\(\int \frac {-1+(1-2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx\) [6427]

Optimal result

Integrand size = 26, antiderivative size = 13 \[ \int \frac {-1+(1-2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=\log \left (\frac {e^{-2 x} x}{\log (\log (x))}\right ) \]

[Out]

ln(x/ln(ln(x))/exp(x)^2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6820, 2339, 29} \[ \int \frac {-1+(1-2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-2 x+\log (x)-\log (\log (\log (x))) \]

[In]

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

[Out]

-2*x + Log[x] - Log[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 (-2+\frac {1}{x}-\frac {1}{x \log (x) \log (\log (x))}\right ) \, dx \\ & = -2 x+\log (x)-\int \frac {1}{x \log (x) \log (\log (x))} \, dx \\ & = -2 x+\log (x)-\text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,\log (x)\right ) \\ & = -2 x+\log (x)-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (\log (x))\right ) \\ & = -2 x+\log (x)-\log (\log (\log (x))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-1+(1-2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-2 x+\log (x)-\log (\log (\log (x))) \]

[In]

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

[Out]

-2*x + Log[x] - Log[Log[Log[x]]]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-1+(1-2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-2 \, x + \log \left (x\right ) - \log \left (\log \left (\log \left (x\right )\right )\right ) \]

[In]

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

[Out]

-2*x + log(x) - log(log(log(x)))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-1+(1-2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=- 2 x + \log {\left (x \right )} - \log {\left (\log {\left (\log {\left (x \right )} \right )} \right )} \]

[In]

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

[Out]

-2*x + log(x) - log(log(log(x)))

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-1+(1-2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-2 \, x + \log \left (x\right ) - \log \left (\log \left (\log \left (x\right )\right )\right ) \]

[In]

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

[Out]

-2*x + log(x) - log(log(log(x)))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-1+(1-2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-2 \, x + \log \left (x\right ) - \log \left (\log \left (\log \left (x\right )\right )\right ) \]

[In]

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

[Out]

-2*x + log(x) - log(log(log(x)))

Mupad [B] (verification not implemented)

Time = 13.89 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-1+(1-2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=\ln \left (x\right )-\ln \left (\ln \left (\ln \left (x\right )\right )\right )-2\,x \]

[In]

int(-(log(log(x))*log(x)*(2*x - 1) + 1)/(x*log(log(x))*log(x)),x)

[Out]

log(x) - log(log(log(x))) - 2*x