\(\int e^{-2 x} \text {sech}^4(x) \, dx\) [601]

Optimal result

Integrand size = 10, antiderivative size = 13 \[ \int e^{-2 x} \text {sech}^4(x) \, dx=-\frac {8}{3 \left (1+e^{2 x}\right )^3} \]

[Out]

-8/3/(1+exp(2*x))^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2320, 12, 267} \[ \int e^{-2 x} \text {sech}^4(x) \, dx=-\frac {8}{3 \left (e^{2 x}+1\right )^3} \]

[In]

Int[Sech[x]^4/E^(2*x),x]

[Out]

-8/(3*(1 + E^(2*x))^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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*} \text {integral}& = \text {Subst}\left (\int \frac {16 x}{\left (1+x^2\right )^4} \, dx,x,e^x\right ) \\ & = 16 \text {Subst}\left (\int \frac {x}{\left (1+x^2\right )^4} \, dx,x,e^x\right ) \\ & = -\frac {8}{3 \left (1+e^{2 x}\right )^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int e^{-2 x} \text {sech}^4(x) \, dx=-\frac {8}{3 \left (1+e^{2 x}\right )^3} \]

[In]

Integrate[Sech[x]^4/E^(2*x),x]

[Out]

-8/(3*(1 + E^(2*x))^3)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {\left (-1+\tanh \left (x \right )\right )^{3}}{3}\)	\(9\)
risch	\(-\frac {8}{3 \left (1+{\mathrm e}^{2 x}\right )^{3}}\)	\(11\)
parallelrisch	\(\frac {\operatorname {sech}\left (x \right )^{2} \left (-1+\tanh \left (x \right )\right ) {\mathrm e}^{-2 x}}{3}\)	\(15\)

[In]

int(1/exp(2*x)/cosh(x)^4,x,method=_RETURNVERBOSE)

[Out]

1/3*(-1+tanh(x))^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (10) = 20\).

Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 7.85 \[ \int e^{-2 x} \text {sech}^4(x) \, dx=-\frac {8}{3 \, {\left (\cosh \left (x\right )^{6} + 6 \, \cosh \left (x\right ) \sinh \left (x\right )^{5} + \sinh \left (x\right )^{6} + 3 \, {\left (5 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{4} + 3 \, \cosh \left (x\right )^{4} + 4 \, {\left (5 \, \cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 3 \, {\left (5 \, \cosh \left (x\right )^{4} + 6 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 3 \, \cosh \left (x\right )^{2} + 6 \, {\left (\cosh \left (x\right )^{5} + 2 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}} \]

[In]

integrate(1/exp(2*x)/cosh(x)^4,x, algorithm="fricas")

[Out]

-8/3/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 3*(5*cosh(x)^2 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(5*cosh(x)
^3 + 3*cosh(x))*sinh(x)^3 + 3*(5*cosh(x)^4 + 6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 + 2*cosh(
x)^3 + cosh(x))*sinh(x) + 1)

Sympy [F]

\[ \int e^{-2 x} \text {sech}^4(x) \, dx=\int \frac {e^{- 2 x}}{\cosh ^{4}{\left (x \right )}}\, dx \]

[In]

integrate(1/exp(2*x)/cosh(x)**4,x)

[Out]

Integral(exp(-2*x)/cosh(x)**4, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (10) = 20\).

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 5.77 \[ \int e^{-2 x} \text {sech}^4(x) \, dx=\frac {8 \, e^{\left (-2 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1} + \frac {8 \, e^{\left (-4 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1} + \frac {8}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} \]

[In]

integrate(1/exp(2*x)/cosh(x)^4,x, algorithm="maxima")

[Out]

8*e^(-2*x)/(3*e^(-2*x) + 3*e^(-4*x) + e^(-6*x) + 1) + 8*e^(-4*x)/(3*e^(-2*x) + 3*e^(-4*x) + e^(-6*x) + 1) + 8/
3/(3*e^(-2*x) + 3*e^(-4*x) + e^(-6*x) + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int e^{-2 x} \text {sech}^4(x) \, dx=-\frac {8}{3 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]

[In]

integrate(1/exp(2*x)/cosh(x)^4,x, algorithm="giac")

[Out]

-8/3/(e^(2*x) + 1)^3

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int e^{-2 x} \text {sech}^4(x) \, dx=-\frac {{\mathrm {e}}^{-3\,x}}{3\,{\left (\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}\right )}^3} \]

[In]

int(exp(-2*x)/cosh(x)^4,x)

[Out]

-exp(-3*x)/(3*(exp(-x)/2 + exp(x)/2)^3)