\(\int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx\) [926]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 15 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {e^{8+\frac {8}{e}} \log (x)}{10000} \]

[Out]

ln(x)/exp(2*ln(10)-4-4/exp(1))^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 29} \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {e^{8+\frac {8}{e}} \log (x)}{10000} \]

[In]

Int[1/(E^((2*(-4 - 4*E + E*Log[100]))/E)*x),x]

[Out]

(E^(8 + 8/E)*Log[x])/10000

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]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{8+\frac {8}{e}} \int \frac {1}{x} \, dx}{10000} \\ & = \frac {e^{8+\frac {8}{e}} \log (x)}{10000} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {e^{8+\frac {8}{e}} \log (x)}{10000} \]

[In]

Integrate[1/(E^((2*(-4 - 4*E + E*Log[100]))/E)*x),x]

[Out]

(E^(8 + 8/E)*Log[x])/10000

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80

method result size
risch \(\frac {\ln \left (x \right ) {\mathrm e}^{8+8 \,{\mathrm e}^{-1}}}{10000}\) \(12\)
norman \(\frac {{\mathrm e}^{8} {\mathrm e}^{8 \,{\mathrm e}^{-1}} \ln \left (x \right )}{10000}\) \(16\)
default \({\mathrm e}^{-4 \left ({\mathrm e} \ln \left (10\right )-2 \,{\mathrm e}-2\right ) {\mathrm e}^{-1}} \ln \left (x \right )\) \(24\)
parallelrisch \({\mathrm e}^{-4 \left ({\mathrm e} \ln \left (10\right )-2 \,{\mathrm e}-2\right ) {\mathrm e}^{-1}} \ln \left (x \right )\) \(24\)

[In]

int(1/x/exp((2*exp(1)*ln(10)-4*exp(1)-4)/exp(1))^2,x,method=_RETURNVERBOSE)

[Out]

1/10000*ln(x)*exp(8+8*exp(-1))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=e^{\left (-4 \, {\left (e \log \left (10\right ) - 2 \, e - 2\right )} e^{\left (-1\right )}\right )} \log \left (x\right ) \]

[In]

integrate(1/x/exp((2*exp(1)*log(10)-4*exp(1)-4)/exp(1))^2,x, algorithm="fricas")

[Out]

e^(-4*(e*log(10) - 2*e - 2)*e^(-1))*log(x)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {e^{8} e^{\frac {8}{e}} \log {\left (x \right )}}{10000} \]

[In]

integrate(1/x/exp((2*exp(1)*ln(10)-4*exp(1)-4)/exp(1))**2,x)

[Out]

exp(8)*exp(8*exp(-1))*log(x)/10000

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=e^{\left (-4 \, {\left (e \log \left (10\right ) - 2 \, e - 2\right )} e^{\left (-1\right )}\right )} \log \left (x\right ) \]

[In]

integrate(1/x/exp((2*exp(1)*log(10)-4*exp(1)-4)/exp(1))^2,x, algorithm="maxima")

[Out]

e^(-4*(e*log(10) - 2*e - 2)*e^(-1))*log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=e^{\left (-4 \, {\left (e \log \left (10\right ) - 2 \, e - 2\right )} e^{\left (-1\right )}\right )} \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/x/exp((2*exp(1)*log(10)-4*exp(1)-4)/exp(1))^2,x, algorithm="giac")

[Out]

e^(-4*(e*log(10) - 2*e - 2)*e^(-1))*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {{\mathrm {e}}^{8\,{\mathrm {e}}^{-1}+8}\,\ln \left (x\right )}{10000} \]

[In]

int(exp(2*exp(-1)*(4*exp(1) - 2*exp(1)*log(10) + 4))/x,x)

[Out]

(exp(8*exp(-1) + 8)*log(x))/10000