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

Optimal. Leaf size=15 \[ \frac {e^{8+\frac {8}{e}} \log (x)}{10000} \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 29} \begin {gather*} \frac {e^{8+\frac {8}{e}} \log (x)}{10000} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} \frac {e^{8+\frac {8}{e}} \log (x)}{10000} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(8 + 8/E)*Log[x])/10000

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fricas [A]  time = 0.47, size = 19, normalized size = 1.27 \begin {gather*} e^{\left (-4 \, {\left (e \log \left (10\right ) - 2 \, e - 2\right )} e^{\left (-1\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.36, size = 20, normalized size = 1.33 \begin {gather*} e^{\left (-4 \, {\left (e \log \left (10\right ) - 2 \, e - 2\right )} e^{\left (-1\right )}\right )} \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.03, size = 12, normalized size = 0.80

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.48, size = 19, normalized size = 1.27 \begin {gather*} e^{\left (-4 \, {\left (e \log \left (10\right ) - 2 \, e - 2\right )} e^{\left (-1\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.07, size = 11, normalized size = 0.73 \begin {gather*} \frac {{\mathrm {e}}^{8\,{\mathrm {e}}^{-1}+8}\,\ln \relax (x)}{10000} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.07, size = 14, normalized size = 0.93 \begin {gather*} \frac {e^{8} e^{\frac {8}{e}} \log {\relax (x )}}{10000} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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