\(\int \frac {-3+x-2 x \log (x)+x \log (x) \log (\log (x))}{x \log (x)} \, dx\) [7933]

Optimal result

Integrand size = 23, antiderivative size = 15 \[ \int \frac {-3+x-2 x \log (x)+x \log (x) \log (\log (x))}{x \log (x)} \, dx=1+x-(-3+x) (3-\log (\log (x))) \]

[Out]

1-(-ln(ln(x))+3)*(-3+x)+x

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6820, 2395, 2335, 2339, 29, 2600} \[ \int \frac {-3+x-2 x \log (x)+x \log (x) \log (\log (x))}{x \log (x)} \, dx=-2 x+x \log (\log (x))-3 \log (\log (x)) \]

[In]

Int[(-3 + x - 2*x*Log[x] + x*Log[x]*Log[Log[x]])/(x*Log[x]),x]

[Out]

-2*x - 3*Log[Log[x]] + x*Log[Log[x]]

Rule 29

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

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2600

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, 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 {-3+x}{x \log (x)}+\log (\log (x))\right ) \, dx \\ & = -2 x+\int \frac {-3+x}{x \log (x)} \, dx+\int \log (\log (x)) \, dx \\ & = -2 x+x \log (\log (x))+\int \left (\frac {1}{\log (x)}-\frac {3}{x \log (x)}\right ) \, dx-\int \frac {1}{\log (x)} \, dx \\ & = -2 x+x \log (\log (x))-\operatorname {LogIntegral}(x)-3 \int \frac {1}{x \log (x)} \, dx+\int \frac {1}{\log (x)} \, dx \\ & = -2 x+x \log (\log (x))-3 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -2 x-3 \log (\log (x))+x \log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {-3+x-2 x \log (x)+x \log (x) \log (\log (x))}{x \log (x)} \, dx=-2 x-3 \log (\log (x))+x \log (\log (x)) \]

[In]

Integrate[(-3 + x - 2*x*Log[x] + x*Log[x]*Log[Log[x]])/(x*Log[x]),x]

[Out]

-2*x - 3*Log[Log[x]] + x*Log[Log[x]]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00

method	result	size
default	\(-2 x -3 \ln \left (\ln \left (x \right )\right )+x \ln \left (\ln \left (x \right )\right )\)	\(15\)
norman	\(-2 x -3 \ln \left (\ln \left (x \right )\right )+x \ln \left (\ln \left (x \right )\right )\)	\(15\)
risch	\(-2 x -3 \ln \left (\ln \left (x \right )\right )+x \ln \left (\ln \left (x \right )\right )\)	\(15\)
parallelrisch	\(-2 x -3 \ln \left (\ln \left (x \right )\right )+x \ln \left (\ln \left (x \right )\right )\)	\(15\)
parts	\(-2 x -3 \ln \left (\ln \left (x \right )\right )+x \ln \left (\ln \left (x \right )\right )\)	\(15\)

[In]

int((x*ln(x)*ln(ln(x))-2*x*ln(x)+x-3)/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

-2*x-3*ln(ln(x))+x*ln(ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-3+x-2 x \log (x)+x \log (x) \log (\log (x))}{x \log (x)} \, dx={\left (x - 3\right )} \log \left (\log \left (x\right )\right ) - 2 \, x \]

[In]

integrate((x*log(x)*log(log(x))-2*x*log(x)+x-3)/x/log(x),x, algorithm="fricas")

[Out]

(x - 3)*log(log(x)) - 2*x

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-3+x-2 x \log (x)+x \log (x) \log (\log (x))}{x \log (x)} \, dx=x \log {\left (\log {\left (x \right )} \right )} - 2 x - 3 \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate((x*ln(x)*ln(ln(x))-2*x*ln(x)+x-3)/x/ln(x),x)

[Out]

x*log(log(x)) - 2*x - 3*log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {-3+x-2 x \log (x)+x \log (x) \log (\log (x))}{x \log (x)} \, dx=x \log \left (\log \left (x\right )\right ) - 2 \, x - 3 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate((x*log(x)*log(log(x))-2*x*log(x)+x-3)/x/log(x),x, algorithm="maxima")

[Out]

x*log(log(x)) - 2*x - 3*log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {-3+x-2 x \log (x)+x \log (x) \log (\log (x))}{x \log (x)} \, dx=x \log \left (\log \left (x\right )\right ) - 2 \, x - 3 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate((x*log(x)*log(log(x))-2*x*log(x)+x-3)/x/log(x),x, algorithm="giac")

[Out]

x*log(log(x)) - 2*x - 3*log(log(x))

Mupad [B] (verification not implemented)

Time = 14.43 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {-3+x-2 x \log (x)+x \log (x) \log (\log (x))}{x \log (x)} \, dx=x\,\ln \left (\ln \left (x\right )\right )-3\,\ln \left (\ln \left (x\right )\right )-2\,x \]

[In]

int((x - 2*x*log(x) + x*log(log(x))*log(x) - 3)/(x*log(x)),x)

[Out]

x*log(log(x)) - 3*log(log(x)) - 2*x