\(\int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx\) [240]

Optimal result

Integrand size = 20, antiderivative size = 17 \[ \int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx=-\log \left (1+e^x\right )+2 \log \left (2+e^x\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2320, 646, 31} \[ \int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx=2 \log \left (e^x+2\right )-\log \left (e^x+1\right ) \]

[In]

Int[E^(2*x)/(2 + 3*E^x + E^(2*x)),x]

[Out]

-Log[1 + E^x] + 2*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 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[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 {x}{2+3 x+x^2} \, dx,x,e^x\right ) \\ & = 2 \text {Subst}\left (\int \frac {1}{2+x} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {1}{1+x} \, dx,x,e^x\right ) \\ & = -\log \left (1+e^x\right )+2 \log \left (2+e^x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx=-\log \left (1+e^x\right )+2 \log \left (2+e^x\right ) \]

[In]

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

[Out]

-Log[1 + E^x] + 2*Log[2 + E^x]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(-\ln \left (1+{\mathrm e}^{x}\right )+2 \ln \left (2+{\mathrm e}^{x}\right )\)	\(16\)
norman	\(-\ln \left (1+{\mathrm e}^{x}\right )+2 \ln \left (2+{\mathrm e}^{x}\right )\)	\(16\)
risch	\(-\ln \left (1+{\mathrm e}^{x}\right )+2 \ln \left (2+{\mathrm e}^{x}\right )\)	\(16\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx=2 \, \log \left (e^{x} + 2\right ) - \log \left (e^{x} + 1\right ) \]

[In]

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

[Out]

2*log(e^x + 2) - log(e^x + 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx=- \log {\left (e^{x} + 1 \right )} + 2 \log {\left (e^{x} + 2 \right )} \]

[In]

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

[Out]

-log(exp(x) + 1) + 2*log(exp(x) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx=2 \, \log \left (e^{x} + 2\right ) - \log \left (e^{x} + 1\right ) \]

[In]

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

[Out]

2*log(e^x + 2) - log(e^x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx=2 \, \log \left (e^{x} + 2\right ) - \log \left (e^{x} + 1\right ) \]

[In]

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

[Out]

2*log(e^x + 2) - log(e^x + 1)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx=2\,\ln \left ({\mathrm {e}}^x+2\right )-\ln \left ({\mathrm {e}}^x+1\right ) \]

[In]

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

[Out]

2*log(exp(x) + 2) - log(exp(x) + 1)