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

Optimal. Leaf size=17 \[ \frac{\log \left (e^{2 a+2 b x}+1\right )}{b} \]

[Out]

Log[1 + E^(2*a + 2*b*x)]/b

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Rubi [A]  time = 0.0173843, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2282, 12, 260} \[ \frac{\log \left (e^{2 a+2 b x}+1\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sech[a + b*x],x]

[Out]

Log[1 + E^(2*a + 2*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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int e^{a+b x} \text{sech}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{2 x}{1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\log \left (1+e^{2 a+2 b x}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0124252, size = 17, normalized size = 1. \[ \frac{\log \left (e^{2 a+2 b x}+1\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + E^(2*a + 2*b*x)]/b

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Maple [A]  time = 0.006, size = 19, normalized size = 1.1 \begin{align*} x+{\frac{\ln \left ( \cosh \left ( bx+a \right ) \right ) }{b}}+{\frac{a}{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+1/b*ln(cosh(b*x+a))+1/b*a

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Maxima [A]  time = 1.59275, size = 22, normalized size = 1.29 \begin{align*} \frac{\log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(2*b*x + 2*a) + 1)/b

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Fricas [A]  time = 1.78251, size = 76, normalized size = 4.47 \begin{align*} \frac{\log \left (\frac{2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(2*cosh(b*x + a)/(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}{\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),x)

[Out]

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

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Giac [A]  time = 1.16051, size = 22, normalized size = 1.29 \begin{align*} \frac{\log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(2*b*x + 2*a) + 1)/b