3.39 \(\int \frac{e^x}{1-e^{2 x}} \, dx\)

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

[Out]

ArcTanh[E^x]

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Rubi [A]  time = 0.0194058, antiderivative size = 4, 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} \[ \tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x/(1 - E^(2*x)),x]

[Out]

ArcTanh[E^x]

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

Mathematica [A]  time = 0.0033732, size = 4, normalized size = 1. \[ \tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/(1 - E^(2*x)),x]

[Out]

ArcTanh[E^x]

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Maple [A]  time = 0.002, size = 4, normalized size = 1. \begin{align*}{\it Artanh} \left ({{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctanh(exp(x))

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Maxima [B]  time = 1.20756, size = 20, normalized size = 5. \begin{align*} \frac{1}{2} \, \log \left (e^{x} + 1\right ) - \frac{1}{2} \, \log \left (e^{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(e^x + 1) - 1/2*log(e^x - 1)

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Fricas [B]  time = 1.50029, size = 50, normalized size = 12.5 \begin{align*} \frac{1}{2} \, \log \left (e^{x} + 1\right ) - \frac{1}{2} \, \log \left (e^{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(e^x + 1) - 1/2*log(e^x - 1)

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Sympy [B]  time = 0.107427, size = 15, normalized size = 3.75 \begin{align*} - \frac{\log{\left (e^{x} - 1 \right )}}{2} + \frac{\log{\left (e^{x} + 1 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(exp(x) - 1)/2 + log(exp(x) + 1)/2

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Giac [B]  time = 1.1911, size = 22, normalized size = 5.5 \begin{align*} \frac{1}{2} \, \log \left (e^{x} + 1\right ) - \frac{1}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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