\(\int \frac {1-\log (x)}{x (x+\log (x))} \, dx\) [292]

Optimal result

Integrand size = 16, antiderivative size = 9 \[ \int \frac {1-\log (x)}{x (x+\log (x))} \, dx=\log \left (1+\frac {\log (x)}{x}\right ) \]

[Out]

ln(1+ln(x)/x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6844, 31} \[ \int \frac {1-\log (x)}{x (x+\log (x))} \, dx=\log \left (\frac {\log (x)}{x}+1\right ) \]

[In]

Int[(1 - Log[x])/(x*(x + Log[x])),x]

[Out]

Log[1 + Log[x]/x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 6844

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, Dist[(-c)*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; Fre
eQ[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11 \[ \int \frac {1-\log (x)}{x (x+\log (x))} \, dx=-\log (x)+\log (x+\log (x)) \]

[In]

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

[Out]

-Log[x] + Log[x + Log[x]]

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.22

method	result	size
default	\(-\ln \left (x \right )+\ln \left (x +\ln \left (x \right )\right )\)	\(11\)
norman	\(-\ln \left (x \right )+\ln \left (x +\ln \left (x \right )\right )\)	\(11\)
risch	\(-\ln \left (x \right )+\ln \left (x +\ln \left (x \right )\right )\)	\(11\)
parallelrisch	\(-\ln \left (x \right )+\ln \left (x +\ln \left (x \right )\right )\)	\(11\)

[In]

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

[Out]

-ln(x)+ln(x+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11 \[ \int \frac {1-\log (x)}{x (x+\log (x))} \, dx=\log \left (x + \log \left (x\right )\right ) - \log \left (x\right ) \]

[In]

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

[Out]

log(x + log(x)) - log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {1-\log (x)}{x (x+\log (x))} \, dx=- \log {\left (x \right )} + \log {\left (x + \log {\left (x \right )} \right )} \]

[In]

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

[Out]

-log(x) + log(x + log(x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11 \[ \int \frac {1-\log (x)}{x (x+\log (x))} \, dx=\log \left (x + \log \left (x\right )\right ) - \log \left (x\right ) \]

[In]

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

[Out]

log(x + log(x)) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.56 \[ \int \frac {1-\log (x)}{x (x+\log (x))} \, dx=-\log \left (x\right ) + \log \left (-x - \log \left (x\right )\right ) \]

[In]

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

[Out]

-log(x) + log(-x - log(x))

Mupad [B] (verification not implemented)

Time = 1.56 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11 \[ \int \frac {1-\log (x)}{x (x+\log (x))} \, dx=\ln \left (x+\ln \left (x\right )\right )-\ln \left (x\right ) \]

[In]

int(-(log(x) - 1)/(x*(x + log(x))),x)

[Out]

log(x + log(x)) - log(x)