3.272 \(\int e^{a+b x} \text {sech}^5(a+b x) \, dx\)

Optimal. Leaf size=60 \[ -\frac {8}{b \left (e^{2 a+2 b x}+1\right )^2}+\frac {32}{3 b \left (e^{2 a+2 b x}+1\right )^3}-\frac {4}{b \left (e^{2 a+2 b x}+1\right )^4} \]

[Out]

-4/b/(1+exp(2*b*x+2*a))^4+32/3/b/(1+exp(2*b*x+2*a))^3-8/b/(1+exp(2*b*x+2*a))^2

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Rubi [A]  time = 0.05, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2282, 12, 266, 43} \[ -\frac {8}{b \left (e^{2 a+2 b x}+1\right )^2}+\frac {32}{3 b \left (e^{2 a+2 b x}+1\right )^3}-\frac {4}{b \left (e^{2 a+2 b x}+1\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sech[a + b*x]^5,x]

[Out]

-4/(b*(1 + E^(2*a + 2*b*x))^4) + 32/(3*b*(1 + E^(2*a + 2*b*x))^3) - 8/(b*(1 + E^(2*a + 2*b*x))^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

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^{a+b x} \text {sech}^5(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {32 x^5}{\left (1+x^2\right )^5} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {32 \operatorname {Subst}\left (\int \frac {x^5}{\left (1+x^2\right )^5} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {16 \operatorname {Subst}\left (\int \frac {x^2}{(1+x)^5} \, dx,x,e^{2 a+2 b x}\right )}{b}\\ &=\frac {16 \operatorname {Subst}\left (\int \left (\frac {1}{(1+x)^5}-\frac {2}{(1+x)^4}+\frac {1}{(1+x)^3}\right ) \, dx,x,e^{2 a+2 b x}\right )}{b}\\ &=-\frac {4}{b \left (1+e^{2 a+2 b x}\right )^4}+\frac {32}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac {8}{b \left (1+e^{2 a+2 b x}\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 44, normalized size = 0.73 \[ -\frac {4 \left (4 e^{2 (a+b x)}+6 e^{4 (a+b x)}+1\right )}{3 b \left (e^{2 (a+b x)}+1\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Sech[a + b*x]^5,x]

[Out]

(-4*(1 + 4*E^(2*(a + b*x)) + 6*E^(4*(a + b*x))))/(3*b*(1 + E^(2*(a + b*x)))^4)

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fricas [B]  time = 0.48, size = 233, normalized size = 3.88 \[ -\frac {4 \, {\left (7 \, \cosh \left (b x + a\right )^{2} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 7 \, \sinh \left (b x + a\right )^{2} + 4\right )}}{3 \, {\left (b \cosh \left (b x + a\right )^{6} + 6 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + b \sinh \left (b x + a\right )^{6} + 4 \, b \cosh \left (b x + a\right )^{4} + {\left (15 \, b \cosh \left (b x + a\right )^{2} + 4 \, b\right )} \sinh \left (b x + a\right )^{4} + 4 \, {\left (5 \, b \cosh \left (b x + a\right )^{3} + 4 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 7 \, b \cosh \left (b x + a\right )^{2} + {\left (15 \, b \cosh \left (b x + a\right )^{4} + 24 \, b \cosh \left (b x + a\right )^{2} + 7 \, b\right )} \sinh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{5} + 8 \, b \cosh \left (b x + a\right )^{3} + 5 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 4 \, b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^5,x, algorithm="fricas")

[Out]

-4/3*(7*cosh(b*x + a)^2 + 10*cosh(b*x + a)*sinh(b*x + a) + 7*sinh(b*x + a)^2 + 4)/(b*cosh(b*x + a)^6 + 6*b*cos
h(b*x + a)*sinh(b*x + a)^5 + b*sinh(b*x + a)^6 + 4*b*cosh(b*x + a)^4 + (15*b*cosh(b*x + a)^2 + 4*b)*sinh(b*x +
a)^4 + 4*(5*b*cosh(b*x + a)^3 + 4*b*cosh(b*x + a))*sinh(b*x + a)^3 + 7*b*cosh(b*x + a)^2 + (15*b*cosh(b*x + a
)^4 + 24*b*cosh(b*x + a)^2 + 7*b)*sinh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^5 + 8*b*cosh(b*x + a)^3 + 5*b*cosh(b*
x + a))*sinh(b*x + a) + 4*b)

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giac [A]  time = 0.12, size = 42, normalized size = 0.70 \[ -\frac {4 \, {\left (6 \, e^{\left (4 \, b x + 4 \, a\right )} + 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{3 \, b {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.22, size = 35, normalized size = 0.58 \[ \frac {-\frac {1}{4 \cosh \left (b x +a \right )^{4}}+\left (\frac {2}{3}+\frac {\mathrm {sech}\left (b x +a \right )^{2}}{3}\right ) \tanh \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*sech(b*x+a)^5,x)

[Out]

1/b*(-1/4/cosh(b*x+a)^4+(2/3+1/3*sech(b*x+a)^2)*tanh(b*x+a))

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maxima [B]  time = 0.33, size = 172, normalized size = 2.87 \[ -\frac {8 \, e^{\left (4 \, b x + 4 \, a\right )}}{b {\left (e^{\left (8 \, b x + 8 \, a\right )} + 4 \, e^{\left (6 \, b x + 6 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} + 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} - \frac {16 \, e^{\left (2 \, b x + 2 \, a\right )}}{3 \, b {\left (e^{\left (8 \, b x + 8 \, a\right )} + 4 \, e^{\left (6 \, b x + 6 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} + 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} - \frac {4}{3 \, b {\left (e^{\left (8 \, b x + 8 \, a\right )} + 4 \, e^{\left (6 \, b x + 6 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} + 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^5,x, algorithm="maxima")

[Out]

-8*e^(4*b*x + 4*a)/(b*(e^(8*b*x + 8*a) + 4*e^(6*b*x + 6*a) + 6*e^(4*b*x + 4*a) + 4*e^(2*b*x + 2*a) + 1)) - 16/
3*e^(2*b*x + 2*a)/(b*(e^(8*b*x + 8*a) + 4*e^(6*b*x + 6*a) + 6*e^(4*b*x + 4*a) + 4*e^(2*b*x + 2*a) + 1)) - 4/3/
(b*(e^(8*b*x + 8*a) + 4*e^(6*b*x + 6*a) + 6*e^(4*b*x + 4*a) + 4*e^(2*b*x + 2*a) + 1))

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mupad [B]  time = 0.95, size = 42, normalized size = 0.70 \[ -\frac {4\,\left (4\,{\mathrm {e}}^{2\,a+2\,b\,x}+6\,{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}{3\,b\,{\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)/cosh(a + b*x)^5,x)

[Out]

-(4*(4*exp(2*a + 2*b*x) + 6*exp(4*a + 4*b*x) + 1))/(3*b*(exp(2*a + 2*b*x) + 1)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a} \int e^{b x} \operatorname {sech}^{5}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)**5,x)

[Out]

exp(a)*Integral(exp(b*x)*sech(a + b*x)**5, x)

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