3.27.15 \(\int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx\)

Optimal. Leaf size=28 \[ \frac {4+\frac {e^{2 x}}{2}}{4 \left (3+e^{\frac {1}{3+e^2}}\right )} \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 2194} \begin {gather*} \frac {e^{2 x}}{8 \left (3+e^{\frac {1}{3+e^2}}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*x)/(12 + 4*E^(3 + E^2)^(-1)),x]

[Out]

E^(2*x)/(8*(3 + E^(3 + E^2)^(-1)))

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^{2 x} \, dx}{4 \left (3+e^{\frac {1}{3+e^2}}\right )}\\ &=\frac {e^{2 x}}{8 \left (3+e^{\frac {1}{3+e^2}}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 24, normalized size = 0.86 \begin {gather*} \frac {e^{2 x}}{2 \left (12+4 e^{\frac {1}{3+e^2}}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(12 + 4*E^(3 + E^2)^(-1)),x]

[Out]

E^(2*x)/(2*(12 + 4*E^(3 + E^2)^(-1)))

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fricas [A]  time = 0.75, size = 17, normalized size = 0.61 \begin {gather*} \frac {e^{\left (2 \, x\right )}}{8 \, {\left (e^{\left (\frac {1}{e^{2} + 3}\right )} + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)^2/(4*exp(1/(exp(2)+3))+12),x, algorithm="fricas")

[Out]

1/8*e^(2*x)/(e^(1/(e^2 + 3)) + 3)

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giac [A]  time = 1.01, size = 17, normalized size = 0.61 \begin {gather*} \frac {e^{\left (2 \, x\right )}}{8 \, {\left (e^{\left (\frac {1}{e^{2} + 3}\right )} + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)^2/(4*exp(1/(exp(2)+3))+12),x, algorithm="giac")

[Out]

1/8*e^(2*x)/(e^(1/(e^2 + 3)) + 3)

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maple [A]  time = 0.03, size = 18, normalized size = 0.64

method	result	size
gosper	\(\frac {{\mathrm e}^{2 x}}{8 \,{\mathrm e}^{\frac {1}{{\mathrm e}^{2}+3}}+24}\)	\(18\)
default	\(\frac {{\mathrm e}^{2 x}}{8 \,{\mathrm e}^{\frac {1}{{\mathrm e}^{2}+3}}+24}\)	\(18\)
norman	\(\frac {{\mathrm e}^{2 x}}{8 \,{\mathrm e}^{\frac {1}{{\mathrm e}^{2}+3}}+24}\)	\(18\)
derivativedivides	\(\frac {{\mathrm e}^{2 x}}{8 \,{\mathrm e}^{\frac {1}{{\mathrm e}^{2}+3}}+24}\)	\(20\)
risch	\(\frac {{\mathrm e}^{2 x}}{8 \,{\mathrm e}^{\frac {1}{{\mathrm e}^{2}+3}}+24}\)	\(20\)
meijerg	\(-\frac {1-{\mathrm e}^{2 x}}{2 \left (4 \,{\mathrm e}^{\frac {1}{{\mathrm e}^{2}+3}}+12\right )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)^2/(4*exp(1/(exp(2)+3))+12),x,method=_RETURNVERBOSE)

[Out]

1/8*exp(x)^2/(exp(1/(exp(2)+3))+3)

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maxima [A]  time = 0.45, size = 17, normalized size = 0.61 \begin {gather*} \frac {e^{\left (2 \, x\right )}}{8 \, {\left (e^{\left (\frac {1}{e^{2} + 3}\right )} + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)^2/(4*exp(1/(exp(2)+3))+12),x, algorithm="maxima")

[Out]

1/8*e^(2*x)/(e^(1/(e^2 + 3)) + 3)

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mupad [B]  time = 1.43, size = 18, normalized size = 0.64 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}}{8\,{\mathrm {e}}^{\frac {1}{{\mathrm {e}}^2+3}}+24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(4*exp(1/(exp(2) + 3)) + 12),x)

[Out]

exp(2*x)/(8*exp(1/(exp(2) + 3)) + 24)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.54 \begin {gather*} \frac {e^{2 x}}{8 e^{\frac {1}{3 + e^{2}}} + 24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)**2/(4*exp(1/(exp(2)+3))+12),x)

[Out]

exp(2*x)/(8*exp(1/(3 + exp(2))) + 24)

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