3.695 \(\int \frac{\coth (x)}{b \cosh (x)+a \sinh (x)} \, dx\)

Optimal. Leaf size=51 \[ \frac{a \tanh ^{-1}\left (\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2}}-\frac{\tanh ^{-1}(\cosh (x))}{b} \]

[Out]

-(ArcTanh[Cosh[x]]/b) + (a*ArcTanh[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2])

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Rubi [A]  time = 0.100152, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3110, 3770, 3074, 206} \[ \frac{a \tanh ^{-1}\left (\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2}}-\frac{\tanh ^{-1}(\cosh (x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/(b*Cosh[x] + a*Sinh[x]),x]

[Out]

-(ArcTanh[Cosh[x]]/b) + (a*ArcTanh[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2])

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3770

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

Rule 3074

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0799109, size = 59, normalized size = 1.16 \[ \frac{\log \left (\tanh \left (\frac{x}{2}\right )\right )-\frac{2 a \tan ^{-1}\left (\frac{a+b \tanh \left (\frac{x}{2}\right )}{\sqrt{b-a} \sqrt{a+b}}\right )}{\sqrt{b-a} \sqrt{a+b}}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 53, normalized size = 1. \begin{align*}{\frac{1}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-2\,{\frac{a}{b\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b+2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*ln(tanh(1/2*x))-2*a/b/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*a)/(-a^2+b^2)^(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(coth(x)/(b*cosh(x)+a*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.04369, size = 675, normalized size = 13.24 \begin{align*} \left [\frac{\sqrt{a^{2} - b^{2}} a \log \left (\frac{{\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a + b\right )} \sinh \left (x\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} + a - b}{{\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a + b\right )} \sinh \left (x\right )^{2} - a + b}\right ) -{\left (a^{2} - b^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) +{\left (a^{2} - b^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{2} b - b^{3}}, -\frac{2 \, \sqrt{-a^{2} + b^{2}} a \arctan \left (\frac{\sqrt{-a^{2} + b^{2}}}{{\left (a + b\right )} \cosh \left (x\right ) +{\left (a + b\right )} \sinh \left (x\right )}\right ) +{\left (a^{2} - b^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) -{\left (a^{2} - b^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{2} b - b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(a^2 - b^2)*a*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + 2*sqrt(a^2 - b^2)
*(cosh(x) + sinh(x)) + a - b)/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - a + b)) - (
a^2 - b^2)*log(cosh(x) + sinh(x) + 1) + (a^2 - b^2)*log(cosh(x) + sinh(x) - 1))/(a^2*b - b^3), -(2*sqrt(-a^2 +
b^2)*a*arctan(sqrt(-a^2 + b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) + (a^2 - b^2)*log(cosh(x) + sinh(x) + 1)
- (a^2 - b^2)*log(cosh(x) + sinh(x) - 1))/(a^2*b - b^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12607, size = 81, normalized size = 1.59 \begin{align*} -\frac{2 \, a \arctan \left (\frac{a e^{x} + b e^{x}}{\sqrt{-a^{2} + b^{2}}}\right )}{\sqrt{-a^{2} + b^{2}} b} - \frac{\log \left (e^{x} + 1\right )}{b} + \frac{\log \left ({\left | e^{x} - 1 \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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