\(\int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx\) [2615]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 28 \[ \int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx=\frac {4+\frac {e^{2 x}}{2}}{4 \left (3+e^{\frac {1}{3+e^2}}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 2225} \[ \int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx=\frac {e^{2 x}}{8 \left (3+e^{\frac {1}{3+e^2}}\right )} \]

[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 2225

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx=\frac {e^{2 x}}{2 \left (12+4 e^{\frac {1}{3+e^2}}\right )} \]

[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)))

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 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 \(\frac {{\mathrm e}^{2 x}}{8 \,{\mathrm e}^{\frac {1}{{\mathrm e}^{2}+3}}+24}\) \(20\)
parallelrisch \(\frac {{\mathrm e}^{2 x}}{8 \,{\mathrm e}^{\frac {1}{{\mathrm e}^{2}+3}}+24}\) \(20\)
meijerg \(-\frac {1-{\mathrm e}^{2 x}}{2 \left (4 \,{\mathrm e}^{\frac {1}{{\mathrm e}^{2}+3}}+12\right )}\) \(24\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx=\frac {e^{\left (2 \, x\right )}}{8 \, {\left (e^{\left (\frac {1}{e^{2} + 3}\right )} + 3\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54 \[ \int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx=\frac {e^{2 x}}{8 e^{\frac {1}{3 + e^{2}}} + 24} \]

[In]

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

[Out]

exp(2*x)/(8*exp(1/(3 + exp(2))) + 24)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx=\frac {e^{\left (2 \, x\right )}}{8 \, {\left (e^{\left (\frac {1}{e^{2} + 3}\right )} + 3\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx=\frac {e^{\left (2 \, x\right )}}{8 \, {\left (e^{\left (\frac {1}{e^{2} + 3}\right )} + 3\right )}} \]

[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)

Mupad [B] (verification not implemented)

Time = 8.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {e^{2 x}}{12+4 e^{\frac {1}{3+e^2}}} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,{\mathrm {e}}^{\frac {1}{{\mathrm {e}}^2+3}}+24} \]

[In]

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

[Out]

exp(2*x)/(8*exp(1/(exp(2) + 3)) + 24)