\(\int \frac {e^x}{1-e^{2 x}} \, dx\) [39]

Optimal result

Integrand size = 15, antiderivative size = 4 \[ \int \frac {e^x}{1-e^{2 x}} \, dx=\text {arctanh}\left (e^x\right ) \]

[Out]

arctanh(exp(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2281, 212} \[ \int \frac {e^x}{1-e^{2 x}} \, dx=\text {arctanh}\left (e^x\right ) \]

[In]

Int[E^x/(1 - E^(2*x)),x]

[Out]

ArcTanh[E^x]

Rule 212

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

Rule 2281

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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{1-e^{2 x}} \, dx=\text {arctanh}\left (e^x\right ) \]

[In]

Integrate[E^x/(1 - E^(2*x)),x]

[Out]

ArcTanh[E^x]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00

method	result	size
default	\(\operatorname {arctanh}\left ({\mathrm e}^{x}\right )\)	\(4\)
norman	\(-\frac {\ln \left (-1+{\mathrm e}^{x}\right )}{2}+\frac {\ln \left (1+{\mathrm e}^{x}\right )}{2}\)	\(16\)
risch	\(-\frac {\ln \left (-1+{\mathrm e}^{x}\right )}{2}+\frac {\ln \left (1+{\mathrm e}^{x}\right )}{2}\)	\(16\)

[In]

int(exp(x)/(1-exp(2*x)),x,method=_RETURNVERBOSE)

[Out]

arctanh(exp(x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (3) = 6\).

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 3.75 \[ \int \frac {e^x}{1-e^{2 x}} \, dx=\frac {1}{2} \, \log \left (e^{x} + 1\right ) - \frac {1}{2} \, \log \left (e^{x} - 1\right ) \]

[In]

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

[Out]

1/2*log(e^x + 1) - 1/2*log(e^x - 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (3) = 6\).

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 3.75 \[ \int \frac {e^x}{1-e^{2 x}} \, dx=- \frac {\log {\left (e^{x} - 1 \right )}}{2} + \frac {\log {\left (e^{x} + 1 \right )}}{2} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (3) = 6\).

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 3.75 \[ \int \frac {e^x}{1-e^{2 x}} \, dx=\frac {1}{2} \, \log \left (e^{x} + 1\right ) - \frac {1}{2} \, \log \left (e^{x} - 1\right ) \]

[In]

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

[Out]

1/2*log(e^x + 1) - 1/2*log(e^x - 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 16 vs. \(2 (3) = 6\).

Time = 0.33 (sec) , antiderivative size = 16, normalized size of antiderivative = 4.00 \[ \int \frac {e^x}{1-e^{2 x}} \, dx=\frac {1}{2} \, \log \left (e^{x} + 1\right ) - \frac {1}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 3.75 \[ \int \frac {e^x}{1-e^{2 x}} \, dx=\frac {\ln \left ({\mathrm {e}}^x+1\right )}{2}-\frac {\ln \left ({\mathrm {e}}^x-1\right )}{2} \]

[In]

int(-exp(x)/(exp(2*x) - 1),x)

[Out]

log(exp(x) + 1)/2 - log(exp(x) - 1)/2