3.193 \(\int \frac{\coth (x)}{a+a \cosh (x)} \, dx\)

Optimal. Leaf size=33 \[ \frac{\text{csch}^2(x)}{2 a}-\frac{\tanh ^{-1}(\cosh (x))}{2 a}-\frac{\coth (x) \text{csch}(x)}{2 a} \]

[Out]

-ArcTanh[Cosh[x]]/(2*a) - (Coth[x]*Csch[x])/(2*a) + Csch[x]^2/(2*a)

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Rubi [A]  time = 0.0651714, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {2706, 2606, 30, 2611, 3770} \[ \frac{\text{csch}^2(x)}{2 a}-\frac{\tanh ^{-1}(\cosh (x))}{2 a}-\frac{\coth (x) \text{csch}(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/(a + a*Cosh[x]),x]

[Out]

-ArcTanh[Cosh[x]]/(2*a) - (Coth[x]*Csch[x])/(2*a) + Csch[x]^2/(2*a)

Rule 2706

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

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

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

Rubi steps

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

Mathematica [A]  time = 0.0371304, size = 42, normalized size = 1.27 \[ -\frac{2 \cosh ^2\left (\frac{x}{2}\right ) \left (\log \left (\cosh \left (\frac{x}{2}\right )\right )-\log \left (\sinh \left (\frac{x}{2}\right )\right )\right )+1}{2 a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]/(a + a*Cosh[x]),x]

[Out]

-(1 + 2*Cosh[x/2]^2*(Log[Cosh[x/2]] - Log[Sinh[x/2]]))/(2*a*(1 + Cosh[x]))

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Maple [A]  time = 0.019, size = 23, normalized size = 0.7 \begin{align*}{\frac{1}{4\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{\frac{1}{2\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a*tanh(1/2*x)^2+1/2/a*ln(tanh(1/2*x))

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Maxima [A]  time = 1.07166, size = 65, normalized size = 1.97 \begin{align*} -\frac{e^{\left (-x\right )}}{2 \, a e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a} - \frac{\log \left (e^{\left (-x\right )} + 1\right )}{2 \, a} + \frac{\log \left (e^{\left (-x\right )} - 1\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+a*cosh(x)),x, algorithm="maxima")

[Out]

-e^(-x)/(2*a*e^(-x) + a*e^(-2*x) + a) - 1/2*log(e^(-x) + 1)/a + 1/2*log(e^(-x) - 1)/a

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Fricas [B]  time = 1.96863, size = 397, normalized size = 12.03 \begin{align*} -\frac{{\left (\cosh \left (x\right )^{2} + 2 \,{\left (\cosh \left (x\right ) + 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) + 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) -{\left (\cosh \left (x\right )^{2} + 2 \,{\left (\cosh \left (x\right ) + 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) + 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )}{2 \,{\left (a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (a \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((cosh(x)^2 + 2*(cosh(x) + 1)*sinh(x) + sinh(x)^2 + 2*cosh(x) + 1)*log(cosh(x) + sinh(x) + 1) - (cosh(x)^
2 + 2*(cosh(x) + 1)*sinh(x) + sinh(x)^2 + 2*cosh(x) + 1)*log(cosh(x) + sinh(x) - 1) + 2*cosh(x) + 2*sinh(x))/(
a*cosh(x)^2 + a*sinh(x)^2 + 2*a*cosh(x) + 2*(a*cosh(x) + a)*sinh(x) + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\coth{\left (x \right )}}{\cosh{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(coth(x)/(cosh(x) + 1), x)/a

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Giac [A]  time = 1.22914, size = 70, normalized size = 2.12 \begin{align*} -\frac{\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \, a} + \frac{\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \, a} + \frac{e^{\left (-x\right )} + e^{x} - 2}{4 \, a{\left (e^{\left (-x\right )} + e^{x} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(e^(-x) + e^x + 2)/a + 1/4*log(e^(-x) + e^x - 2)/a + 1/4*(e^(-x) + e^x - 2)/(a*(e^(-x) + e^x + 2))