3.140 \(\int \text{sech}(c-b x) \text{sech}(a+b x) \, dx\)

Optimal. Leaf size=33 \[ \frac{\text{csch}(a+c) \log (\cosh (a+b x))}{b}-\frac{\text{csch}(a+c) \log (\cosh (c-b x))}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0234886, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5644, 3475} \[ \frac{\text{csch}(a+c) \log (\cosh (a+b x))}{b}-\frac{\text{csch}(a+c) \log (\cosh (c-b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c - b*x]*Sech[a + b*x],x]

[Out]

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

Rule 5644

Int[Sech[(a_.) + (b_.)*(x_)]*Sech[(c_) + (d_.)*(x_)], x_Symbol] :> -Dist[Csch[(b*c - a*d)/d], Int[Tanh[a + b*x
], x], x] + Dist[Csch[(b*c - a*d)/b], Int[Tanh[c + d*x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b^2 - d^2, 0]
&& NeQ[b*c - a*d, 0]

Rule 3475

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

Rubi steps

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

Mathematica [A]  time = 0.199106, size = 27, normalized size = 0.82 \[ -\frac{\text{csch}(a+c) (\log (\cosh (c-b x))-\log (\cosh (a+b x)))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.033, size = 75, normalized size = 2.3 \begin{align*} -2\,{\frac{\ln \left ({{\rm e}^{2\,a+2\,c}}+{{\rm e}^{2\,bx+2\,a}} \right ){{\rm e}^{a+c}}}{b \left ({{\rm e}^{2\,a+2\,c}}-1 \right ) }}+2\,{\frac{\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ){{\rm e}^{a+c}}}{b \left ({{\rm e}^{2\,a+2\,c}}-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.61928, size = 90, normalized size = 2.73 \begin{align*} \frac{2 \, e^{\left (a + c\right )} \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b{\left (e^{\left (2 \, a + 2 \, c\right )} - 1\right )}} - \frac{2 \, e^{\left (a + c\right )} \log \left (e^{\left (-2 \, b x + 2 \, c\right )} + 1\right )}{b{\left (e^{\left (2 \, a + 2 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.14801, size = 471, normalized size = 14.27 \begin{align*} \frac{2 \,{\left ({\left (\cosh \left (a + c\right ) - \sinh \left (a + c\right )\right )} \log \left (\frac{2 \,{\left (\cosh \left (b x + a\right ) \cosh \left (a + c\right ) - \sinh \left (b x + a\right ) \sinh \left (a + c\right )\right )}}{\cosh \left (b x + a\right ) \cosh \left (a + c\right ) -{\left (\cosh \left (a + c\right ) + \sinh \left (a + c\right )\right )} \sinh \left (b x + a\right ) + \cosh \left (b x + a\right ) \sinh \left (a + c\right )}\right ) -{\left (\cosh \left (a + c\right ) - \sinh \left (a + c\right )\right )} \log \left (\frac{2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )\right )}}{b \cosh \left (a + c\right )^{2} - 2 \, b \cosh \left (a + c\right ) \sinh \left (a + c\right ) + b \sinh \left (a + c\right )^{2} - b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((cosh(a + c) - sinh(a + c))*log(2*(cosh(b*x + a)*cosh(a + c) - sinh(b*x + a)*sinh(a + c))/(cosh(b*x + a)*co
sh(a + c) - (cosh(a + c) + sinh(a + c))*sinh(b*x + a) + cosh(b*x + a)*sinh(a + c))) - (cosh(a + c) - sinh(a +
c))*log(2*cosh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))))/(b*cosh(a + c)^2 - 2*b*cosh(a + c)*sinh(a + c) + b*s
inh(a + c)^2 - b)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.17126, size = 95, normalized size = 2.88 \begin{align*} -\frac{2 \,{\left (\frac{e^{\left (a + c\right )} \log \left (e^{\left (2 \, b x\right )} + e^{\left (2 \, c\right )}\right )}{e^{\left (2 \, a + 2 \, c\right )} - 1} + \frac{e^{\left (3 \, a + c\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{e^{\left (2 \, a\right )} - e^{\left (4 \, a + 2 \, c\right )}}\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(e^(a + c)*log(e^(2*b*x) + e^(2*c))/(e^(2*a + 2*c) - 1) + e^(3*a + c)*log(e^(2*b*x + 2*a) + 1)/(e^(2*a) - e
^(4*a + 2*c)))/b