\(\int \frac {x^{-1-\frac {1}{\log (\log (x))}} (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x)))}{\log ^2(\log (x))} \, dx\) [7816]

Optimal result

Integrand size = 45, antiderivative size = 15 \[ \int \frac {x^{-1-\frac {1}{\log (\log (x))}} \left (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x))\right )}{\log ^2(\log (x))} \, dx=e^{2+x} x^{-\frac {1}{\log (\log (x))}} \]

[Out]

exp(x)/exp(-2)/exp(ln(x)/ln(ln(x)))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6820, 2326} \[ \int \frac {x^{-1-\frac {1}{\log (\log (x))}} \left (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x))\right )}{\log ^2(\log (x))} \, dx=e^{x+2} x^{-\frac {1}{\log (\log (x))}} \]

[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 2326

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 6820

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x^{-1-\frac {1}{\log (\log (x))}} \left (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x))\right )}{\log ^2(\log (x))} \, dx=e^{2+x} x^{-\frac {1}{\log (\log (x))}} \]

[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)

Maple [A] (verified)

Time = 52.88 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00

method	result	size
risch	\(x^{-\frac {1}{\ln \left (\ln \left (x \right )\right )}} {\mathrm e}^{2+x}\)	\(15\)
parallelrisch	\({\mathrm e}^{2} {\mathrm e}^{x} {\mathrm e}^{-\frac {\ln \left (x \right )}{\ln \left (\ln \left (x \right )\right )}}\)	\(17\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {x^{-1-\frac {1}{\log (\log (x))}} \left (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x))\right )}{\log ^2(\log (x))} \, dx=\frac {e^{\left (x + 2\right )}}{x^{\left (\frac {1}{\log \left (\log \left (x\right )\right )}\right )}} \]

[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)))

Sympy [A] (verification not implemented)

Time = 13.36 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x^{-1-\frac {1}{\log (\log (x))}} \left (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x))\right )}{\log ^2(\log (x))} \, dx=e^{2} e^{x} e^{- \frac {\log {\left (x \right )}}{\log {\left (\log {\left (x \right )} \right )}}} \]

[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)))

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^{-1-\frac {1}{\log (\log (x))}} \left (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x))\right )}{\log ^2(\log (x))} \, dx=\text {Exception raised: RuntimeError} \]

[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

Giac [F]

\[ \int \frac {x^{-1-\frac {1}{\log (\log (x))}} \left (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x))\right )}{\log ^2(\log (x))} \, dx=\int { \frac {x e^{\left (x + 2\right )} \log \left (\log \left (x\right )\right )^{2} - e^{\left (x + 2\right )} \log \left (\log \left (x\right )\right ) + e^{\left (x + 2\right )}}{x x^{\left (\frac {1}{\log \left (\log \left (x\right )\right )}\right )} \log \left (\log \left (x\right )\right )^{2}} \,d x } \]

[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

Mupad [B] (verification not implemented)

Time = 14.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {x^{-1-\frac {1}{\log (\log (x))}} \left (e^{2+x}-e^{2+x} \log (\log (x))+e^{2+x} x \log ^2(\log (x))\right )}{\log ^2(\log (x))} \, dx=\frac {{\mathrm {e}}^2\,{\mathrm {e}}^x}{x^{\frac {1}{\ln \left (\ln \left (x\right )\right )}}} \]

[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)))