3.8.21 \(\int \frac {1}{(-e^{-x}+e^x)^2} \, dx\) [721]

Optimal. Leaf size=15 \[ \frac {1}{2 \left (1-e^{2 x}\right )} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 15, 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, 267} \begin {gather*} \frac {1}{2 \left (1-e^{2 x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^(-x) + E^x)^(-2),x]

[Out]

1/(2*(1 - E^(2*x)))

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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

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Mathematica [A]
time = 0.03, size = 11, normalized size = 0.73 \begin {gather*} \frac {1}{2-2 e^{2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^(-x) + E^x)^(-2),x]

[Out]

(2 - 2*E^(2*x))^(-1)

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Maple [A]
time = 0.01, size = 11, normalized size = 0.73

method	result	size
derivativedivides	\(-\frac {1}{2 \left (-1+{\mathrm e}^{2 x}\right )}\)	\(11\)
default	\(-\frac {1}{2 \left (-1+{\mathrm e}^{2 x}\right )}\)	\(11\)
norman	\(-\frac {1}{2 \left (-1+{\mathrm e}^{2 x}\right )}\)	\(11\)
risch	\(-\frac {1}{2 \left (-1+{\mathrm e}^{2 x}\right )}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(exp(x)^2-1)

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Maxima [A]
time = 0.28, size = 10, normalized size = 0.67 \begin {gather*} \frac {1}{2 \, {\left (e^{\left (-2 \, x\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(e^(-2*x) - 1)

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Fricas [A]
time = 0.38, size = 10, normalized size = 0.67 \begin {gather*} -\frac {1}{2 \, {\left (e^{\left (2 \, x\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(e^(2*x) - 1)

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Sympy [A]
time = 0.02, size = 10, normalized size = 0.67 \begin {gather*} - \frac {1}{2 e^{2 x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*exp(2*x) - 2)

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Giac [A]
time = 3.52, size = 10, normalized size = 0.67 \begin {gather*} -\frac {1}{2 \, {\left (e^{\left (2 \, x\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1/exp(x)+exp(x))^2,x, algorithm="giac")

[Out]

-1/2/(e^(2*x) - 1)

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Mupad [B]
time = 3.40, size = 12, normalized size = 0.80 \begin {gather*} -\frac {1}{2\,\left ({\mathrm {e}}^{2\,x}-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(exp(-x) - exp(x))^2,x)

[Out]

-1/(2*(exp(2*x) - 1))

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