3.583 \(\int \frac{1}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=41 \[ \frac{2 \tanh ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}} \]

[Out]

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

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Rubi [A]  time = 0.0491715, antiderivative size = 42, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2659, 208} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[x])^(-1),x]

[Out]

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

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

Mathematica [A]  time = 0.0316551, size = 41, normalized size = 1. \[ -\frac{2 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[x])^(-1),x]

[Out]

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

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Maple [A]  time = 0.01, size = 36, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(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))

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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(1/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10459, size = 460, normalized size = 11.22 \begin{align*} \left [\frac{\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 )}{\sqrt{a^{2} - b^{2}}}, -\frac{2 \, \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 )}{a^{2} - b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 10.6283, size = 126, normalized size = 3.07 \begin{align*} \begin{cases} \tilde{\infty } \operatorname{atan}{\left (\tanh{\left (\frac{x}{2} \right )} \right )} & \text{for}\: a = 0 \wedge b = 0 \\- \frac{1}{b \tanh{\left (\frac{x}{2} \right )}} & \text{for}\: a = - b \\\frac{\tanh{\left (\frac{x}{2} \right )}}{b} & \text{for}\: a = b \\- \frac{\log{\left (- \sqrt{\frac{a}{a - b} + \frac{b}{a - b}} + \tanh{\left (\frac{x}{2} \right )} \right )}}{a \sqrt{\frac{a}{a - b} + \frac{b}{a - b}} - b \sqrt{\frac{a}{a - b} + \frac{b}{a - b}}} + \frac{\log{\left (\sqrt{\frac{a}{a - b} + \frac{b}{a - b}} + \tanh{\left (\frac{x}{2} \right )} \right )}}{a \sqrt{\frac{a}{a - b} + \frac{b}{a - b}} - b \sqrt{\frac{a}{a - b} + \frac{b}{a - b}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(x)),x)

[Out]

Piecewise((zoo*atan(tanh(x/2)), Eq(a, 0) & Eq(b, 0)), (-1/(b*tanh(x/2)), Eq(a, -b)), (tanh(x/2)/b, Eq(a, b)),
(-log(-sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*sqrt(a/(a - b) + b/(a - b)) - b*sqrt(a/(a - b) + b/(a - b))
) + log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*sqrt(a/(a - b) + b/(a - b)) - b*sqrt(a/(a - b) + b/(a - b)
)), True))

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Giac [A]  time = 1.1686, size = 43, normalized size = 1.05 \begin{align*} \frac{2 \, \arctan \left (\frac{b e^{x} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{\sqrt{-a^{2} + b^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(x)),x, algorithm="giac")

[Out]

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