\(\int \frac {-2-2 x+x \log (x)}{(x+x^2) \log (x)} \, dx\) [2496]

Optimal result

Integrand size = 21, antiderivative size = 18 \[ \int \frac {-2-2 x+x \log (x)}{\left (x+x^2\right ) \log (x)} \, dx=e^2-\log \left (\frac {16 \log ^2(x)}{1+x}\right ) \]

[Out]

exp(2)-ln(16*ln(x)^2/(1+x))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1607, 6820, 2339, 29} \[ \int \frac {-2-2 x+x \log (x)}{\left (x+x^2\right ) \log (x)} \, dx=\log (x+1)-2 \log (\log (x)) \]

[In]

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

[Out]

Log[1 + x] - 2*Log[Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 6820

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {-2-2 x+x \log (x)}{\left (x+x^2\right ) \log (x)} \, dx=\log (1+x)-2 \log (\log (x)) \]

[In]

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

[Out]

Log[1 + x] - 2*Log[Log[x]]

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61

method	result	size
default	\(\ln \left (1+x \right )-2 \ln \left (\ln \left (x \right )\right )\)	\(11\)
norman	\(\ln \left (1+x \right )-2 \ln \left (\ln \left (x \right )\right )\)	\(11\)
risch	\(\ln \left (1+x \right )-2 \ln \left (\ln \left (x \right )\right )\)	\(11\)
parallelrisch	\(\ln \left (1+x \right )-2 \ln \left (\ln \left (x \right )\right )\)	\(11\)
parts	\(\ln \left (1+x \right )-2 \ln \left (\ln \left (x \right )\right )\)	\(11\)

[In]

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

[Out]

ln(1+x)-2*ln(ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {-2-2 x+x \log (x)}{\left (x+x^2\right ) \log (x)} \, dx=\log \left (x + 1\right ) - 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

log(x + 1) - 2*log(log(x))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {-2-2 x+x \log (x)}{\left (x+x^2\right ) \log (x)} \, dx=\log {\left (x + 1 \right )} - 2 \log {\left (\log {\left (x \right )} \right )} \]

[In]

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

[Out]

log(x + 1) - 2*log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {-2-2 x+x \log (x)}{\left (x+x^2\right ) \log (x)} \, dx=\log \left (x + 1\right ) - 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

log(x + 1) - 2*log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {-2-2 x+x \log (x)}{\left (x+x^2\right ) \log (x)} \, dx=\log \left (x + 1\right ) - 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

log(x + 1) - 2*log(log(x))

Mupad [B] (verification not implemented)

Time = 10.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {-2-2 x+x \log (x)}{\left (x+x^2\right ) \log (x)} \, dx=\ln \left (x+1\right )-2\,\ln \left (\ln \left (x\right )\right ) \]

[In]

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

[Out]

log(x + 1) - 2*log(log(x))