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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.00, 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 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]

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}^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 = 0.02, size = 29, normalized size = 1.00 \[ \frac {2 e^{4 a+4 b x}}{b \left (e^{2 a+2 b x}+1\right )^2} \]

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)

________________________________________________________________________________________

fricas [B]  time = 0.68, size = 86, normalized size = 2.97 \[ -\frac {2 \, {\left (3 \, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b \sinh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right ) + {\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )} \]

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))

________________________________________________________________________________________

giac [A]  time = 0.12, size = 31, normalized size = 1.07 \[ -\frac {2 \, {\left (2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{b {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.21, size = 22, normalized size = 0.76 \[ \frac {-\frac {1}{2 \cosh \left (b x +a \right )^{2}}+\tanh \left (b x +a \right )}{b} \]

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

________________________________________________________________________________________

maxima [B]  time = 0.31, size = 68, normalized size = 2.34 \[ -\frac {4 \, e^{\left (2 \, b x + 2 \, a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} - \frac {2}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]

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))

________________________________________________________________________________________

mupad [B]  time = 0.93, size = 31, normalized size = 1.07 \[ -\frac {2\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}{b\,{\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)/cosh(a + b*x)^3,x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a} \int e^{b x} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________