### 3.149 $$\int \text{sech}(c+b x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=26 $\frac{\cosh (a-c) \log (\cosh (b x+c))}{b}+x \sinh (a-c)$

[Out]

(Cosh[a - c]*Log[Cosh[c + b*x]])/b + x*Sinh[a - c]

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Rubi [A]  time = 0.0158705, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.231, Rules used = {5624, 3475, 8} $\frac{\cosh (a-c) \log (\cosh (b x+c))}{b}+x \sinh (a-c)$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[c + b*x]*Sinh[a + b*x],x]

[Out]

(Cosh[a - c]*Log[Cosh[c + b*x]])/b + x*Sinh[a - c]

Rule 5624

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 8

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

Rubi steps

\begin{align*} \int \text{sech}(c+b x) \sinh (a+b x) \, dx &=\cosh (a-c) \int \tanh (c+b x) \, dx+\sinh (a-c) \int 1 \, dx\\ &=\frac{\cosh (a-c) \log (\cosh (c+b x))}{b}+x \sinh (a-c)\\ \end{align*}

Mathematica [A]  time = 0.11894, size = 26, normalized size = 1. $\frac{\cosh (a-c) \log (\cosh (b x+c))}{b}+x \sinh (a-c)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[c + b*x]*Sinh[a + b*x],x]

[Out]

(Cosh[a - c]*Log[Cosh[c + b*x]])/b + x*Sinh[a - c]

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Maple [B]  time = 0.036, size = 148, normalized size = 5.7 \begin{align*} x{{\rm e}^{a-c}}-{{\rm e}^{-a-c}}{{\rm e}^{2\,a}}x-{{\rm e}^{-a-c}}{{\rm e}^{2\,c}}x-{\frac{{{\rm e}^{-a-c}}{{\rm e}^{2\,a}}a}{b}}-{\frac{{{\rm e}^{-a-c}}{{\rm e}^{2\,c}}a}{b}}+{\frac{\ln \left ({{\rm e}^{2\,bx+2\,a}}+{{\rm e}^{2\,a-2\,c}} \right ){{\rm e}^{-a-c}}{{\rm e}^{2\,a}}}{2\,b}}+{\frac{\ln \left ({{\rm e}^{2\,bx+2\,a}}+{{\rm e}^{2\,a-2\,c}} \right ){{\rm e}^{-a-c}}{{\rm e}^{2\,c}}}{2\,b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+c)*sinh(b*x+a),x)

[Out]

x*exp(a-c)-exp(-a-c)*exp(2*a)*x-exp(-a-c)*exp(2*c)*x-1/b*exp(-a-c)*exp(2*a)*a-1/b*exp(-a-c)*exp(2*c)*a+1/2/b*l
n(exp(2*b*x+2*a)+exp(2*a-2*c))*exp(-a-c)*exp(2*a)+1/2/b*ln(exp(2*b*x+2*a)+exp(2*a-2*c))*exp(-a-c)*exp(2*c)

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Maxima [A]  time = 1.26344, size = 66, normalized size = 2.54 \begin{align*} \frac{{\left (e^{\left (2 \, a\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (-a - c\right )} \log \left (e^{\left (-2 \, b x\right )} + e^{\left (2 \, c\right )}\right )}{2 \, b} + \frac{{\left (b x + a\right )} e^{\left (a - c\right )}}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+c)*sinh(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.15378, size = 231, normalized size = 8.88 \begin{align*} -\frac{2 \, b x -{\left (\cosh \left (-a + c\right )^{2} - 2 \, \cosh \left (-a + c\right ) \sinh \left (-a + c\right ) + \sinh \left (-a + c\right )^{2} + 1\right )} \log \left (\frac{2 \, \cosh \left (b x + c\right )}{\cosh \left (b x + c\right ) - \sinh \left (b x + c\right )}\right )}{2 \,{\left (b \cosh \left (-a + c\right ) - b \sinh \left (-a + c\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+c)*sinh(b*x+a),x, algorithm="fricas")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+c)*sinh(b*x+a),x)

[Out]

Integral(sinh(a + b*x)*sech(b*x + c), x)

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Giac [A]  time = 1.14728, size = 66, normalized size = 2.54 \begin{align*} -\frac{2 \, b x e^{\left (-a + c\right )} -{\left (e^{\left (2 \, a + c\right )} + e^{\left (3 \, c\right )}\right )} e^{\left (-a - 2 \, c\right )} \log \left (e^{\left (2 \, b x + 2 \, c\right )} + 1\right )}{2 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+c)*sinh(b*x+a),x, algorithm="giac")

[Out]

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