3.1.19 \(\int \frac {e^x}{-1+e^{2 x}} \, dx\) [19]

Optimal. Leaf size=6 \[ -\tanh ^{-1}\left (e^x\right ) \]

[Out]

-arctanh(exp(x))

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Rubi [A]
time = 0.01, antiderivative size = 6, normalized size of antiderivative = 1.00, 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 = {2281, 213} \begin {gather*} -\tanh ^{-1}\left (e^x\right ) \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.00, size = 6, normalized size = 1.00 \begin {gather*} -\tanh ^{-1}\left (e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^x/(-1 + E^(2*x)),x]

[Out]

-ArcTanh[E^x]

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Mathics [B] Leaf count is larger than twice the leaf count of optimal. \(21\) vs. \(2(6)=12\).
time = 1.84, size = 17, normalized size = 2.83 \begin {gather*} -\frac {\text {Log}\left [1+E^x\right ]}{2}+\frac {\text {Log}\left [-1+E^x\right ]}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[E^x/(-1 + E^(2*x)),x]')

[Out]

-Log[1 + E ^ x] / 2 + Log[-1 + E ^ x] / 2

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Maple [A]
time = 0.01, size = 6, normalized size = 1.00

method	result	size
default	\(-\arctanh \left ({\mathrm e}^{x}\right )\)	\(6\)
norman	\(\frac {\ln \left (-1+{\mathrm e}^{x}\right )}{2}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{2}\)	\(16\)
risch	\(\frac {\ln \left (-1+{\mathrm e}^{x}\right )}{2}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{2}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctanh(exp(x))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (5) = 10\).
time = 0.26, size = 15, normalized size = 2.50 \begin {gather*} -\frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left (e^{x} - 1\right ) \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (5) = 10\).
time = 0.33, size = 15, normalized size = 2.50 \begin {gather*} -\frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left (e^{x} - 1\right ) \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (5) = 10\)
time = 0.05, size = 15, normalized size = 2.50 \begin {gather*} \frac {\log {\left (e^{x} - 1 \right )}}{2} - \frac {\log {\left (e^{x} + 1 \right )}}{2} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 18 vs. \(2 (5) = 10\).
time = 0.00, size = 19, normalized size = 3.17 \begin {gather*} \frac {\ln \left |\mathrm {e}^{x}-1\right |}{2}-\frac {\ln \left (\mathrm {e}^{x}+1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.00, size = 15, normalized size = 2.50 \begin {gather*} \frac {\ln \left ({\mathrm {e}}^x-1\right )}{2}-\frac {\ln \left ({\mathrm {e}}^x+1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/(exp(2*x) - 1),x)

[Out]

log(exp(x) - 1)/2 - log(exp(x) + 1)/2

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