3.644 \(\int \frac{e^x}{-4+e^{2 x}} \, dx\)

Optimal. Leaf size=12 \[ -\frac{1}{2} \tanh ^{-1}\left (\frac{e^x}{2}\right ) \]

[Out]

-ArcTanh[E^x/2]/2

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Rubi [A]  time = 0.0198792, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2249, 207} \[ -\frac{1}{2} \tanh ^{-1}\left (\frac{e^x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x/(-4 + E^(2*x)),x]

[Out]

-ArcTanh[E^x/2]/2

Rule 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{e^x}{-4+e^{2 x}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{-4+x^2} \, dx,x,e^x\right )\\ &=-\frac{1}{2} \tanh ^{-1}\left (\frac{e^x}{2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0027812, size = 12, normalized size = 1. \[ -\frac{1}{2} \tanh ^{-1}\left (\frac{e^x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/(-4 + E^(2*x)),x]

[Out]

-ArcTanh[E^x/2]/2

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Maple [B]  time = 0.027, size = 16, normalized size = 1.3 \begin{align*} -{\frac{\ln \left ( 2+{{\rm e}^{x}} \right ) }{4}}+{\frac{\ln \left ( -2+{{\rm e}^{x}} \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/(-4+exp(2*x)),x)

[Out]

-1/4*ln(2+exp(x))+1/4*ln(-2+exp(x))

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Maxima [B]  time = 0.961811, size = 20, normalized size = 1.67 \begin{align*} -\frac{1}{4} \, \log \left (e^{x} + 2\right ) + \frac{1}{4} \, \log \left (e^{x} - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(e^x + 2) + 1/4*log(e^x - 2)

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Fricas [B]  time = 0.740969, size = 51, normalized size = 4.25 \begin{align*} -\frac{1}{4} \, \log \left (e^{x} + 2\right ) + \frac{1}{4} \, \log \left (e^{x} - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(e^x + 2) + 1/4*log(e^x - 2)

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Sympy [A]  time = 0.101817, size = 15, normalized size = 1.25 \begin{align*} \frac{\log{\left (e^{x} - 2 \right )}}{4} - \frac{\log{\left (e^{x} + 2 \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(exp(x) - 2)/4 - log(exp(x) + 2)/4

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Giac [B]  time = 1.24851, size = 22, normalized size = 1.83 \begin{align*} -\frac{1}{4} \, \log \left (e^{x} + 2\right ) + \frac{1}{4} \, \log \left ({\left | e^{x} - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(e^x + 2) + 1/4*log(abs(e^x - 2))