3.64.35 \(\int -\frac {e^{-4+x}}{-e^3+\log ^2(3) (i \pi +\log (3-\log (\frac {2}{\log (4)})))} \, dx\)

Optimal. Leaf size=36 \[ \frac {e^{-4+x}}{e^3-\log ^2(3) \left (i \pi +\log \left (3-\log \left (\frac {2}{\log (4)}\right )\right )\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 0.83, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 2194} \begin {gather*} \frac {e^{x-4}}{e^3-\log ^2(3) (\log (3+\log (\log (2)))+i \pi )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-(E^(-4 + x)/(-E^3 + Log[3]^2*(I*Pi + Log[3 - Log[2/Log[4]]]))),x]

[Out]

E^(-4 + x)/(E^3 - Log[3]^2*(I*Pi + Log[3 + Log[Log[2]]]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int e^{-4+x} \, dx}{e^3-\log ^2(3) (i \pi +\log (3+\log (\log (2))))}\\ &=\frac {e^{-4+x}}{e^3-\log ^2(3) (i \pi +\log (3+\log (\log (2))))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.86 \begin {gather*} \frac {e^{-4+x}}{e^3+\log ^2(3) (-i \pi -\log (3+\log (\log (2))))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-(E^(-4 + x)/(-E^3 + Log[3]^2*(I*Pi + Log[3 - Log[2/Log[4]]]))),x]

[Out]

E^(-4 + x)/(E^3 + Log[3]^2*((-I)*Pi - Log[3 + Log[Log[2]]]))

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fricas [A]  time = 0.77, size = 26, normalized size = 0.72 \begin {gather*} -\frac {e^{\left (x - 4\right )}}{\log \relax (3)^{2} \log \left (-\log \left (\log \relax (2)\right ) - 3\right ) - e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x-4)/(log(3)^2*log(-log(log(2))-3)-exp(3)),x, algorithm="fricas")

[Out]

-e^(x - 4)/(log(3)^2*log(-log(log(2)) - 3) - e^3)

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giac [A]  time = 0.17, size = 26, normalized size = 0.72 \begin {gather*} -\frac {e^{\left (x - 4\right )}}{\log \relax (3)^{2} \log \left (-\log \left (\log \relax (2)\right ) - 3\right ) - e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x-4)/(log(3)^2*log(-log(log(2))-3)-exp(3)),x, algorithm="giac")

[Out]

-e^(x - 4)/(log(3)^2*log(-log(log(2)) - 3) - e^3)

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maple [A]  time = 0.06, size = 25, normalized size = 0.69

method	result	size
gosper	\(\frac {{\mathrm e}^{x -4}}{{\mathrm e}^{3}-\ln \relax (3)^{2} \ln \left (-\ln \left (\ln \relax (2)\right )-3\right )}\)	\(25\)
norman	\(\frac {{\mathrm e}^{x -4}}{{\mathrm e}^{3}-\ln \relax (3)^{2} \ln \left (-\ln \left (\ln \relax (2)\right )-3\right )}\)	\(25\)
derivativedivides	\(-\frac {{\mathrm e}^{x -4}}{\ln \relax (3)^{2} \ln \left (-\ln \left (\ln \relax (2)\right )-3\right )-{\mathrm e}^{3}}\)	\(27\)
default	\(-\frac {{\mathrm e}^{x -4}}{\ln \relax (3)^{2} \ln \left (-\ln \left (\ln \relax (2)\right )-3\right )-{\mathrm e}^{3}}\)	\(27\)
risch	\(-\frac {{\mathrm e}^{x -4}}{\ln \relax (3)^{2} \ln \left (-\ln \left (\ln \relax (2)\right )-3\right )-{\mathrm e}^{3}}\)	\(27\)
meijerg	\(\frac {{\mathrm e}^{-4} \left (1-{\mathrm e}^{x}\right )}{\ln \relax (3)^{2} \ln \left (-\ln \left (\ln \relax (2)\right )-3\right )-{\mathrm e}^{3}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x-4)/(ln(3)^2*ln(-ln(ln(2))-3)-exp(3)),x,method=_RETURNVERBOSE)

[Out]

exp(x-4)/(exp(3)-ln(3)^2*ln(-ln(ln(2))-3))

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maxima [A]  time = 0.37, size = 26, normalized size = 0.72 \begin {gather*} -\frac {e^{\left (x - 4\right )}}{\log \relax (3)^{2} \log \left (-\log \left (\log \relax (2)\right ) - 3\right ) - e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x-4)/(log(3)^2*log(-log(log(2))-3)-exp(3)),x, algorithm="maxima")

[Out]

-e^(x - 4)/(log(3)^2*log(-log(log(2)) - 3) - e^3)

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mupad [B]  time = 0.05, size = 24, normalized size = 0.67 \begin {gather*} \frac {{\mathrm {e}}^{-4}\,{\mathrm {e}}^x}{{\mathrm {e}}^3-\ln \left (-\ln \left (\ln \relax (2)\right )-3\right )\,{\ln \relax (3)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x - 4)/(exp(3) - log(- log(log(2)) - 3)*log(3)^2),x)

[Out]

(exp(-4)*exp(x))/(exp(3) - log(- log(log(2)) - 3)*log(3)^2)

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sympy [A]  time = 0.16, size = 36, normalized size = 1.00 \begin {gather*} - \frac {e^{x}}{- e^{7} + e^{4} \log {\relax (3 )}^{2} \log {\left (\log {\left (\log {\relax (2 )} \right )} + 3 \right )} + i \pi e^{4} \log {\relax (3 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x-4)/(ln(3)**2*ln(-ln(ln(2))-3)-exp(3)),x)

[Out]

-exp(x)/(-exp(7) + exp(4)*log(3)**2*log(log(log(2)) + 3) + I*pi*exp(4)*log(3)**2)

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