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

Optimal. Leaf size=13 \[ -\frac {8}{3 \left (e^{2 x}+1\right )^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2282, 12, 261} \[ -\frac {8}{3 \left (e^{2 x}+1\right )^3} \]

Antiderivative was successfully verified.

[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 261

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 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 e^{-2 x} \text {sech}^4(x) \, dx &=\operatorname {Subst}\left (\int \frac {16 x}{\left (1+x^2\right )^4} \, dx,x,e^x\right )\\ &=16 \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 13, normalized size = 1.00 \[ -\frac {8}{3 \left (e^{2 x}+1\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-8/(3*(1 + E^(2*x))^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{-2 x} \text {sech}^4(x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [B]  time = 0.77, size = 102, normalized size = 7.85 \[ -\frac {8}{3 \, {\left (\cosh \relax (x)^{6} + 6 \, \cosh \relax (x) \sinh \relax (x)^{5} + \sinh \relax (x)^{6} + 3 \, {\left (5 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{4} + 3 \, \cosh \relax (x)^{4} + 4 \, {\left (5 \, \cosh \relax (x)^{3} + 3 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 3 \, {\left (5 \, \cosh \relax (x)^{4} + 6 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{2} + 3 \, \cosh \relax (x)^{2} + 6 \, {\left (\cosh \relax (x)^{5} + 2 \, \cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.59, size = 10, normalized size = 0.77 \[ -\frac {8}{3 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/3/(e^(2*x) + 1)^3

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maple [A]  time = 0.35, size = 11, normalized size = 0.85

method	result	size
risch	\(-\frac {8}{3 \left (1+{\mathrm e}^{2 x}\right )^{3}}\)	\(11\)
default	\(2 \tanh \relax (x )+\frac {1}{\cosh \relax (x )^{2}}-\left (\frac {2}{3}+\frac {\mathrm {sech}\relax (x )^{2}}{3}\right ) \tanh \relax (x )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.46, size = 75, normalized size = 5.77 \[ \frac {8 \, e^{\left (-2 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1} + \frac {8 \, e^{\left (-4 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1} + \frac {8}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.31, size = 19, normalized size = 1.46 \[ -\frac {{\mathrm {e}}^{-3\,x}}{3\,{\left (\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{- 2 x}}{\cosh ^{4}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(-2*x)/cosh(x)**4, x)

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