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

Optimal. Leaf size=17 \[ \frac {e^{x/4}}{1-\frac {5}{e^2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2194} \begin {gather*} -\frac {e^{\frac {x}{4}+2}}{5-e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^(2 + x/4)/(5 - 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 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 {1}{4} \int e^{\frac {1}{4} \left (x-4 \log \left (\frac {-5+e^2}{e^2}\right )\right )} \, dx\\ &=-\frac {e^{2+\frac {x}{4}}}{5-e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^{2+\frac {x}{4}}}{-5+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2 + x/4)/(-5 + E^2)

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fricas [A]  time = 0.80, size = 15, normalized size = 0.88 \begin {gather*} e^{\left (\frac {1}{4} \, x - \log \left ({\left (e^{2} - 5\right )} e^{\left (-2\right )}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*exp(-log((exp(1)^2-5)/exp(1)^2)+1/4*x),x, algorithm="fricas")

[Out]

e^(1/4*x - log((e^2 - 5)*e^(-2)))

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giac [A]  time = 0.18, size = 15, normalized size = 0.88 \begin {gather*} e^{\left (\frac {1}{4} \, x - \log \left ({\left (e^{2} - 5\right )} e^{\left (-2\right )}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*exp(-log((exp(1)^2-5)/exp(1)^2)+1/4*x),x, algorithm="giac")

[Out]

e^(1/4*x - log((e^2 - 5)*e^(-2)))

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maple [A]  time = 0.02, size = 14, normalized size = 0.82

method	result	size
risch	\(\frac {{\mathrm e}^{2+\frac {x}{4}}}{{\mathrm e}^{2}-5}\)	\(14\)
meijerg	\(-\frac {{\mathrm e}^{2} \left (1-{\mathrm e}^{\frac {x}{4}}\right )}{{\mathrm e}^{2}-5}\)	\(19\)
gosper	\({\mathrm e}^{-\ln \left (\left ({\mathrm e}^{2}-5\right ) {\mathrm e}^{-2}\right )+\frac {x}{4}}\)	\(20\)
derivativedivides	\({\mathrm e}^{-\ln \left (\left ({\mathrm e}^{2}-5\right ) {\mathrm e}^{-2}\right )+\frac {x}{4}}\)	\(20\)
default	\({\mathrm e}^{-\ln \left (\left ({\mathrm e}^{2}-5\right ) {\mathrm e}^{-2}\right )+\frac {x}{4}}\)	\(20\)
norman	\({\mathrm e}^{-\ln \left (\left ({\mathrm e}^{2}-5\right ) {\mathrm e}^{-2}\right )+\frac {x}{4}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*exp(-ln((exp(1)^2-5)/exp(1)^2)+1/4*x),x,method=_RETURNVERBOSE)

[Out]

1/(exp(2)-5)*exp(2+1/4*x)

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maxima [A]  time = 0.34, size = 13, normalized size = 0.76 \begin {gather*} \frac {e^{\left (\frac {1}{4} \, x + 2\right )}}{e^{2} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*exp(-log((exp(1)^2-5)/exp(1)^2)+1/4*x),x, algorithm="maxima")

[Out]

e^(1/4*x + 2)/(e^2 - 5)

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mupad [B]  time = 0.04, size = 14, normalized size = 0.82 \begin {gather*} -\frac {{\mathrm {e}}^{x/4}}{5\,{\mathrm {e}}^{-2}-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x/4 - log(exp(-2)*(exp(2) - 5)))/4,x)

[Out]

-exp(x/4)/(5*exp(-2) - 1)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.71 \begin {gather*} \frac {e^{2} e^{\frac {x}{4}}}{-5 + e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*exp(-ln((exp(1)**2-5)/exp(1)**2)+1/4*x),x)

[Out]

exp(2)*exp(x/4)/(-5 + exp(2))

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