\(\int \frac {-1+4 x \log (x)}{x \log (x)} \, dx\) [8022]

Optimal result

Integrand size = 15, antiderivative size = 10 \[ \int \frac {-1+4 x \log (x)}{x \log (x)} \, dx=23+4 x-\log (\log (x)) \]

[Out]

23+4*x-ln(ln(x))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90, 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} \[ \int \frac {-1+4 x \log (x)}{x \log (x)} \, dx=4 x-\log (\log (x)) \]

[In]

Int[(-1 + 4*x*Log[x])/(x*Log[x]),x]

[Out]

4*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 (4-\frac {1}{x \log (x)}\right ) \, dx \\ & = 4 x-\int \frac {1}{x \log (x)} \, dx \\ & = 4 x-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = 4 x-\log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {-1+4 x \log (x)}{x \log (x)} \, dx=4 x-\log (\log (x)) \]

[In]

Integrate[(-1 + 4*x*Log[x])/(x*Log[x]),x]

[Out]

4*x - Log[Log[x]]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00

method	result	size
default	\(4 x -\ln \left (\ln \left (x \right )\right )\)	\(10\)
norman	\(4 x -\ln \left (\ln \left (x \right )\right )\)	\(10\)
risch	\(4 x -\ln \left (\ln \left (x \right )\right )\)	\(10\)
parallelrisch	\(4 x -\ln \left (\ln \left (x \right )\right )\)	\(10\)
parts	\(4 x -\ln \left (\ln \left (x \right )\right )\)	\(10\)

[In]

int((4*x*ln(x)-1)/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

4*x-ln(ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {-1+4 x \log (x)}{x \log (x)} \, dx=4 \, x - \log \left (\log \left (x\right )\right ) \]

[In]

integrate((4*x*log(x)-1)/x/log(x),x, algorithm="fricas")

[Out]

4*x - log(log(x))

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int \frac {-1+4 x \log (x)}{x \log (x)} \, dx=4 x - \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate((4*x*ln(x)-1)/x/ln(x),x)

[Out]

4*x - log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {-1+4 x \log (x)}{x \log (x)} \, dx=4 \, x - \log \left (\log \left (x\right )\right ) \]

[In]

integrate((4*x*log(x)-1)/x/log(x),x, algorithm="maxima")

[Out]

4*x - log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {-1+4 x \log (x)}{x \log (x)} \, dx=4 \, x - \log \left (\log \left (x\right )\right ) \]

[In]

integrate((4*x*log(x)-1)/x/log(x),x, algorithm="giac")

[Out]

4*x - log(log(x))

Mupad [B] (verification not implemented)

Time = 13.05 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {-1+4 x \log (x)}{x \log (x)} \, dx=4\,x-\ln \left (\ln \left (x\right )\right ) \]

[In]

int((4*x*log(x) - 1)/(x*log(x)),x)

[Out]

4*x - log(log(x))