3.79.16 \(\int \frac {x^{-1-\frac {1}{\log (\log (x))}} (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x)))}{\log ^2(\log (x))} \, dx\)

Optimal. Leaf size=15 \[ e^{2+x} x^{-\frac {1}{\log (\log (x))}} \]

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Rubi [A]  time = 0.20, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6688, 2288} \begin {gather*} e^{x+2} x^{-\frac {1}{\log (\log (x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2 + x)/x^Log[Log[x]]^(-1)

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 15, normalized size = 1.00 \begin {gather*} e^{2+x} x^{-\frac {1}{\log (\log (x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2 + x)/x^Log[Log[x]]^(-1)

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fricas [A]  time = 0.51, size = 14, normalized size = 0.93 \begin {gather*} \frac {e^{\left (x + 2\right )}}{x^{\left (\frac {1}{\log \left (\log \relax (x)\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x + 2)/x^(1/log(log(x)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.04, size = 15, normalized size = 1.00

method	result	size
risch	\(x^{-\frac {1}{\ln \left (\ln \relax (x )\right )}} {\mathrm e}^{2+x}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(2)*exp(x)*ln(ln(x))^2-exp(2)*exp(x)*ln(ln(x))+exp(2)*exp(x))/x/ln(ln(x))^2/exp(ln(x)/ln(ln(x))),x,m
ethod=_RETURNVERBOSE)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 5.55, size = 14, normalized size = 0.93 \begin {gather*} \frac {{\mathrm {e}}^2\,{\mathrm {e}}^x}{x^{\frac {1}{\ln \left (\ln \relax (x)\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(2)*exp(x))/x^(1/log(log(x)))

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sympy [A]  time = 23.86, size = 15, normalized size = 1.00 \begin {gather*} e^{2} e^{x} e^{- \frac {\log {\relax (x )}}{\log {\left (\log {\relax (x )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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