\(\int \frac {e^x}{-1+e^{2 x}} \, dx\) [652]

Optimal result

Integrand size = 13, antiderivative size = 6 \[ \int \frac {e^x}{-1+e^{2 x}} \, dx=-\text {arctanh}\left (e^x\right ) \]

[Out]

-arctanh(exp(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2281, 213} \[ \int \frac {e^x}{-1+e^{2 x}} \, dx=-\text {arctanh}\left (e^x\right ) \]

[In]

Int[E^x/(-1 + E^(2*x)),x]

[Out]

-ArcTanh[E^x]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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 = 0.02 (sec) , antiderivative size = 6, 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 = 6, normalized size of antiderivative = 1.00

method	result	size
default	\(-\operatorname {arctanh}\left ({\mathrm e}^{x}\right )\)	\(6\)
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. \(2 (5) = 10\).

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 2.50 \[ \int \frac {e^x}{-1+e^{2 x}} \, dx=-\frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left (e^{x} - 1\right ) \]

[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. \(2 (5) = 10\).

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 2.50 \[ \int \frac {e^x}{-1+e^{2 x}} \, dx=\frac {\log {\left (e^{x} - 1 \right )}}{2} - \frac {\log {\left (e^{x} + 1 \right )}}{2} \]

[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. \(2 (5) = 10\).

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 2.50 \[ \int \frac {e^x}{-1+e^{2 x}} \, dx=-\frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left (e^{x} - 1\right ) \]

[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. \(2 (5) = 10\).

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 2.67 \[ \int \frac {e^x}{-1+e^{2 x}} \, dx=-\frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

[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 = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 2.50 \[ \int \frac {e^x}{-1+e^{2 x}} \, dx=\frac {\ln \left ({\mathrm {e}}^x-1\right )}{2}-\frac {\ln \left ({\mathrm {e}}^x+1\right )}{2} \]

[In]

int(exp(x)/(exp(2*x) - 1),x)

[Out]

log(exp(x) - 1)/2 - log(exp(x) + 1)/2