3.204 \(\int \frac{A+B \text{sech}(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=62 \[ \frac{2 (a A-b B) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a \sqrt{a-b} \sqrt{a+b}}+\frac{B \tan ^{-1}(\sinh (x))}{a} \]

[Out]

(B*ArcTan[Sinh[x]])/a + (2*(a*A - b*B)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b
])

________________________________________________________________________________________

Rubi [A]  time = 0.133207, antiderivative size = 62, 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 = {2828, 3001, 3770, 2659, 208} \[ \frac{2 (a A-b B) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a \sqrt{a-b} \sqrt{a+b}}+\frac{B \tan ^{-1}(\sinh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sech[x])/(a + b*Cosh[x]),x]

[Out]

(B*ArcTan[Sinh[x]])/a + (2*(a*A - b*B)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b
])

Rule 2828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[((a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n)/Sin[e + f*x]^n, x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B \text{sech}(x)}{a+b \cosh (x)} \, dx &=\int \frac{(B+A \cosh (x)) \text{sech}(x)}{a+b \cosh (x)} \, dx\\ &=\frac{B \int \text{sech}(x) \, dx}{a}+\frac{(a A-b B) \int \frac{1}{a+b \cosh (x)} \, dx}{a}\\ &=\frac{B \tan ^{-1}(\sinh (x))}{a}+\frac{(2 (a A-b B)) \operatorname{Subst}\left (\int \frac{1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a}\\ &=\frac{B \tan ^{-1}(\sinh (x))}{a}+\frac{2 (a A-b B) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a \sqrt{a-b} \sqrt{a+b}}\\ \end{align*}

Mathematica [A]  time = 0.113905, size = 63, normalized size = 1.02 \[ \frac{2 \left (\frac{(b B-a A) \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+B \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sech[x])/(a + b*Cosh[x]),x]

[Out]

(2*(B*ArcTan[Tanh[x/2]] + ((-(a*A) + b*B)*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2]))/a

________________________________________________________________________________________

Maple [A]  time = 0.03, size = 89, normalized size = 1.4 \begin{align*} 2\,{\frac{A}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{Bb}{a\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{B\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sech(x))/(a+b*cosh(x)),x)

[Out]

2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))*A-2*B/a*b/((a+b)*(a-b))^(1/2)*arctanh((a-
b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))+2*B/a*arctan(tanh(1/2*x))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sech(x))/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 4.64704, size = 643, normalized size = 10.37 \begin{align*} \left [-\frac{{\left (A a - B b\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} - b^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + b}\right ) - 2 \,{\left (B a^{2} - B b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}{a^{3} - a b^{2}}, -\frac{2 \,{\left ({\left (A a - B b\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{a^{2} - b^{2}}\right ) -{\left (B a^{2} - B b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )\right )}}{a^{3} - a b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sech(x))/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

[-((A*a - B*b)*sqrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(
x) + a*b)*sinh(x) + 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) +
2*(b*cosh(x) + a)*sinh(x) + b)) - 2*(B*a^2 - B*b^2)*arctan(cosh(x) + sinh(x)))/(a^3 - a*b^2), -2*((A*a - B*b)*
sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) - (B*a^2 - B*b^2)*arctan(co
sh(x) + sinh(x)))/(a^3 - a*b^2)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sech(x))/(a+b*cosh(x)),x)

[Out]

Integral((A + B*sech(x))/(a + b*cosh(x)), x)

________________________________________________________________________________________

Giac [A]  time = 1.21048, size = 72, normalized size = 1.16 \begin{align*} \frac{2 \, B \arctan \left (e^{x}\right )}{a} + \frac{2 \,{\left (A a - B b\right )} \arctan \left (\frac{b e^{x} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{\sqrt{-a^{2} + b^{2}} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sech(x))/(a+b*cosh(x)),x, algorithm="giac")

[Out]

2*B*arctan(e^x)/a + 2*(A*a - B*b)*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*a)