3.3.91 \(\int \frac {1}{-e^{-x}+e^x} \, dx\) [291]

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 = {2320, 213} \begin {gather*} -\tanh ^{-1}\left (e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^(-x) + E^x)^(-1),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 2320

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 &=\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 = 0.02, size = 6, normalized size = 1.00 \begin {gather*} -\tanh ^{-1}\left (e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^(-x) + E^x)^(-1),x]

[Out]

-ArcTanh[E^x]

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

method	result	size
derivativedivides	\(-\arctanh \left ({\mathrm e}^{x}\right )\)	\(6\)
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(1/(-1/exp(x)+exp(x)),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. 19 vs. \(2 (5) = 10\).
time = 4.16, size = 19, normalized size = 3.17 \begin {gather*} -\frac {1}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac {1}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (5) = 10\).
time = 0.52, 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(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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (5) = 10\).
time = 0.04, 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(1/(-1/exp(x)+exp(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. 16 vs. \(2 (5) = 10\).
time = 1.24, size = 16, normalized size = 2.67 \begin {gather*} -\frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end {gather*}

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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Mupad [B]
time = 0.10, 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(-1/(exp(-x) - exp(x)),x)

[Out]

log(exp(x) - 1)/2 - log(exp(x) + 1)/2

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