3.163 $$\int \cosh (a+b x) \text{sech}^3(c+b x) \, dx$$

Optimal. Leaf size=38 $\frac{\cosh (a-c) \tanh (b x+c)}{b}-\frac{\sinh (a-c) \text{sech}^2(b x+c)}{2 b}$

[Out]

-(Sech[c + b*x]^2*Sinh[a - c])/(2*b) + (Cosh[a - c]*Tanh[c + b*x])/b

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Rubi [A]  time = 0.0418814, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {5627, 2606, 30, 3767, 8} $\frac{\cosh (a-c) \tanh (b x+c)}{b}-\frac{\sinh (a-c) \text{sech}^2(b x+c)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]*Sech[c + b*x]^3,x]

[Out]

-(Sech[c + b*x]^2*Sinh[a - c])/(2*b) + (Cosh[a - c]*Tanh[c + b*x])/b

Rule 5627

Int[Cosh[v_]*Sech[w_]^(n_.), x_Symbol] :> Dist[Sinh[v - w], Int[Tanh[w]*Sech[w]^(n - 1), x], x] + Dist[Cosh[v
- w], Int[Sech[w]^(n - 1), x], x] /; GtQ[n, 0] && NeQ[w, v] && FreeQ[v - w, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \cosh (a+b x) \text{sech}^3(c+b x) \, dx &=\cosh (a-c) \int \text{sech}^2(c+b x) \, dx+\sinh (a-c) \int \text{sech}^2(c+b x) \tanh (c+b x) \, dx\\ &=\frac{(i \cosh (a-c)) \operatorname{Subst}(\int 1 \, dx,x,-i \tanh (c+b x))}{b}-\frac{\sinh (a-c) \operatorname{Subst}(\int x \, dx,x,\text{sech}(c+b x))}{b}\\ &=-\frac{\text{sech}^2(c+b x) \sinh (a-c)}{2 b}+\frac{\cosh (a-c) \tanh (c+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.167805, size = 35, normalized size = 0.92 $-\frac{\text{sech}(c) \text{sech}^2(b x+c) (\sinh (a)-\cosh (a-c) \sinh (2 b x+c))}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[a + b*x]*Sech[c + b*x]^3,x]

[Out]

-(Sech[c]*Sech[c + b*x]^2*(Sinh[a] - Cosh[a - c]*Sinh[c + 2*b*x]))/(2*b)

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Maple [A]  time = 0.03, size = 56, normalized size = 1.5 \begin{align*} -{\frac{ \left ( 2\,{{\rm e}^{2\,bx+2\,a+2\,c}}+{{\rm e}^{2\,a}}+{{\rm e}^{2\,c}} \right ){{\rm e}^{3\,a-c}}}{b \left ({{\rm e}^{2\,bx+2\,a+2\,c}}+{{\rm e}^{2\,a}} \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*exp(2*b*x+2*a+2*c)+exp(2*a)+exp(2*c))/b/(exp(2*b*x+2*a+2*c)+exp(2*a))^2*exp(3*a-c)

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Maxima [B]  time = 1.101, size = 161, normalized size = 4.24 \begin{align*} \frac{2 \, e^{\left (-2 \, b x + 3 \, c\right )}}{b{\left (2 \, e^{\left (-2 \, b x + a + 2 \, c\right )} + e^{\left (-4 \, b x + a\right )} + e^{\left (a + 4 \, c\right )}\right )}} + \frac{e^{\left (2 \, a + 3 \, c\right )}}{b{\left (2 \, e^{\left (-2 \, b x + a + 2 \, c\right )} + e^{\left (-4 \, b x + a\right )} + e^{\left (a + 4 \, c\right )}\right )}} + \frac{e^{\left (5 \, c\right )}}{b{\left (2 \, e^{\left (-2 \, b x + a + 2 \, c\right )} + e^{\left (-4 \, b x + a\right )} + e^{\left (a + 4 \, c\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sech(b*x+c)^3,x, algorithm="maxima")

[Out]

2*e^(-2*b*x + 3*c)/(b*(2*e^(-2*b*x + a + 2*c) + e^(-4*b*x + a) + e^(a + 4*c))) + e^(2*a + 3*c)/(b*(2*e^(-2*b*x
+ a + 2*c) + e^(-4*b*x + a) + e^(a + 4*c))) + e^(5*c)/(b*(2*e^(-2*b*x + a + 2*c) + e^(-4*b*x + a) + e^(a + 4*
c)))

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Fricas [B]  time = 1.82322, size = 644, normalized size = 16.95 \begin{align*} -\frac{2 \,{\left (2 \, \cosh \left (b x + c\right ) \cosh \left (-a + c\right ) - \cosh \left (b x + c\right ) \sinh \left (-a + c\right ) - \sinh \left (b x + c\right ) \sinh \left (-a + c\right )\right )}}{b \cosh \left (b x + c\right )^{3} \cosh \left (-a + c\right )^{2} + 3 \, b \cosh \left (b x + c\right ) \cosh \left (-a + c\right )^{2} +{\left (b \cosh \left (-a + c\right )^{2} - b \sinh \left (-a + c\right )^{2}\right )} \sinh \left (b x + c\right )^{3} + 3 \,{\left (b \cosh \left (b x + c\right ) \cosh \left (-a + c\right )^{2} - b \cosh \left (b x + c\right ) \sinh \left (-a + c\right )^{2}\right )} \sinh \left (b x + c\right )^{2} -{\left (b \cosh \left (b x + c\right )^{3} + 3 \, b \cosh \left (b x + c\right )\right )} \sinh \left (-a + c\right )^{2} +{\left (3 \, b \cosh \left (b x + c\right )^{2} \cosh \left (-a + c\right )^{2} + b \cosh \left (-a + c\right )^{2} -{\left (3 \, b \cosh \left (b x + c\right )^{2} + b\right )} \sinh \left (-a + c\right )^{2}\right )} \sinh \left (b x + c\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh{\left (a + b x \right )} \operatorname{sech}^{3}{\left (b x + c \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sech(b*x+c)**3,x)

[Out]

Integral(cosh(a + b*x)*sech(b*x + c)**3, x)

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Giac [A]  time = 1.18868, size = 66, normalized size = 1.74 \begin{align*} -\frac{{\left (2 \, e^{\left (2 \, b x + 2 \, a + 2 \, c\right )} + e^{\left (2 \, a\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (-a - c\right )}}{b{\left (e^{\left (2 \, b x + 2 \, c\right )} + 1\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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