\(\int \frac {2+x-2 \log (x)}{(2 x+x^2) \log (x)} \, dx\) [5280]

Optimal result

Integrand size = 21, antiderivative size = 15 \[ \int \frac {2+x-2 \log (x)}{\left (2 x+x^2\right ) \log (x)} \, dx=\log \left (\frac {3 (-2-x) \log (x)}{50 x}\right ) \]

[Out]

ln(3/50*ln(x)*(-2-x)/x)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1607, 6874, 36, 29, 31, 2339} \[ \int \frac {2+x-2 \log (x)}{\left (2 x+x^2\right ) \log (x)} \, dx=-\log (x)+\log (x+2)+\log (\log (x)) \]

[In]

Int[(2 + x - 2*Log[x])/((2*x + x^2)*Log[x]),x]

[Out]

-Log[x] + Log[2 + x] + Log[Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {2+x-2 \log (x)}{x (2+x) \log (x)} \, dx \\ & = \int \left (-\frac {2}{x (2+x)}+\frac {1}{x \log (x)}\right ) \, dx \\ & = -\left (2 \int \frac {1}{x (2+x)} \, dx\right )+\int \frac {1}{x \log (x)} \, dx \\ & = -\int \frac {1}{x} \, dx+\int \frac {1}{2+x} \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -\log (x)+\log (2+x)+\log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {2+x-2 \log (x)}{\left (2 x+x^2\right ) \log (x)} \, dx=-\log (x)+\log (2+x)+\log (\log (x)) \]

[In]

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

[Out]

-Log[x] + Log[2 + x] + Log[Log[x]]

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
default	\(\ln \left (\ln \left (x \right )\right )-\ln \left (x \right )+\ln \left (2+x \right )\)	\(13\)
norman	\(\ln \left (\ln \left (x \right )\right )-\ln \left (x \right )+\ln \left (2+x \right )\)	\(13\)
risch	\(\ln \left (\ln \left (x \right )\right )-\ln \left (x \right )+\ln \left (2+x \right )\)	\(13\)
parallelrisch	\(\ln \left (\ln \left (x \right )\right )-\ln \left (x \right )+\ln \left (2+x \right )\)	\(13\)
parts	\(\ln \left (\ln \left (x \right )\right )-\ln \left (x \right )+\ln \left (2+x \right )\)	\(13\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {2+x-2 \log (x)}{\left (2 x+x^2\right ) \log (x)} \, dx=\log \left (x + 2\right ) - \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

log(x + 2) - log(x) + log(log(x))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {2+x-2 \log (x)}{\left (2 x+x^2\right ) \log (x)} \, dx=- \log {\left (x \right )} + \log {\left (x + 2 \right )} + \log {\left (\log {\left (x \right )} \right )} \]

[In]

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

[Out]

-log(x) + log(x + 2) + log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {2+x-2 \log (x)}{\left (2 x+x^2\right ) \log (x)} \, dx=\log \left (x + 2\right ) - \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

log(x + 2) - log(x) + log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {2+x-2 \log (x)}{\left (2 x+x^2\right ) \log (x)} \, dx=\log \left (x + 2\right ) - \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

log(x + 2) - log(x) + log(log(x))

Mupad [B] (verification not implemented)

Time = 14.44 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {2+x-2 \log (x)}{\left (2 x+x^2\right ) \log (x)} \, dx=\ln \left (x+2\right )+\ln \left (\ln \left (x\right )\right )-\ln \left (x\right ) \]

[In]

int((x - 2*log(x) + 2)/(log(x)*(2*x + x^2)),x)

[Out]

log(x + 2) + log(log(x)) - log(x)