\(\int \frac {e^x}{1+2 e^x+e^{2 x}} \, dx\) [676]

   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 = 18, antiderivative size = 9 \[ \int \frac {e^x}{1+2 e^x+e^{2 x}} \, dx=-\frac {1}{1+e^x} \]

[Out]

-1/(1+exp(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2320, 32} \[ \int \frac {e^x}{1+2 e^x+e^{2 x}} \, dx=-\frac {1}{e^x+1} \]

[In]

Int[E^x/(1 + 2*E^x + E^(2*x)),x]

[Out]

-(1 + E^x)^(-1)

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}& = \text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,e^x\right ) \\ & = -\frac {1}{1+e^x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{1+2 e^x+e^{2 x}} \, dx=-\frac {1}{1+e^x} \]

[In]

Integrate[E^x/(1 + 2*E^x + E^(2*x)),x]

[Out]

-(1 + E^x)^(-1)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00

method result size
default \(-\frac {1}{1+{\mathrm e}^{x}}\) \(9\)
norman \(-\frac {1}{1+{\mathrm e}^{x}}\) \(9\)
risch \(-\frac {1}{1+{\mathrm e}^{x}}\) \(9\)

[In]

int(exp(x)/(1+2*exp(x)+exp(2*x)),x,method=_RETURNVERBOSE)

[Out]

-1/(1+exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {e^x}{1+2 e^x+e^{2 x}} \, dx=-\frac {1}{e^{x} + 1} \]

[In]

integrate(exp(x)/(1+2*exp(x)+exp(2*x)),x, algorithm="fricas")

[Out]

-1/(e^x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {e^x}{1+2 e^x+e^{2 x}} \, dx=- \frac {1}{e^{x} + 1} \]

[In]

integrate(exp(x)/(1+2*exp(x)+exp(2*x)),x)

[Out]

-1/(exp(x) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {e^x}{1+2 e^x+e^{2 x}} \, dx=-\frac {1}{e^{x} + 1} \]

[In]

integrate(exp(x)/(1+2*exp(x)+exp(2*x)),x, algorithm="maxima")

[Out]

-1/(e^x + 1)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {e^x}{1+2 e^x+e^{2 x}} \, dx=-\frac {1}{e^{x} + 1} \]

[In]

integrate(exp(x)/(1+2*exp(x)+exp(2*x)),x, algorithm="giac")

[Out]

-1/(e^x + 1)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {e^x}{1+2 e^x+e^{2 x}} \, dx=-\frac {1}{{\mathrm {e}}^x+1} \]

[In]

int(exp(x)/(exp(2*x) + 2*exp(x) + 1),x)

[Out]

-1/(exp(x) + 1)