3.184 \(\int \frac{\coth ^2(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=77 \[ -\frac{a \coth (x)}{a^2-b^2}+\frac{b \text{csch}(x)}{a^2-b^2}+\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

(2*a^2*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)) - (a*Coth[x])/(a^2 - b^2) +
(b*Csch[x])/(a^2 - b^2)

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Rubi [A]  time = 0.0935763, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2727, 3767, 8, 2606, 2659, 208} \[ -\frac{a \coth (x)}{a^2-b^2}+\frac{b \text{csch}(x)}{a^2-b^2}+\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^2/(a + b*Cosh[x]),x]

[Out]

(2*a^2*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)) - (a*Coth[x])/(a^2 - b^2) +
(b*Csch[x])/(a^2 - b^2)

Rule 2727

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a/(a^2 - b^
2), Int[(g*Tan[e + f*x])^p/Sin[e + f*x]^2, x], x] + (-Dist[(b*g)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 1)/Cos
[e + f*x], x], x] - Dist[(a^2*g^2)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 2)/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*p] && GtQ[p, 1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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{\coth ^2(x)}{a+b \cosh (x)} \, dx &=\frac{a \int \text{csch}^2(x) \, dx}{a^2-b^2}+\frac{a^2 \int \frac{1}{a+b \cosh (x)} \, dx}{a^2-b^2}-\frac{b \int \coth (x) \text{csch}(x) \, dx}{a^2-b^2}\\ &=-\frac{(i a) \operatorname{Subst}(\int 1 \, dx,x,-i \coth (x))}{a^2-b^2}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a^2-b^2}+\frac{(i b) \operatorname{Subst}(\int 1 \, dx,x,-i \text{csch}(x))}{a^2-b^2}\\ &=\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}-\frac{a \coth (x)}{a^2-b^2}+\frac{b \text{csch}(x)}{a^2-b^2}\\ \end{align*}

Mathematica [A]  time = 0.20125, size = 77, normalized size = 1. \[ \frac{2 a^2 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac{\tanh \left (\frac{x}{2}\right )}{2 (a-b)}-\frac{\coth \left (\frac{x}{2}\right )}{2 (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) - Coth[x/2]/(2*(a + b)) - Tanh[x/2]/(2
*(a - b))

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Maple [A]  time = 0.024, size = 78, normalized size = 1. \begin{align*} -{\frac{1}{2\,a-2\,b}\tanh \left ({\frac{x}{2}} \right ) }+2\,{\frac{{a}^{2}}{ \left ( a+b \right ) \left ( a-b \right ) \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 ) }-{\frac{1}{2\,a+2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.96197, size = 1172, normalized size = 15.22 \begin{align*} \left [\frac{2 \, a^{3} - 2 \, a b^{2} +{\left (a^{2} \cosh \left (x\right )^{2} + 2 \, a^{2} \cosh \left (x\right ) \sinh \left (x\right ) + a^{2} \sinh \left (x\right )^{2} - a^{2}\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 (a^{2} b - b^{3}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{2} b - b^{3}\right )} \sinh \left (x\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4} -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}}, \frac{2 \,{\left (a^{3} - a b^{2} +{\left (a^{2} \cosh \left (x\right )^{2} + 2 \, a^{2} \cosh \left (x\right ) \sinh \left (x\right ) + a^{2} \sinh \left (x\right )^{2} - a^{2}\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 (a^{2} b - b^{3}\right )} \cosh \left (x\right ) -{\left (a^{2} b - b^{3}\right )} \sinh \left (x\right )\right )}}{a^{4} - 2 \, a^{2} b^{2} + b^{4} -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a^3 - 2*a*b^2 + (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((b^2*cos
h(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*co
sh(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*(a^2*b
- b^3)*cosh(x) - 2*(a^2*b - b^3)*sinh(x))/(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 - 2*(a^4
- 2*a^2*b^2 + b^4)*cosh(x)*sinh(x) - (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^2), 2*(a^3 - a*b^2 + (a^2*cosh(x)^2 + 2*a
^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 - a^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 - b^3)*cosh(x) - (a^2*b - b^3)*sinh(x))/(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^
4)*cosh(x)^2 - 2*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x) - (a^4 - 2*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{\coth ^{2}{\left (x \right )}}{a + b \cosh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**2/(a+b*cosh(x)),x)

[Out]

Integral(coth(x)**2/(a + b*cosh(x)), x)

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Giac [A]  time = 1.19427, size = 103, normalized size = 1.34 \begin{align*} \frac{2 \, a^{2} \arctan \left (\frac{b e^{x} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \,{\left (b e^{x} - a\right )}}{{\left (a^{2} - b^{2}\right )}{\left (e^{\left (2 \, x\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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