\(\int \frac {e^{\frac {1}{3} (6+e^2)+e^{\frac {1}{3} (6+e^2)} (259-x)}}{4+4 e^{e^{\frac {1}{3} (6+e^2)} (259-x)}+e^{2 e^{\frac {1}{3} (6+e^2)} (259-x)}} \, dx\) [10113]

Optimal result

Integrand size = 75, antiderivative size = 23 \[ \int \frac {e^{\frac {1}{3} \left (6+e^2\right )+e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}}{4+4 e^{e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}+e^{2 e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}} \, dx=\frac {1}{2+e^{e^{2+\frac {e^2}{3}} (259-x)}} \]

[Out]

1/(exp((-x+259)*exp(1/3*exp(2)+2))+2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2320, 12, 32} \[ \int \frac {e^{\frac {1}{3} \left (6+e^2\right )+e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}}{4+4 e^{e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}+e^{2 e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}} \, dx=\frac {1}{e^{259 e^{2+\frac {e^2}{3}}-e^{2+\frac {e^2}{3}} x}+2} \]

[In]

Int[E^((6 + E^2)/3 + E^((6 + E^2)/3)*(259 - x))/(4 + 4*E^(E^((6 + E^2)/3)*(259 - x)) + E^(2*E^((6 + E^2)/3)*(2
59 - x))),x]

[Out]

(2 + E^(259*E^(2 + E^2/3) - E^(2 + E^2/3)*x))^(-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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = -\left (e^{-2-\frac {e^2}{3}} \text {Subst}\left (\int \frac {e^{2+\frac {e^2}{3}+259 e^{2+\frac {e^2}{3}}}}{\left (2+e^{259 e^{2+\frac {e^2}{3}}} x\right )^2} \, dx,x,e^{-e^{\frac {1}{3} \left (6+e^2\right )} x}\right )\right ) \\ & = -\left (e^{259 e^{2+\frac {e^2}{3}}} \text {Subst}\left (\int \frac {1}{\left (2+e^{259 e^{2+\frac {e^2}{3}}} x\right )^2} \, dx,x,e^{-e^{\frac {1}{3} \left (6+e^2\right )} x}\right )\right ) \\ & = \frac {1}{2+e^{259 e^{2+\frac {e^2}{3}}-e^{2+\frac {e^2}{3}} x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {1}{3} \left (6+e^2\right )+e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}}{4+4 e^{e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}+e^{2 e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}} \, dx=-\frac {1}{2 \left (1+2 e^{e^{2+\frac {e^2}{3}} (-259+x)}\right )} \]

[In]

Integrate[E^((6 + E^2)/3 + E^((6 + E^2)/3)*(259 - x))/(4 + 4*E^(E^((6 + E^2)/3)*(259 - x)) + E^(2*E^((6 + E^2)
/3)*(259 - x))),x]

[Out]

-1/2*1/(1 + 2*E^(E^(2 + E^2/3)*(-259 + x)))

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {1}{{\mathrm e}^{-\left (x -259\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{3}+2}}+2}\)	\(18\)
derivativedivides	\(\frac {1}{{\mathrm e}^{\left (-x +259\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{3}+2}}+2}\)	\(19\)
default	\(\frac {1}{{\mathrm e}^{\left (-x +259\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{3}+2}}+2}\)	\(19\)
norman	\(\frac {1}{{\mathrm e}^{\left (-x +259\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{3}+2}}+2}\)	\(19\)
parallelrisch	\(\frac {1}{{\mathrm e}^{\left (-x +259\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{3}+2}}+2}\)	\(19\)

[In]

int(exp(1/3*exp(2)+2)*exp((-x+259)*exp(1/3*exp(2)+2))/(exp((-x+259)*exp(1/3*exp(2)+2))^2+4*exp((-x+259)*exp(1/
3*exp(2)+2))+4),x,method=_RETURNVERBOSE)

[Out]

1/(exp(-(x-259)*exp(1/3*exp(2)+2))+2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (17) = 34\).

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {1}{3} \left (6+e^2\right )+e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}}{4+4 e^{e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}+e^{2 e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}} \, dx=\frac {e^{\left (\frac {1}{3} \, e^{2} + 2\right )}}{e^{\left (-{\left (x - 259\right )} e^{\left (\frac {1}{3} \, e^{2} + 2\right )} + \frac {1}{3} \, e^{2} + 2\right )} + 2 \, e^{\left (\frac {1}{3} \, e^{2} + 2\right )}} \]

[In]

integrate(exp(1/3*exp(2)+2)*exp((-x+259)*exp(1/3*exp(2)+2))/(exp((-x+259)*exp(1/3*exp(2)+2))^2+4*exp((-x+259)*
exp(1/3*exp(2)+2))+4),x, algorithm="fricas")

[Out]

e^(1/3*e^2 + 2)/(e^(-(x - 259)*e^(1/3*e^2 + 2) + 1/3*e^2 + 2) + 2*e^(1/3*e^2 + 2))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {1}{3} \left (6+e^2\right )+e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}}{4+4 e^{e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}+e^{2 e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}} \, dx=\frac {1}{e^{\left (259 - x\right ) e^{2 + \frac {e^{2}}{3}}} + 2} \]

[In]

integrate(exp(1/3*exp(2)+2)*exp((-x+259)*exp(1/3*exp(2)+2))/(exp((-x+259)*exp(1/3*exp(2)+2))**2+4*exp((-x+259)
*exp(1/3*exp(2)+2))+4),x)

[Out]

1/(exp((259 - x)*exp(2 + exp(2)/3)) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {1}{3} \left (6+e^2\right )+e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}}{4+4 e^{e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}+e^{2 e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}} \, dx=\frac {1}{e^{\left (-x e^{\left (\frac {1}{3} \, e^{2} + 2\right )} + 259 \, e^{\left (\frac {1}{3} \, e^{2} + 2\right )}\right )} + 2} \]

[In]

integrate(exp(1/3*exp(2)+2)*exp((-x+259)*exp(1/3*exp(2)+2))/(exp((-x+259)*exp(1/3*exp(2)+2))^2+4*exp((-x+259)*
exp(1/3*exp(2)+2))+4),x, algorithm="maxima")

[Out]

1/(e^(-x*e^(1/3*e^2 + 2) + 259*e^(1/3*e^2 + 2)) + 2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (17) = 34\).

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {1}{3} \left (6+e^2\right )+e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}}{4+4 e^{e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}+e^{2 e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}} \, dx=\frac {e^{\left (\frac {1}{3} \, e^{2} + 2\right )}}{e^{\left (-x e^{\left (\frac {1}{3} \, e^{2} + 2\right )} + \frac {1}{3} \, e^{2} + 259 \, e^{\left (\frac {1}{3} \, e^{2} + 2\right )} + 2\right )} + 2 \, e^{\left (\frac {1}{3} \, e^{2} + 2\right )}} \]

[In]

integrate(exp(1/3*exp(2)+2)*exp((-x+259)*exp(1/3*exp(2)+2))/(exp((-x+259)*exp(1/3*exp(2)+2))^2+4*exp((-x+259)*
exp(1/3*exp(2)+2))+4),x, algorithm="giac")

[Out]

e^(1/3*e^2 + 2)/(e^(-x*e^(1/3*e^2 + 2) + 1/3*e^2 + 259*e^(1/3*e^2 + 2) + 2) + 2*e^(1/3*e^2 + 2))

Mupad [B] (verification not implemented)

Time = 16.84 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {1}{3} \left (6+e^2\right )+e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}}{4+4 e^{e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}+e^{2 e^{\frac {1}{3} \left (6+e^2\right )} (259-x)}} \, dx=-\frac {1}{2\,\left (2\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{3}}\,{\mathrm {e}}^2\,\left (x-259\right )}+1\right )} \]

[In]

int((exp(-exp(exp(2)/3 + 2)*(x - 259))*exp(exp(2)/3 + 2))/(4*exp(-exp(exp(2)/3 + 2)*(x - 259)) + exp(-2*exp(ex
p(2)/3 + 2)*(x - 259)) + 4),x)

[Out]

-1/(2*(2*exp(exp(exp(2)/3)*exp(2)*(x - 259)) + 1))