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]

ln(1+exp(2*b*x+2*a))/b

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, 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 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]

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}(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*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \[ \frac {\log \left (e^{2 a+2 b x}+1\right )}{b} \]

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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fricas [A]  time = 0.43, size = 30, normalized size = 1.76 \[ \frac {\log \left (\frac {2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )}{b} \]

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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giac [A]  time = 0.11, size = 16, normalized size = 0.94 \[ \frac {\log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{b} \]

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

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maple [A]  time = 0.03, size = 19, normalized size = 1.12 \[ x +\frac {\ln \left (\cosh \left (b x +a \right )\right )}{b}+\frac {a}{b} \]

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))+a/b

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maxima [A]  time = 0.42, size = 16, normalized size = 0.94 \[ \frac {\log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{b} \]

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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mupad [B]  time = 0.92, size = 16, normalized size = 0.94 \[ \frac {\ln \left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)/cosh(a + b*x),x)

[Out]

log(exp(2*a + 2*b*x) + 1)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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