3.82.53 \(\int \frac {e^{-\frac {2}{e^4}}}{x \log (x)} \, dx\)

Optimal. Leaf size=22 \[ e^{-\frac {2}{e^4}} \log \left (\frac {\log (x)}{\left (-\frac {13}{4}-\log (2)\right )^2}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 11, normalized size of antiderivative = 0.50, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2302, 29} \begin {gather*} e^{-\frac {2}{e^4}} \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(2/E^4)*x*Log[x]),x]

[Out]

Log[Log[x]]/E^(2/E^4)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

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

Rule 2302

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{-\frac {2}{e^4}} \int \frac {1}{x \log (x)} \, dx\\ &=e^{-\frac {2}{e^4}} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=e^{-\frac {2}{e^4}} \log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 11, normalized size = 0.50 \begin {gather*} e^{-\frac {2}{e^4}} \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2/E^4)*x*Log[x]),x]

[Out]

Log[Log[x]]/E^(2/E^4)

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fricas [A]  time = 1.05, size = 9, normalized size = 0.41 \begin {gather*} e^{\left (-2 \, e^{\left (-4\right )}\right )} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/exp(2/exp(4))/log(x),x, algorithm="fricas")

[Out]

e^(-2*e^(-4))*log(log(x))

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giac [A]  time = 0.23, size = 10, normalized size = 0.45 \begin {gather*} e^{\left (-2 \, e^{\left (-4\right )}\right )} \log \left ({\left | \log \relax (x) \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/exp(2/exp(4))/log(x),x, algorithm="giac")

[Out]

e^(-2*e^(-4))*log(abs(log(x)))

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maple [A]  time = 0.04, size = 10, normalized size = 0.45

method	result	size
risch	\({\mathrm e}^{-2 \,{\mathrm e}^{-4}} \ln \left (\ln \relax (x )\right )\)	\(10\)
norman	\({\mathrm e}^{-2 \,{\mathrm e}^{-4}} \ln \left (\ln \relax (x )\right )\)	\(12\)
derivativedivides	\({\mathrm e}^{-2 \,{\mathrm e}^{-4}} \ln \left (\ln \relax (x )\right )\)	\(14\)
default	\({\mathrm e}^{-2 \,{\mathrm e}^{-4}} \ln \left (\ln \relax (x )\right )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/exp(2/exp(4))/ln(x),x,method=_RETURNVERBOSE)

[Out]

exp(-2*exp(-4))*ln(ln(x))

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maxima [A]  time = 0.36, size = 9, normalized size = 0.41 \begin {gather*} e^{\left (-2 \, e^{\left (-4\right )}\right )} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/exp(2/exp(4))/log(x),x, algorithm="maxima")

[Out]

e^(-2*e^(-4))*log(log(x))

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mupad [B]  time = 4.84, size = 9, normalized size = 0.41 \begin {gather*} \ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{-4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-2*exp(-4))/(x*log(x)),x)

[Out]

log(log(x))*exp(-2*exp(-4))

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sympy [A]  time = 0.11, size = 10, normalized size = 0.45 \begin {gather*} \frac {\log {\left (\log {\relax (x )} \right )}}{e^{\frac {2}{e^{4}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/exp(2/exp(4))/ln(x),x)

[Out]

exp(-2*exp(-4))*log(log(x))

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