\(\int \frac {1}{x (2+3 x)} \, dx\) [255]

Optimal result

Integrand size = 11, antiderivative size = 17 \[ \int \frac {1}{x (2+3 x)} \, dx=\frac {\log (x)}{2}-\frac {1}{2} \log (2+3 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {36, 29, 31} \[ \int \frac {1}{x (2+3 x)} \, dx=\frac {\log (x)}{2}-\frac {1}{2} \log (3 x+2) \]

[In]

Int[1/(x*(2 + 3*x)),x]

[Out]

Log[x]/2 - Log[2 + 3*x]/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {1}{x} \, dx-\frac {3}{2} \int \frac {1}{2+3 x} \, dx \\ & = \frac {\log (x)}{2}-\frac {1}{2} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (2+3 x)} \, dx=\frac {\log (x)}{2}-\frac {1}{2} \log (2+3 x) \]

[In]

Integrate[1/(x*(2 + 3*x)),x]

[Out]

Log[x]/2 - Log[2 + 3*x]/2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(\frac {\ln \left (x \right )}{2}-\frac {\ln \left (\frac {2}{3}+x \right )}{2}\)	\(12\)
default	\(\frac {\ln \left (x \right )}{2}-\frac {\ln \left (2+3 x \right )}{2}\)	\(14\)
norman	\(\frac {\ln \left (x \right )}{2}-\frac {\ln \left (2+3 x \right )}{2}\)	\(14\)
risch	\(\frac {\ln \left (x \right )}{2}-\frac {\ln \left (2+3 x \right )}{2}\)	\(14\)
meijerg	\(\frac {\ln \left (x \right )}{2}+\frac {\ln \left (3\right )}{2}-\frac {\ln \left (2\right )}{2}-\frac {\ln \left (1+\frac {3 x}{2}\right )}{2}\)	\(22\)

[In]

int(1/x/(2+3*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x (2+3 x)} \, dx=-\frac {1}{2} \, \log \left (3 \, x + 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]

[In]

integrate(1/x/(2+3*x),x, algorithm="fricas")

[Out]

-1/2*log(3*x + 2) + 1/2*log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x (2+3 x)} \, dx=\frac {\log {\left (x \right )}}{2} - \frac {\log {\left (x + \frac {2}{3} \right )}}{2} \]

[In]

integrate(1/x/(2+3*x),x)

[Out]

log(x)/2 - log(x + 2/3)/2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x (2+3 x)} \, dx=-\frac {1}{2} \, \log \left (3 \, x + 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]

[In]

integrate(1/x/(2+3*x),x, algorithm="maxima")

[Out]

-1/2*log(3*x + 2) + 1/2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x (2+3 x)} \, dx=-\frac {1}{2} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/x/(2+3*x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x (2+3 x)} \, dx=-\frac {\ln \left (\frac {2}{x}+3\right )}{2} \]

[In]

int(1/(x*(3*x + 2)),x)

[Out]

-log(2/x + 3)/2