\(\int \frac {1}{x (a+b \log (x))} \, dx\) [620]

Optimal result

Integrand size = 12, antiderivative size = 11 \[ \int \frac {1}{x (a+b \log (x))} \, dx=\frac {\log (a+b \log (x))}{b} \]

[Out]

ln(a+b*ln(x))/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2339, 29} \[ \int \frac {1}{x (a+b \log (x))} \, dx=\frac {\log (a+b \log (x))}{b} \]

[In]

Int[1/(x*(a + b*Log[x])),x]

[Out]

Log[a + b*Log[x]]/b

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (a+b \log (x))} \, dx=\frac {\log (a+b \log (x))}{b} \]

[In]

Integrate[1/(x*(a + b*Log[x])),x]

[Out]

Log[a + b*Log[x]]/b

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\ln \left (a +b \ln \left (x \right )\right )}{b}\)	\(12\)
default	\(\frac {\ln \left (a +b \ln \left (x \right )\right )}{b}\)	\(12\)
norman	\(\frac {\ln \left (a +b \ln \left (x \right )\right )}{b}\)	\(12\)
parallelrisch	\(\frac {\ln \left (a +b \ln \left (x \right )\right )}{b}\)	\(12\)
risch	\(\frac {\ln \left (\ln \left (x \right )+\frac {a}{b}\right )}{b}\)	\(14\)

[In]

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

[Out]

ln(a+b*ln(x))/b

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (a+b \log (x))} \, dx=\frac {\log \left (b \log \left (x\right ) + a\right )}{b} \]

[In]

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

[Out]

log(b*log(x) + a)/b

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x (a+b \log (x))} \, dx=\frac {\log {\left (\frac {a}{b} + \log {\left (x \right )} \right )}}{b} \]

[In]

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

[Out]

log(a/b + log(x))/b

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (a+b \log (x))} \, dx=\frac {\log \left (b \log \left (x\right ) + a\right )}{b} \]

[In]

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

[Out]

log(b*log(x) + a)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (11) = 22\).

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.73 \[ \int \frac {1}{x (a+b \log (x))} \, dx=\frac {\log \left (\frac {1}{4} \, \pi ^{2} b^{2} {\left (\mathrm {sgn}\left (x\right ) - 1\right )}^{2} + {\left (b \log \left ({\left | x \right |}\right ) + a\right )}^{2}\right )}{2 \, b} \]

[In]

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

[Out]

1/2*log(1/4*pi^2*b^2*(sgn(x) - 1)^2 + (b*log(abs(x)) + a)^2)/b

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (a+b \log (x))} \, dx=\frac {\ln \left (a+b\,\ln \left (x\right )\right )}{b} \]

[In]

int(1/(x*(a + b*log(x))),x)

[Out]

log(a + b*log(x))/b