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*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

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Rubi [A]  time = 0.0467283, antiderivative size = 95, normalized size of antiderivative = 1., 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 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]]]

Rule 12

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

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 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])

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 = 0.0828558, 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

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Maple [A]  time = 0.04, size = 83, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\,b \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}}+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\,b\cosh \left ( bx+a \right ) }}-{\frac{\cosh \left ( bx+a \right ) }{3\,b}}+{\frac{{\rm sech} \left (bx+a\right )\tanh \left ( bx+a \right ) }{2\,b}}+{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}} \end{align*}

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

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Maxima [A]  time = 1.56712, size = 112, normalized size = 1.18 \begin{align*} \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 )}} \end{align*}

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))

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Fricas [B]  time = 1.8413, size = 1424, normalized size = 14.99 \begin{align*} \text{result too large to display} \end{align*}

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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \operatorname{sech}^{4}{\left (a + b x \right )}\, dx \end{align*}

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)

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Giac [A]  time = 1.25156, size = 82, normalized size = 0.86 \begin{align*} \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 (2 \, b x + 2 \, a\right )} + 1\right )}^{3}} \end{align*}

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]

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^(2*b*x + 2*a) + 1)^3
)