}+\log (-6 x)-\log (4+6 x)\right ) \]

[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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {\frac {3 x}{32}+\frac {3}{32}}{\left (2+3 x \right )^{2}}+\frac {\ln \left (x \right )}{64}-\frac {\ln \left (2+3 x \right )}{64}\)	\(28\)
norman	\(\frac {-\frac {3}{16} x -\frac {27}{128} x^{2}}{\left (2+3 x \right )^{2}}+\frac {\ln \left (x \right )}{64}-\frac {\ln \left (2+3 x \right )}{64}\)	\(31\)
default	\(\frac {1}{32 \left (2+3 x \right )^{2}}+\frac {1}{64+96 x}+\frac {\ln \left (x \right )}{64}-\frac {\ln \left (2+3 x \right )}{64}\)	\(32\)
meijerg	\(\frac {3}{128}+\frac {\ln \left (x \right )}{64}+\frac {\ln \left (3\right )}{64}-\frac {\ln \left (2\right )}{64}-\frac {3 x \left (4+\frac {9 x}{2}\right )}{256 \left (1+\frac {3 x}{2}\right )^{2}}-\frac {\ln \left (1+\frac {3 x}{2}\right )}{64}\)	\(38\)
parallelrisch	\(\frac {18 \ln \left (x \right ) x^{2}-18 \ln \left (\frac {2}{3}+x \right ) x^{2}+24 \ln \left (x \right ) x -24 \ln \left (\frac {2}{3}+x \right ) x -27 x^{2}+8 \ln \left (x \right )-8 \ln \left (\frac {2}{3}+x \right )-24 x}{128 \left (2+3 x \right )^{2}}\)	\(57\)

[In]

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

[Out]

9*(1/96*x+1/96)/(2+3*x)^2+1/64*ln(x)-1/64*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x (4+6 x)^3} \, dx=-\frac {{\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) - {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (x\right ) - 6 \, x - 6}{64 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x (4+6 x)^3} \, dx=\frac {3 x + 3}{288 x^{2} + 384 x + 128} + \frac {\log {\left (x \right )}}{64} - \frac {\log {\left (x + \frac {2}{3} \right )}}{64} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x (4+6 x)^3} \, dx=\frac {3 \, {\left (x + 1\right )}}{32 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {1}{64} \, \log \left (3 \, x + 2\right ) + \frac {1}{64} \, \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x (4+6 x)^3} \, dx=\frac {3 \, {\left (x + 1\right )}}{32 \, {\left (3 \, x + 2\right )}^{2}} - \frac {1}{64} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {1}{64} \, \log \left ({\left | x \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x (4+6 x)^3} \, dx=\frac {1}{32\,\left (3\,x+2\right )}-\frac {\ln \left (\frac {6\,x+4}{x}\right )}{64}+\frac {1}{8\,{\left (6\,x+4\right )}^2} \]

[In]

int(1/(x*(6*x + 4)^3),x)

[Out]

1/(32*(3*x + 2)) - log((6*x + 4)/x)/64 + 1/(8*(6*x + 4)^2)