3.270 \(\int e^{a+b x} \text{sech}^3(a+b x) \, dx\)

Optimal. Leaf size=29 \[ \frac{2 e^{4 a+4 b x}}{b \left (e^{2 a+2 b x}+1\right )^2} \]

[Out]

(2*E^(4*a + 4*b*x))/(b*(1 + E^(2*a + 2*b*x))^2)

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Rubi [A]  time = 0.0265666, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2282, 12, 264} \[ \frac{2 e^{4 a+4 b x}}{b \left (e^{2 a+2 b x}+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sech[a + b*x]^3,x]

[Out]

(2*E^(4*a + 4*b*x))/(b*(1 + E^(2*a + 2*b*x))^2)

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 264

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

Rubi steps

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

Mathematica [A]  time = 0.0177166, size = 29, normalized size = 1. \[ \frac{2 e^{4 a+4 b x}}{b \left (e^{2 a+2 b x}+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Sech[a + b*x]^3,x]

[Out]

(2*E^(4*a + 4*b*x))/(b*(1 + E^(2*a + 2*b*x))^2)

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Maple [A]  time = 0.037, size = 30, normalized size = 1. \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{2\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+\tanh \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*sech(b*x+a)^3,x)

[Out]

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

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Maxima [B]  time = 1.06647, size = 92, normalized size = 3.17 \begin{align*} -\frac{4 \, e^{\left (2 \, b x + 2 \, a\right )}}{b{\left (e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} - \frac{2}{b{\left (e^{\left (4 \, b x + 4 \, a\right )} + 2 \, 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)^3,x, algorithm="maxima")

[Out]

-4*e^(2*b*x + 2*a)/(b*(e^(4*b*x + 4*a) + 2*e^(2*b*x + 2*a) + 1)) - 2/(b*(e^(4*b*x + 4*a) + 2*e^(2*b*x + 2*a) +
1))

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Fricas [B]  time = 1.65953, size = 238, normalized size = 8.21 \begin{align*} -\frac{2 \,{\left (3 \, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b \sinh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right ) +{\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^3,x, algorithm="fricas")

[Out]

-2*(3*cosh(b*x + a) + sinh(b*x + a))/(b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a)*sinh(b*x + a)^2 + b*sinh(b*x + a)^
3 + 3*b*cosh(b*x + a) + (3*b*cosh(b*x + a)^2 + b)*sinh(b*x + a))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \operatorname{sech}^{3}{\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)**3,x)

[Out]

exp(a)*Integral(exp(b*x)*sech(a + b*x)**3, x)

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Giac [A]  time = 1.28117, size = 42, normalized size = 1.45 \begin{align*} -\frac{2 \,{\left (2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{b{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^3,x, algorithm="giac")

[Out]

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