\(\int \frac {1}{x (1+\log (\frac {x}{a}))} \, dx\) [23]

Optimal result

Integrand size = 14, antiderivative size = 9 \[ \int \frac {1}{x \left (1+\log \left (\frac {x}{a}\right )\right )} \, dx=\log \left (1+\log \left (\frac {x}{a}\right )\right ) \]

[Out]

ln(1+ln(x/a))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2339, 29} \[ \int \frac {1}{x \left (1+\log \left (\frac {x}{a}\right )\right )} \, dx=\log \left (\log \left (\frac {x}{a}\right )+1\right ) \]

[In]

Int[1/(x*(1 + Log[x/a])),x]

[Out]

Log[1 + Log[x/a]]

Rule 29

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

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{x} \, dx,x,1+\log \left (\frac {x}{a}\right )\right ) \\ & = \log \left (1+\log \left (\frac {x}{a}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log \left (\frac {x}{a}\right )\right )} \, dx=\log \left (1+\log \left (\frac {x}{a}\right )\right ) \]

[In]

Integrate[1/(x*(1 + Log[x/a])),x]

[Out]

Log[1 + Log[x/a]]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\ln \left (1+\ln \left (\frac {x}{a}\right )\right )\)	\(10\)
default	\(\ln \left (1+\ln \left (\frac {x}{a}\right )\right )\)	\(10\)
norman	\(\ln \left (1+\ln \left (\frac {x}{a}\right )\right )\)	\(10\)
risch	\(\ln \left (1+\ln \left (\frac {x}{a}\right )\right )\)	\(10\)
parallelrisch	\(\ln \left (1+\ln \left (\frac {x}{a}\right )\right )\)	\(10\)

[In]

int(1/x/(1+ln(x/a)),x,method=_RETURNVERBOSE)

[Out]

ln(1+ln(x/a))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log \left (\frac {x}{a}\right )\right )} \, dx=\log \left (\log \left (\frac {x}{a}\right ) + 1\right ) \]

[In]

integrate(1/x/(1+log(x/a)),x, algorithm="fricas")

[Out]

log(log(x/a) + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \left (1+\log \left (\frac {x}{a}\right )\right )} \, dx=\log {\left (\log {\left (\frac {x}{a} \right )} + 1 \right )} \]

[In]

integrate(1/x/(1+ln(x/a)),x)

[Out]

log(log(x/a) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log \left (\frac {x}{a}\right )\right )} \, dx=\log \left (\log \left (\frac {x}{a}\right ) + 1\right ) \]

[In]

integrate(1/x/(1+log(x/a)),x, algorithm="maxima")

[Out]

log(log(x/a) + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log \left (\frac {x}{a}\right )\right )} \, dx=\log \left (\log \left (\frac {x}{a}\right ) + 1\right ) \]

[In]

integrate(1/x/(1+log(x/a)),x, algorithm="giac")

[Out]

log(log(x/a) + 1)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+\log \left (\frac {x}{a}\right )\right )} \, dx=\ln \left (\ln \left (\frac {x}{a}\right )+1\right ) \]

[In]

int(1/(x*(log(x/a) + 1)),x)

[Out]

log(log(x/a) + 1)