3.101 \(\int \frac{\text{sech}^3(x)}{a+b \text{sech}(x)} \, dx\)

Optimal. Leaf size=64 \[ \frac{2 a^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^2 \sqrt{a-b} \sqrt{a+b}}-\frac{a \tan ^{-1}(\sinh (x))}{b^2}+\frac{\tanh (x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.14027, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3790, 3789, 3770, 3831, 2659, 205} \[ \frac{2 a^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^2 \sqrt{a-b} \sqrt{a+b}}-\frac{a \tan ^{-1}(\sinh (x))}{b^2}+\frac{\tanh (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + b*Sech[x]),x]

[Out]

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

Rule 3790

Int[csc[(e_.) + (f_.)*(x_)]^3/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(b*f), x
] - Dist[a/b, Int[Csc[e + f*x]^2/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

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}^3(x)}{a+b \text{sech}(x)} \, dx &=\frac{\tanh (x)}{b}-\frac{a \int \frac{\text{sech}^2(x)}{a+b \text{sech}(x)} \, dx}{b}\\ &=\frac{\tanh (x)}{b}-\frac{a \int \text{sech}(x) \, dx}{b^2}+\frac{a^2 \int \frac{\text{sech}(x)}{a+b \text{sech}(x)} \, dx}{b^2}\\ &=-\frac{a \tan ^{-1}(\sinh (x))}{b^2}+\frac{\tanh (x)}{b}+\frac{a^2 \int \frac{1}{1+\frac{a \cosh (x)}{b}} \, dx}{b^3}\\ &=-\frac{a \tan ^{-1}(\sinh (x))}{b^2}+\frac{\tanh (x)}{b}+\frac{\left (2 a^2\right ) \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^3}\\ &=-\frac{a \tan ^{-1}(\sinh (x))}{b^2}+\frac{2 a^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^2 \sqrt{a+b}}+\frac{\tanh (x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.105394, size = 63, normalized size = 0.98 \[ \frac{-\frac{2 a^2 \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-2 a \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+b \tanh (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(a + b*Sech[x]),x]

[Out]

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

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Maple [A]  time = 0.024, size = 73, normalized size = 1.1 \begin{align*} 2\,{\frac{\tanh \left ( x/2 \right ) }{b \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{a\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{{b}^{2}}}+2\,{\frac{{a}^{2}}{{b}^{2}\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)^3/(a+b*sech(x)),x)

[Out]

2/b*tanh(1/2*x)/(tanh(1/2*x)^2+1)-2/b^2*a*arctan(tanh(1/2*x))+2*a^2/b^2/((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)^3/(a+b*sech(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.70445, size = 1285, normalized size = 20.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*sech(x)),x, algorithm="fricas")

[Out]

[-(2*a^2*b - 2*b^3 + (a^2*cosh(x)^2 + 2*a^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 + a^2)*sqrt(-a^2 + b^2)*log((a^2*c
osh(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*cosh(x) + b)*sinh(x) + a)) + 2*(a^3
- a*b^2 + (a^3 - a*b^2)*cosh(x)^2 + 2*(a^3 - a*b^2)*cosh(x)*sinh(x) + (a^3 - a*b^2)*sinh(x)^2)*arctan(cosh(x)
+ sinh(x)))/(a^2*b^2 - b^4 + (a^2*b^2 - b^4)*cosh(x)^2 + 2*(a^2*b^2 - b^4)*cosh(x)*sinh(x) + (a^2*b^2 - b^4)*
sinh(x)^2), -2*(a^2*b - b^3 + (a^2*cosh(x)^2 + 2*a^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 + a^2)*sqrt(a^2 - b^2)*ar
ctan(-(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2)) + (a^3 - a*b^2 + (a^3 - a*b^2)*cosh(x)^2 + 2*(a^3 - a*b^2)*
cosh(x)*sinh(x) + (a^3 - a*b^2)*sinh(x)^2)*arctan(cosh(x) + sinh(x)))/(a^2*b^2 - b^4 + (a^2*b^2 - b^4)*cosh(x)
^2 + 2*(a^2*b^2 - b^4)*cosh(x)*sinh(x) + (a^2*b^2 - b^4)*sinh(x)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{3}{\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)**3/(a+b*sech(x)),x)

[Out]

Integral(sech(x)**3/(a + b*sech(x)), x)

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Giac [A]  time = 1.16892, size = 82, normalized size = 1.28 \begin{align*} \frac{2 \, a^{2} \arctan \left (\frac{a e^{x} + b}{\sqrt{a^{2} - b^{2}}}\right )}{\sqrt{a^{2} - b^{2}} b^{2}} - \frac{2 \, a \arctan \left (e^{x}\right )}{b^{2}} - \frac{2}{b{\left (e^{\left (2 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*sech(x)),x, algorithm="giac")

[Out]

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