3.665 \(\int \frac{e^x}{16-e^{2 x}} \, dx\)

Optimal. Leaf size=12 \[ \frac{1}{4} \tanh ^{-1}\left (\frac{e^x}{4}\right ) \]

[Out]

ArcTanh[E^x/4]/4

________________________________________________________________________________________

Rubi [A]  time = 0.0221998, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2249, 206} \[ \frac{1}{4} \tanh ^{-1}\left (\frac{e^x}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x/(16 - E^(2*x)),x]

[Out]

ArcTanh[E^x/4]/4

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 206

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

Rubi steps

\begin{align*} \int \frac{e^x}{16-e^{2 x}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{16-x^2} \, dx,x,e^x\right )\\ &=\frac{1}{4} \tanh ^{-1}\left (\frac{e^x}{4}\right )\\ \end{align*}

Mathematica [A]  time = 0.0031379, size = 12, normalized size = 1. \[ \frac{1}{4} \tanh ^{-1}\left (\frac{e^x}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/(16 - E^(2*x)),x]

[Out]

ArcTanh[E^x/4]/4

________________________________________________________________________________________

Maple [B]  time = 0.023, size = 16, normalized size = 1.3 \begin{align*}{\frac{\ln \left ({{\rm e}^{x}}+4 \right ) }{8}}-{\frac{\ln \left ( -4+{{\rm e}^{x}} \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/(16-exp(2*x)),x)

[Out]

1/8*ln(exp(x)+4)-1/8*ln(-4+exp(x))

________________________________________________________________________________________

Maxima [B]  time = 0.966717, size = 20, normalized size = 1.67 \begin{align*} \frac{1}{8} \, \log \left (e^{x} + 4\right ) - \frac{1}{8} \, \log \left (e^{x} - 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(16-exp(2*x)),x, algorithm="maxima")

[Out]

1/8*log(e^x + 4) - 1/8*log(e^x - 4)

________________________________________________________________________________________

Fricas [B]  time = 0.7882, size = 50, normalized size = 4.17 \begin{align*} \frac{1}{8} \, \log \left (e^{x} + 4\right ) - \frac{1}{8} \, \log \left (e^{x} - 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(16-exp(2*x)),x, algorithm="fricas")

[Out]

1/8*log(e^x + 4) - 1/8*log(e^x - 4)

________________________________________________________________________________________

Sympy [B]  time = 0.105618, size = 15, normalized size = 1.25 \begin{align*} - \frac{\log{\left (e^{x} - 4 \right )}}{8} + \frac{\log{\left (e^{x} + 4 \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(16-exp(2*x)),x)

[Out]

-log(exp(x) - 4)/8 + log(exp(x) + 4)/8

________________________________________________________________________________________

Giac [B]  time = 1.21672, size = 22, normalized size = 1.83 \begin{align*} \frac{1}{8} \, \log \left (e^{x} + 4\right ) - \frac{1}{8} \, \log \left ({\left | e^{x} - 4 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(16-exp(2*x)),x, algorithm="giac")

[Out]

1/8*log(e^x + 4) - 1/8*log(abs(e^x - 4))