\(\int \frac {1+\log (x)}{-2+x \log (x)} \, dx\) [10130]

Optimal result

Integrand size = 13, antiderivative size = 11 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log (2)+\log (2-x \log (x)) \]

[Out]

ln(2)+ln(2-x*ln(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6816} \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log (2-x \log (x)) \]

[In]

Int[(1 + Log[x])/(-2 + x*Log[x]),x]

[Out]

Log[2 - x*Log[x]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps \begin{align*} \text {integral}& = \log (2-x \log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log (-2+x \log (x)) \]

[In]

Integrate[(1 + Log[x])/(-2 + x*Log[x]),x]

[Out]

Log[-2 + x*Log[x]]

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\ln \left (x \ln \left (x \right )-2\right )\)	\(8\)
default	\(\ln \left (x \ln \left (x \right )-2\right )\)	\(8\)
norman	\(\ln \left (x \ln \left (x \right )-2\right )\)	\(8\)
parallelrisch	\(\ln \left (x \ln \left (x \right )-2\right )\)	\(8\)
risch	\(\ln \left (x \right )+\ln \left (\ln \left (x \right )-\frac {2}{x}\right )\)	\(13\)

[In]

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

[Out]

ln(x*ln(x)-2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log \left (x\right ) + \log \left (\frac {x \log \left (x\right ) - 2}{x}\right ) \]

[In]

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

[Out]

log(x) + log((x*log(x) - 2)/x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (x \right )} - \frac {2}{x} \right )} \]

[In]

integrate((ln(x)+1)/(x*ln(x)-2),x)

[Out]

log(x) + log(log(x) - 2/x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log \left (x \log \left (x\right ) - 2\right ) \]

[In]

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

[Out]

log(x*log(x) - 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\log \left (x \log \left (x\right ) - 2\right ) \]

[In]

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

[Out]

log(x*log(x) - 2)

Mupad [B] (verification not implemented)

Time = 15.45 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int \frac {1+\log (x)}{-2+x \log (x)} \, dx=\ln \left (x\,\ln \left (x\right )-2\right ) \]

[In]

int((log(x) + 1)/(x*log(x) - 2),x)

[Out]

log(x*log(x) - 2)