3.271 \(\int e^{a+b x} \text {sech}^4(a+b x) \, dx\)

Optimal. Leaf size=95 \[ \frac {e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac {2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )^2}-\frac {8 e^{3 a+3 b x}}{3 b \left (e^{2 a+2 b x}+1\right )^3}+\frac {\tan ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2282, 12, 288, 199, 203} \[ \frac {e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac {2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )^2}-\frac {8 e^{3 a+3 b x}}{3 b \left (e^{2 a+2 b x}+1\right )^3}+\frac {\tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sech[a + b*x]^4,x]

[Out]

(-8*E^(3*a + 3*b*x))/(3*b*(1 + E^(2*a + 2*b*x))^3) - (2*E^(a + b*x))/(b*(1 + E^(2*a + 2*b*x))^2) + E^(a + b*x)
/(b*(1 + E^(2*a + 2*b*x))) + ArcTan[E^(a + b*x)]/b

Rule 12

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 64, normalized size = 0.67 \[ \frac {e^{a+b x} \left (-8 e^{2 (a+b x)}+3 e^{4 (a+b x)}-3\right )}{3 b \left (e^{2 (a+b x)}+1\right )^3}+\frac {\tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Sech[a + b*x]^4,x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.65, size = 513, normalized size = 5.40 \[ \frac {3 \, \cosh \left (b x + a\right )^{5} + 15 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 3 \, \sinh \left (b x + a\right )^{5} + 2 \, {\left (15 \, \cosh \left (b x + a\right )^{2} - 4\right )} \sinh \left (b x + a\right )^{3} - 8 \, \cosh \left (b x + a\right )^{3} + 6 \, {\left (5 \, \cosh \left (b x + a\right )^{3} - 4 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 3 \, {\left (\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{4} + 3 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 3 \, \cosh \left (b x + a\right )^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{5} + 2 \, \cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} - 8 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 3 \, \cosh \left (b x + a\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} + 3 \, b \cosh \left (b x + a\right )^{4} + 3 \, {\left (5 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{4} + 4 \, {\left (5 \, b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )^{2} + 3 \, {\left (5 \, b \cosh \left (b x + a\right )^{4} + 6 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 6 \, {\left (b \cosh \left (b x + a\right )^{5} + 2 \, b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^4,x, algorithm="fricas")

[Out]

1/3*(3*cosh(b*x + a)^5 + 15*cosh(b*x + a)*sinh(b*x + a)^4 + 3*sinh(b*x + a)^5 + 2*(15*cosh(b*x + a)^2 - 4)*sin
h(b*x + a)^3 - 8*cosh(b*x + a)^3 + 6*(5*cosh(b*x + a)^3 - 4*cosh(b*x + a))*sinh(b*x + a)^2 + 3*(cosh(b*x + a)^
6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^4 + 3*cosh(b*x
+ a)^4 + 4*(5*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 + 6*cosh(b*x + a)^2 +
1)*sinh(b*x + a)^2 + 3*cosh(b*x + a)^2 + 6*(cosh(b*x + a)^5 + 2*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a
) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + 3*(5*cosh(b*x + a)^4 - 8*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - 3
*cosh(b*x + a))/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*sinh(b*x + a)^5 + b*sinh(b*x + a)^6 + 3*b*cosh(b*x + a)
^4 + 3*(5*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^4 + 4*(5*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^3
+ 3*b*cosh(b*x + a)^2 + 3*(5*b*cosh(b*x + a)^4 + 6*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 6*(b*cosh(b*x + a
)^5 + 2*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a) + b)

________________________________________________________________________________________

giac [A]  time = 0.12, size = 60, normalized size = 0.63 \[ \frac {\frac {3 \, e^{\left (5 \, b x + 5 \, a\right )} - 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{3}} + 3 \, \arctan \left (e^{\left (b x + a\right )}\right )}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^4,x, algorithm="giac")

[Out]

1/3*((3*e^(5*b*x + 5*a) - 8*e^(3*b*x + 3*a) - 3*e^(b*x + a))/(e^(2*b*x + 2*a) + 1)^3 + 3*arctan(e^(b*x + a)))/
b

________________________________________________________________________________________

maple [A]  time = 0.22, size = 43, normalized size = 0.45 \[ -\frac {1}{3 b \cosh \left (b x +a \right )^{3}}+\frac {\mathrm {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2 b}+\frac {\arctan \left ({\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*sech(b*x+a)^4,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 83, normalized size = 0.87 \[ \frac {\arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac {3 \, e^{\left (5 \, b x + 5 \, a\right )} - 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a\right )} + 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^4,x, algorithm="maxima")

[Out]

arctan(e^(b*x + a))/b + 1/3*(3*e^(5*b*x + 5*a) - 8*e^(3*b*x + 3*a) - 3*e^(b*x + a))/(b*(e^(6*b*x + 6*a) + 3*e^
(4*b*x + 4*a) + 3*e^(2*b*x + 2*a) + 1))

________________________________________________________________________________________

mupad [B]  time = 0.96, size = 130, normalized size = 1.37 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}-\frac {8\,{\mathrm {e}}^{3\,a+3\,b\,x}}{3\,b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1\right )}+\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)/cosh(a + b*x)^4,x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a} \int e^{b x} \operatorname {sech}^{4}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)**4,x)

[Out]

exp(a)*Integral(exp(b*x)*sech(a + b*x)**4, x)

________________________________________________________________________________________