3.153 \(\int \frac {1}{x \log (e^x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {\log \left (\log \left (e^x\right )\right )}{x-\log \left (e^x\right )}-\frac {\log (x)}{x-\log \left (e^x\right )} \]

[Out]

-ln(x)/(x-ln(exp(x)))+ln(ln(exp(x)))/(x-ln(exp(x)))

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2160, 2157, 29} \[ \frac {\log \left (\log \left (e^x\right )\right )}{x-\log \left (e^x\right )}-\frac {\log (x)}{x-\log \left (e^x\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[x]/(x - Log[E^x])) + Log[Log[E^x]]/(x - Log[E^x])

Rule 29

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

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2160

Int[1/((u_)*(v_)), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Dist[b/(b*u - a*v), Int[1
/v, x], x] - Dist[a/(b*u - a*v), Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rubi steps

\begin {align*} \int \frac {1}{x \log \left (e^x\right )} \, dx &=-\frac {\int \frac {1}{x} \, dx}{x-\log \left (e^x\right )}+\frac {\int \frac {1}{\log \left (e^x\right )} \, dx}{x-\log \left (e^x\right )}\\ &=-\frac {\log (x)}{x-\log \left (e^x\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (e^x\right )\right )}{x-\log \left (e^x\right )}\\ &=-\frac {\log (x)}{x-\log \left (e^x\right )}+\frac {\log \left (\log \left (e^x\right )\right )}{x-\log \left (e^x\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.68 \[ \frac {\log \left (\log \left (e^x\right )\right )-\log (x)}{x-\log \left (e^x\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-Log[x] + Log[Log[E^x]])/(x - Log[E^x])

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fricas [A]  time = 0.53, size = 5, normalized size = 0.16 \[ -\frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x

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giac [A]  time = 0.16, size = 5, normalized size = 0.16 \[ -\frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x

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maple [A]  time = 0.07, size = 29, normalized size = 0.94 \[ \frac {\ln \relax (x )}{-x +\ln \left ({\mathrm e}^{x}\right )}-\frac {\ln \left (\ln \left ({\mathrm e}^{x}\right )\right )}{-x +\ln \left ({\mathrm e}^{x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/ln(exp(x)),x)

[Out]

-1/(ln(exp(x))-x)*ln(ln(exp(x)))+1/(ln(exp(x))-x)*ln(x)

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maxima [A]  time = 0.70, size = 5, normalized size = 0.16 \[ -\frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x

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mupad [B]  time = 0.36, size = 5, normalized size = 0.16 \[ -\frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*log(exp(x))),x)

[Out]

-1/x

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sympy [A]  time = 0.07, size = 3, normalized size = 0.10 \[ - \frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x

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