3.100 \(\int \frac{\text{sech}^2(x)}{a+b \text{sech}(x)} \, dx\)

Optimal. Leaf size=54 \[ \frac{\tan ^{-1}(\sinh (x))}{b}-\frac{2 a \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b \sqrt{a-b} \sqrt{a+b}} \]

[Out]

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

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Rubi [A]  time = 0.0995088, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3789, 3770, 3831, 2659, 205} \[ \frac{\tan ^{-1}(\sinh (x))}{b}-\frac{2 a \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b \sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(a + b*Sech[x]),x]

[Out]

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

Rule 3789

Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[Csc[e + f*x],
x], x] - Dist[a/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3770

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

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 205

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

Rubi steps

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

Mathematica [A]  time = 0.0501867, size = 54, normalized size = 1. \[ \frac{2 \left (\frac{a \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(a + b*Sech[x]),x]

[Out]

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

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Maple [A]  time = 0.014, size = 51, normalized size = 0.9 \begin{align*} 2\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{b}}-2\,{\frac{a}{b\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(a+b*sech(x)),x)

[Out]

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

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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(sech(x)^2/(a+b*sech(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.60518, size = 587, normalized size = 10.87 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} a \log \left (\frac{a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) - a^{2} + 2 \, b^{2} + 2 \,{\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}}{a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) + 2 \,{\left (a \cosh \left (x\right ) + b\right )} \sinh \left (x\right ) + a}\right ) - 2 \,{\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}{a^{2} b - b^{3}}, \frac{2 \,{\left (\sqrt{a^{2} - b^{2}} a \arctan \left (-\frac{a \cosh \left (x\right ) + a \sinh \left (x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right ) +{\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )\right )}}{a^{2} b - b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+b*sech(x)),x, algorithm="fricas")

[Out]

[-(sqrt(-a^2 + b^2)*a*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) - a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)
*sinh(x) + 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cos
h(x) + b)*sinh(x) + a)) - 2*(a^2 - b^2)*arctan(cosh(x) + sinh(x)))/(a^2*b - b^3), 2*(sqrt(a^2 - b^2)*a*arctan(
-(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2)) + (a^2 - b^2)*arctan(cosh(x) + sinh(x)))/(a^2*b - b^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{2}{\left (x \right )}}{a + b \operatorname{sech}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(a+b*sech(x)),x)

[Out]

Integral(sech(x)**2/(a + b*sech(x)), x)

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Giac [A]  time = 1.15848, size = 61, normalized size = 1.13 \begin{align*} -\frac{2 \, a \arctan \left (\frac{a e^{x} + b}{\sqrt{a^{2} - b^{2}}}\right )}{\sqrt{a^{2} - b^{2}} b} + \frac{2 \, \arctan \left (e^{x}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+b*sech(x)),x, algorithm="giac")

[Out]

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