\(\int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx\) [686]

   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 = 15 \[ \int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx=\log \left (1+e^x\right )-\log \left (2+e^x\right ) \]

[Out]

ln(1+exp(x))-ln(2+exp(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2320, 630, 31} \[ \int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx=\log \left (e^x+1\right )-\log \left (e^x+2\right ) \]

[In]

Int[E^x/(2 + 3*E^x + E^(2*x)),x]

[Out]

Log[1 + E^x] - Log[2 + E^x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

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}{2+3 x+x^2} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {1}{2+x} \, dx,x,e^x\right ) \\ & = \log \left (1+e^x\right )-\log \left (2+e^x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx=-2 \text {arctanh}\left (3+2 e^x\right ) \]

[In]

Integrate[E^x/(2 + 3*E^x + E^(2*x)),x]

[Out]

-2*ArcTanh[3 + 2*E^x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method result size
default \(\ln \left (1+{\mathrm e}^{x}\right )-\ln \left (2+{\mathrm e}^{x}\right )\) \(14\)
norman \(\ln \left (1+{\mathrm e}^{x}\right )-\ln \left (2+{\mathrm e}^{x}\right )\) \(14\)
risch \(\ln \left (1+{\mathrm e}^{x}\right )-\ln \left (2+{\mathrm e}^{x}\right )\) \(14\)

[In]

int(exp(x)/(2+3*exp(x)+exp(2*x)),x,method=_RETURNVERBOSE)

[Out]

ln(1+exp(x))-ln(2+exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx=-\log \left (e^{x} + 2\right ) + \log \left (e^{x} + 1\right ) \]

[In]

integrate(exp(x)/(2+3*exp(x)+exp(2*x)),x, algorithm="fricas")

[Out]

-log(e^x + 2) + log(e^x + 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx=\log {\left (e^{x} + 1 \right )} - \log {\left (e^{x} + 2 \right )} \]

[In]

integrate(exp(x)/(2+3*exp(x)+exp(2*x)),x)

[Out]

log(exp(x) + 1) - log(exp(x) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx=-\log \left (e^{x} + 2\right ) + \log \left (e^{x} + 1\right ) \]

[In]

integrate(exp(x)/(2+3*exp(x)+exp(2*x)),x, algorithm="maxima")

[Out]

-log(e^x + 2) + log(e^x + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx=-\log \left (e^{x} + 2\right ) + \log \left (e^{x} + 1\right ) \]

[In]

integrate(exp(x)/(2+3*exp(x)+exp(2*x)),x, algorithm="giac")

[Out]

-log(e^x + 2) + log(e^x + 1)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx=\ln \left ({\mathrm {e}}^x+1\right )-\ln \left ({\mathrm {e}}^x+2\right ) \]

[In]

int(exp(x)/(exp(2*x) + 3*exp(x) + 2),x)

[Out]

log(exp(x) + 1) - log(exp(x) + 2)