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]

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

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Rubi [A]  time = 0.04, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 8

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

Rule 30

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

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 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 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]

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*}

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Mathematica [A]  time = 0.18, 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]

-1/2*(Sech[c]*Sech[c + b*x]^2*(Sinh[a] - Cosh[a - c]*Sinh[c + 2*b*x]))/b

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fricas [B]  time = 0.39, size = 248, normalized size = 6.53 \[ -\frac {2 \, {\left (2 \, \cosh \left (b x + c\right ) \cosh \left (-a + c\right ) - \cosh \left (b x + c\right ) \sinh \left (-a + c\right ) - \sinh \left (b x + c\right ) \sinh \left (-a + c\right )\right )}}{b \cosh \left (b x + c\right )^{3} \cosh \left (-a + c\right )^{2} + 3 \, b \cosh \left (b x + c\right ) \cosh \left (-a + c\right )^{2} + {\left (b \cosh \left (-a + c\right )^{2} - b \sinh \left (-a + c\right )^{2}\right )} \sinh \left (b x + c\right )^{3} + 3 \, {\left (b \cosh \left (b x + c\right ) \cosh \left (-a + c\right )^{2} - b \cosh \left (b x + c\right ) \sinh \left (-a + c\right )^{2}\right )} \sinh \left (b x + c\right )^{2} - {\left (b \cosh \left (b x + c\right )^{3} + 3 \, b \cosh \left (b x + c\right )\right )} \sinh \left (-a + c\right )^{2} + {\left (3 \, b \cosh \left (b x + c\right )^{2} \cosh \left (-a + c\right )^{2} + b \cosh \left (-a + c\right )^{2} - {\left (3 \, b \cosh \left (b x + c\right )^{2} + b\right )} \sinh \left (-a + c\right )^{2}\right )} \sinh \left (b x + c\right )} \]

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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giac [A]  time = 0.12, size = 49, normalized size = 1.29 \[ -\frac {{\left (2 \, e^{\left (2 \, b x + 2 \, a + 2 \, c\right )} + e^{\left (2 \, a\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (-a - c\right )}}{b {\left (e^{\left (2 \, b x + 2 \, c\right )} + 1\right )}^{2}} \]

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)

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maple [A]  time = 0.22, size = 56, normalized size = 1.47 \[ -\frac {\left (2 \,{\mathrm e}^{2 b x +2 a +2 c}+{\mathrm e}^{2 a}+{\mathrm e}^{2 c}\right ) {\mathrm e}^{3 a -c}}{\left ({\mathrm e}^{2 b x +2 a +2 c}+{\mathrm e}^{2 a}\right )^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(exp(2*b*x+2*a+2*c)+exp(2*a))^2/b*(2*exp(2*b*x+2*a+2*c)+exp(2*a)+exp(2*c))*exp(3*a-c)

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maxima [B]  time = 0.32, size = 119, normalized size = 3.13 \[ \frac {2 \, e^{\left (-2 \, b x + 3 \, c\right )}}{b {\left (2 \, e^{\left (-2 \, b x + a + 2 \, c\right )} + e^{\left (-4 \, b x + a\right )} + e^{\left (a + 4 \, c\right )}\right )}} + \frac {e^{\left (2 \, a + 3 \, c\right )}}{b {\left (2 \, e^{\left (-2 \, b x + a + 2 \, c\right )} + e^{\left (-4 \, b x + a\right )} + e^{\left (a + 4 \, c\right )}\right )}} + \frac {e^{\left (5 \, c\right )}}{b {\left (2 \, e^{\left (-2 \, b x + a + 2 \, c\right )} + e^{\left (-4 \, b x + a\right )} + e^{\left (a + 4 \, c\right )}\right )}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {cosh}\left (a+b\,x\right )}{{\mathrm {cosh}\left (c+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)/cosh(c + b*x)^3,x)

[Out]

int(cosh(a + b*x)/cosh(c + b*x)^3, x)

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

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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