3.46.37 \(\int \frac {e^{e^2+\frac {e^{e^2} (x+\log (4))}{-e^{\frac {1}{e^{26}}}+5 e^{e^2}}}}{-e^{\frac {1}{e^{26}}}+5 e^{e^2}} \, dx\)

Optimal. Leaf size=24 \[ e^{\frac {x+\log (4)}{5-e^{\frac {1}{e^{26}}-e^2}}} \]

________________________________________________________________________________________

Rubi [B]  time = 0.07, antiderivative size = 50, normalized size of antiderivative = 2.08, number of steps used = 2, number of rules used = 2, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {12, 2203} \begin {gather*} 4^{-\frac {e^{e^2}}{e^{\frac {1}{e^{26}}}-5 e^{e^2}}} e^{-\frac {e^{e^2} x}{e^{\frac {1}{e^{26}}}-5 e^{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(E^2 + (E^E^2*(x + Log[4]))/(-E^E^(-26) + 5*E^E^2))/(-E^E^(-26) + 5*E^E^2),x]

[Out]

1/(4^(E^E^2/(E^E^(-26) - 5*E^E^2))*E^((E^E^2*x)/(E^E^(-26) - 5*E^E^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 2203

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {\int \exp \left (e^2+\frac {e^{e^2} (x+\log (4))}{-e^{\frac {1}{e^{26}}}+5 e^{e^2}}\right ) \, dx}{e^{\frac {1}{e^{26}}}-5 e^{e^2}}\\ &=4^{-\frac {e^{e^2}}{e^{\frac {1}{e^{26}}}-5 e^{e^2}}} e^{-\frac {e^{e^2} x}{e^{\frac {1}{e^{26}}}-5 e^{e^2}}}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.09, size = 82, normalized size = 3.42 \begin {gather*} -\frac {e^{-\frac {e^{e^2} x}{e^{\frac {1}{e^{26}}}-5 e^{e^2}}-\frac {e^{e^2} \log (4)}{e^{\frac {1}{e^{26}}}-5 e^{e^2}}} \left (e^{\frac {1}{e^{26}}}-5 e^{e^2}\right )}{-e^{\frac {1}{e^{26}}}+5 e^{e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(E^2 + (E^E^2*(x + Log[4]))/(-E^E^(-26) + 5*E^E^2))/(-E^E^(-26) + 5*E^E^2),x]

[Out]

-((E^(-((E^E^2*x)/(E^E^(-26) - 5*E^E^2)) - (E^E^2*Log[4])/(E^E^(-26) - 5*E^E^2))*(E^E^(-26) - 5*E^E^2))/(-E^E^
(-26) + 5*E^E^2))

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 42, normalized size = 1.75 \begin {gather*} e^{\left (\frac {{\left (x + 5 \, e^{2} + 2 \, \log \relax (2)\right )} e^{\left (e^{2}\right )} - e^{\left (e^{\left (-26\right )} + 2\right )}}{5 \, e^{\left (e^{2}\right )} - e^{\left (e^{\left (-26\right )}\right )}} - e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(exp(2))*exp((x+2*log(2))*exp(exp(2))/(5*exp(exp(2))-exp(1/exp(13)^2)))/(5*exp(exp(2))-exp(1/exp(
13)^2)),x, algorithm="fricas")

[Out]

e^(((x + 5*e^2 + 2*log(2))*e^(e^2) - e^(e^(-26) + 2))/(5*e^(e^2) - e^(e^(-26))) - e^2)

________________________________________________________________________________________

giac [A]  time = 0.23, size = 24, normalized size = 1.00 \begin {gather*} e^{\left (\frac {{\left (x + 2 \, \log \relax (2)\right )} e^{\left (e^{2}\right )}}{5 \, e^{\left (e^{2}\right )} - e^{\left (e^{\left (-26\right )}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(exp(2))*exp((x+2*log(2))*exp(exp(2))/(5*exp(exp(2))-exp(1/exp(13)^2)))/(5*exp(exp(2))-exp(1/exp(
13)^2)),x, algorithm="giac")

[Out]

e^((x + 2*log(2))*e^(e^2)/(5*e^(e^2) - e^(e^(-26))))

________________________________________________________________________________________

maple [A]  time = 0.07, size = 24, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{-\frac {\left (x +2 \ln \relax (2)\right ) {\mathrm e}^{{\mathrm e}^{2}}}{-5 \,{\mathrm e}^{{\mathrm e}^{2}}+{\mathrm e}^{{\mathrm e}^{-26}}}}\)	\(24\)
gosper	\({\mathrm e}^{\frac {\left (x +2 \ln \relax (2)\right ) {\mathrm e}^{{\mathrm e}^{2}}}{5 \,{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{{\mathrm e}^{-26}}}}\)	\(27\)
derivativedivides	\({\mathrm e}^{\frac {\left (x +2 \ln \relax (2)\right ) {\mathrm e}^{{\mathrm e}^{2}}}{5 \,{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{{\mathrm e}^{-26}}}}\)	\(27\)
default	\({\mathrm e}^{\frac {\left (x +2 \ln \relax (2)\right ) {\mathrm e}^{{\mathrm e}^{2}}}{5 \,{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{{\mathrm e}^{-26}}}}\)	\(27\)
norman	\({\mathrm e}^{\frac {\left (x +2 \ln \relax (2)\right ) {\mathrm e}^{{\mathrm e}^{2}}}{5 \,{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{{\mathrm e}^{-26}}}}\)	\(27\)
meijerg	\(-{\mathrm e}^{\frac {2 \,{\mathrm e}^{{\mathrm e}^{2}} \ln \relax (2)}{5 \,{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{{\mathrm e}^{-26}}}} \left (1-{\mathrm e}^{\frac {x \,{\mathrm e}^{{\mathrm e}^{2}}}{5 \,{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{{\mathrm e}^{-26}}}}\right )\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(exp(2))*exp((x+2*ln(2))*exp(exp(2))/(5*exp(exp(2))-exp(1/exp(13)^2)))/(5*exp(exp(2))-exp(1/exp(13)^2))
,x,method=_RETURNVERBOSE)

[Out]

exp(-(x+2*ln(2))*exp(exp(2))/(-5*exp(exp(2))+exp(exp(-26))))

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 24, normalized size = 1.00 \begin {gather*} e^{\left (\frac {{\left (x + 2 \, \log \relax (2)\right )} e^{\left (e^{2}\right )}}{5 \, e^{\left (e^{2}\right )} - e^{\left (e^{\left (-26\right )}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(exp(2))*exp((x+2*log(2))*exp(exp(2))/(5*exp(exp(2))-exp(1/exp(13)^2)))/(5*exp(exp(2))-exp(1/exp(
13)^2)),x, algorithm="maxima")

[Out]

e^((x + 2*log(2))*e^(e^2)/(5*e^(e^2) - e^(e^(-26))))

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 24, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^2}\,\left (x+2\,\ln \relax (2)\right )}{5\,{\mathrm {e}}^{{\mathrm {e}}^2}-{\mathrm {e}}^{{\mathrm {e}}^{-26}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(exp(2))*(x + 2*log(2)))/(5*exp(exp(2)) - exp(exp(-26))))*exp(exp(2)))/(5*exp(exp(2)) - exp(exp(-
26))),x)

[Out]

exp((exp(exp(2))*(x + 2*log(2)))/(5*exp(exp(2)) - exp(exp(-26))))

________________________________________________________________________________________

sympy [A]  time = 0.12, size = 26, normalized size = 1.08 \begin {gather*} e^{\frac {\left (x + 2 \log {\relax (2 )}\right ) e^{e^{2}}}{- e^{e^{-26}} + 5 e^{e^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(exp(2))*exp((x+2*ln(2))*exp(exp(2))/(5*exp(exp(2))-exp(1/exp(13)**2)))/(5*exp(exp(2))-exp(1/exp(
13)**2)),x)

[Out]

exp((x + 2*log(2))*exp(exp(2))/(-exp(exp(-26)) + 5*exp(exp(2))))

________________________________________________________________________________________