\(\int \frac {1-x \log (x)}{x \log (x)} \, dx\) [1274]

Optimal result

Integrand size = 15, antiderivative size = 7 \[ \int \frac {1-x \log (x)}{x \log (x)} \, dx=-x+\log (\log (x)) \]

[Out]

ln(ln(x))-x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6820, 2339, 29} \[ \int \frac {1-x \log (x)}{x \log (x)} \, dx=\log (\log (x))-x \]

[In]

Int[(1 - x*Log[x])/(x*Log[x]),x]

[Out]

-x + Log[Log[x]]

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]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {1-x \log (x)}{x \log (x)} \, dx=-x+\log (\log (x)) \]

[In]

Integrate[(1 - x*Log[x])/(x*Log[x]),x]

[Out]

-x + Log[Log[x]]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.14

method	result	size
default	\(\ln \left (\ln \left (x \right )\right )-x\)	\(8\)
norman	\(\ln \left (\ln \left (x \right )\right )-x\)	\(8\)
risch	\(\ln \left (\ln \left (x \right )\right )-x\)	\(8\)
parallelrisch	\(\ln \left (\ln \left (x \right )\right )-x\)	\(8\)
parts	\(\ln \left (\ln \left (x \right )\right )-x\)	\(8\)

[In]

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

[Out]

ln(ln(x))-x

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {1-x \log (x)}{x \log (x)} \, dx=-x + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

-x + log(log(x))

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.71 \[ \int \frac {1-x \log (x)}{x \log (x)} \, dx=- x + \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate((-x*ln(x)+1)/x/ln(x),x)

[Out]

-x + log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {1-x \log (x)}{x \log (x)} \, dx=-x + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

-x + log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {1-x \log (x)}{x \log (x)} \, dx=-x + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

-x + log(log(x))

Mupad [B] (verification not implemented)

Time = 8.42 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {1-x \log (x)}{x \log (x)} \, dx=\ln \left (\ln \left (x\right )\right )-x \]

[In]

int(-(x*log(x) - 1)/(x*log(x)),x)

[Out]

log(log(x)) - x