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)*ln(cosh(b*x+c))/b+x*sinh(a-c)

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Rubi [A]  time = 0.02, antiderivative size = 26, normalized size of antiderivative = 1.00, 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 8

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

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*}

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Mathematica [A]  time = 0.13, size = 26, normalized size = 1.00 \[ \frac {\cosh (a-c) \log (\cosh (b x+c))}{b}+x \sinh (a-c) \]

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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fricas [B]  time = 0.43, size = 87, normalized size = 3.35 \[ -\frac {2 \, b x - {\left (\cosh \left (-a + c\right )^{2} - 2 \, \cosh \left (-a + c\right ) \sinh \left (-a + c\right ) + \sinh \left (-a + c\right )^{2} + 1\right )} \log \left (\frac {2 \, \cosh \left (b x + c\right )}{\cosh \left (b x + c\right ) - \sinh \left (b x + c\right )}\right )}{2 \, {\left (b \cosh \left (-a + c\right ) - b \sinh \left (-a + c\right )\right )}} \]

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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giac [A]  time = 0.11, size = 49, normalized size = 1.88 \[ -\frac {2 \, b x e^{\left (-a + c\right )} - {\left (e^{\left (2 \, a + c\right )} + e^{\left (3 \, c\right )}\right )} e^{\left (-a - 2 \, c\right )} \log \left (e^{\left (2 \, b x + 2 \, c\right )} + 1\right )}{2 \, b} \]

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

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maple [B]  time = 0.14, size = 148, normalized size = 5.69 \[ x \,{\mathrm e}^{a -c}-{\mathrm e}^{-a -c} {\mathrm e}^{2 a} x -{\mathrm e}^{-a -c} {\mathrm e}^{2 c} x -\frac {{\mathrm e}^{-a -c} {\mathrm e}^{2 a} a}{b}-\frac {{\mathrm e}^{-a -c} {\mathrm e}^{2 c} a}{b}+\frac {\ln \left ({\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 a -2 c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 a}}{2 b}+\frac {\ln \left ({\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 a -2 c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 c}}{2 b} \]

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

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maxima [A]  time = 0.32, size = 49, normalized size = 1.88 \[ \frac {{\left (e^{\left (2 \, a\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (-a - c\right )} \log \left (e^{\left (-2 \, b x\right )} + e^{\left (2 \, c\right )}\right )}{2 \, b} + \frac {{\left (b x + a\right )} e^{\left (a - c\right )}}{b} \]

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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mupad [B]  time = 0.22, size = 65, normalized size = 2.50 \[ \frac {{\mathrm {e}}^{2\,c-2\,a}\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-2\,c}\right )\,\left (2\,b\,{\mathrm {e}}^{3\,a-3\,c}+2\,b\,{\mathrm {e}}^{a-c}\right )}{4\,b^2}-x\,{\mathrm {e}}^{c-a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)/cosh(c + b*x),x)

[Out]

(exp(2*c - 2*a)*log(exp(2*a)*exp(2*b*x) + exp(2*a)*exp(-2*c))*(2*b*exp(3*a - 3*c) + 2*b*exp(a - c)))/(4*b^2) -
x*exp(c - a)

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

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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