3.291 \(\int \frac {1}{-e^{-x}+e^x} \, dx\)

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 = {2282, 207} \[ -\tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-E^(-x) + E^x)^(-1),x]

[Out]

-ArcTanh[E^x]

Rule 207

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {align*} \int \frac {1}{-e^{-x}+e^x} \, dx &=\operatorname {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 \[ -\tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-E^(-x) + E^x)^(-1),x]

[Out]

-ArcTanh[E^x]

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fricas [B]  time = 0.42, size = 15, normalized size = 2.50 \[ -\frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left (e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(e^x + 1) + 1/2*log(e^x - 1)

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giac [B]  time = 0.79, size = 16, normalized size = 2.67 \[ -\frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 6, normalized size = 1.00 \[ -\arctanh \left ({\mathrm e}^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-1/exp(x)+exp(x)),x)

[Out]

-arctanh(exp(x))

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maxima [B]  time = 0.58, size = 19, normalized size = 3.17 \[ -\frac {1}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac {1}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.10, size = 15, normalized size = 2.50 \[ \frac {\ln \left ({\mathrm {e}}^x-1\right )}{2}-\frac {\ln \left ({\mathrm {e}}^x+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(exp(-x) - exp(x)),x)

[Out]

log(exp(x) - 1)/2 - log(exp(x) + 1)/2

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sympy [B]  time = 0.11, size = 15, normalized size = 2.50 \[ \frac {\log {\left (e^{x} - 1 \right )}}{2} - \frac {\log {\left (e^{x} + 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(exp(x) - 1)/2 - log(exp(x) + 1)/2

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