3.269 \(\int e^{a+b x} \text{sech}^2(a+b x) \, dx\)

Optimal. Leaf size=40 \[ \frac{2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )} \]

[Out]

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

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Rubi [A]  time = 0.0293319, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2282, 12, 288, 203} \[ \frac{2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sech[a + b*x]^2,x]

[Out]

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

Mathematica [A]  time = 0.0610293, size = 36, normalized size = 0.9 \[ \frac{2 \left (\tan ^{-1}\left (e^{a+b x}\right )-\frac{e^{a+b x}}{e^{2 (a+b x)}+1}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Sech[a + b*x]^2,x]

[Out]

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

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Maple [A]  time = 0.011, size = 45, normalized size = 1.1 \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{b\cosh \left ( bx+a \right ) }}-{\frac{\cosh \left ( bx+a \right ) }{b}}+2\,{\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)^2,x)

[Out]

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

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Maxima [A]  time = 1.65203, size = 50, normalized size = 1.25 \begin{align*} \frac{2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} - \frac{2 \, e^{\left (b x + a\right )}}{b{\left (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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.73908, size = 304, normalized size = 7.6 \begin{align*} \frac{2 \,{\left ({\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - \cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^2,x, algorithm="fricas")

[Out]

2*((cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*arctan(cosh(b*x + a) + sinh(b*x + a
)) - cosh(b*x + a) - sinh(b*x + a))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2 +
b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \operatorname{sech}^{2}{\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)**2,x)

[Out]

exp(a)*Integral(exp(b*x)*sech(a + b*x)**2, x)

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Giac [A]  time = 1.30845, size = 50, normalized size = 1.25 \begin{align*} \frac{2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} - \frac{2 \, e^{\left (b x + a\right )}}{b{\left (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)^2,x, algorithm="giac")

[Out]

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