3.195 \(\int \frac{\coth ^3(x)}{a+a \cosh (x)} \, dx\)

Optimal. Leaf size=46 \[ \frac{\coth ^4(x)}{4 a}-\frac{3 \tanh ^{-1}(\cosh (x))}{8 a}-\frac{\coth ^3(x) \text{csch}(x)}{4 a}-\frac{3 \coth (x) \text{csch}(x)}{8 a} \]

[Out]

(-3*ArcTanh[Cosh[x]])/(8*a) + Coth[x]^4/(4*a) - (3*Coth[x]*Csch[x])/(8*a) - (Coth[x]^3*Csch[x])/(4*a)

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Rubi [A]  time = 0.106865, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2706, 2607, 30, 2611, 3770} \[ \frac{\coth ^4(x)}{4 a}-\frac{3 \tanh ^{-1}(\cosh (x))}{8 a}-\frac{\coth ^3(x) \text{csch}(x)}{4 a}-\frac{3 \coth (x) \text{csch}(x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(a + a*Cosh[x]),x]

[Out]

(-3*ArcTanh[Cosh[x]])/(8*a) + Coth[x]^4/(4*a) - (3*Coth[x]*Csch[x])/(8*a) - (Coth[x]^3*Csch[x])/(4*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 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 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 ^3(x)}{a+a \cosh (x)} \, dx &=\frac{\int \coth ^4(x) \text{csch}(x) \, dx}{a}-\frac{\int \coth ^3(x) \text{csch}^2(x) \, dx}{a}\\ &=-\frac{\coth ^3(x) \text{csch}(x)}{4 a}+\frac{3 \int \coth ^2(x) \text{csch}(x) \, dx}{4 a}+\frac{\operatorname{Subst}\left (\int x^3 \, dx,x,i \coth (x)\right )}{a}\\ &=\frac{\coth ^4(x)}{4 a}-\frac{3 \coth (x) \text{csch}(x)}{8 a}-\frac{\coth ^3(x) \text{csch}(x)}{4 a}+\frac{3 \int \text{csch}(x) \, dx}{8 a}\\ &=-\frac{3 \tanh ^{-1}(\cosh (x))}{8 a}+\frac{\coth ^4(x)}{4 a}-\frac{3 \coth (x) \text{csch}(x)}{8 a}-\frac{\coth ^3(x) \text{csch}(x)}{4 a}\\ \end{align*}

Mathematica [A]  time = 0.0990674, size = 60, normalized size = 1.3 \[ \frac{-2 \coth ^2\left (\frac{x}{2}\right )+\text{sech}^2\left (\frac{x}{2}\right )-12 \cosh ^2\left (\frac{x}{2}\right ) \left (\log \left (\cosh \left (\frac{x}{2}\right )\right )-\log \left (\sinh \left (\frac{x}{2}\right )\right )\right )-8}{16 a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^3/(a + a*Cosh[x]),x]

[Out]

(-8 - 2*Coth[x/2]^2 - 12*Cosh[x/2]^2*(Log[Cosh[x/2]] - Log[Sinh[x/2]]) + Sech[x/2]^2)/(16*a*(1 + Cosh[x]))

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Maple [A]  time = 0.027, size = 45, normalized size = 1. \begin{align*}{\frac{1}{32\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{4}}+{\frac{3}{16\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{\frac{3}{8\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{1}{16\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^3/(a+a*cosh(x)),x)

[Out]

1/32/a*tanh(1/2*x)^4+3/16/a*tanh(1/2*x)^2+3/8/a*ln(tanh(1/2*x))-1/16/a/tanh(1/2*x)^2

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Maxima [B]  time = 1.08625, size = 139, normalized size = 3.02 \begin{align*} -\frac{5 \, e^{\left (-x\right )} + 2 \, e^{\left (-2 \, x\right )} + 2 \, e^{\left (-3 \, x\right )} + 2 \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )}}{4 \,{\left (2 \, a e^{\left (-x\right )} - a e^{\left (-2 \, x\right )} - 4 \, a e^{\left (-3 \, x\right )} - a e^{\left (-4 \, x\right )} + 2 \, a e^{\left (-5 \, x\right )} + a e^{\left (-6 \, x\right )} + a\right )}} - \frac{3 \, \log \left (e^{\left (-x\right )} + 1\right )}{8 \, a} + \frac{3 \, \log \left (e^{\left (-x\right )} - 1\right )}{8 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^3/(a+a*cosh(x)),x, algorithm="maxima")

[Out]

-1/4*(5*e^(-x) + 2*e^(-2*x) + 2*e^(-3*x) + 2*e^(-4*x) + 5*e^(-5*x))/(2*a*e^(-x) - a*e^(-2*x) - 4*a*e^(-3*x) -
a*e^(-4*x) + 2*a*e^(-5*x) + a*e^(-6*x) + a) - 3/8*log(e^(-x) + 1)/a + 3/8*log(e^(-x) - 1)/a

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Fricas [B]  time = 2.01018, size = 2071, normalized size = 45.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^3/(a+a*cosh(x)),x, algorithm="fricas")

[Out]

-1/8*(10*cosh(x)^5 + 2*(25*cosh(x) + 2)*sinh(x)^4 + 10*sinh(x)^5 + 4*cosh(x)^4 + 4*(25*cosh(x)^2 + 4*cosh(x) +
1)*sinh(x)^3 + 4*cosh(x)^3 + 4*(25*cosh(x)^3 + 6*cosh(x)^2 + 3*cosh(x) + 1)*sinh(x)^2 + 4*cosh(x)^2 + 3*(cosh
(x)^6 + 2*(3*cosh(x) + 1)*sinh(x)^5 + sinh(x)^6 + 2*cosh(x)^5 + (15*cosh(x)^2 + 10*cosh(x) - 1)*sinh(x)^4 - co
sh(x)^4 + 4*(5*cosh(x)^3 + 5*cosh(x)^2 - cosh(x) - 1)*sinh(x)^3 - 4*cosh(x)^3 + (15*cosh(x)^4 + 20*cosh(x)^3 -
6*cosh(x)^2 - 12*cosh(x) - 1)*sinh(x)^2 - cosh(x)^2 + 2*(3*cosh(x)^5 + 5*cosh(x)^4 - 2*cosh(x)^3 - 6*cosh(x)^
2 - cosh(x) + 1)*sinh(x) + 2*cosh(x) + 1)*log(cosh(x) + sinh(x) + 1) - 3*(cosh(x)^6 + 2*(3*cosh(x) + 1)*sinh(x
)^5 + sinh(x)^6 + 2*cosh(x)^5 + (15*cosh(x)^2 + 10*cosh(x) - 1)*sinh(x)^4 - cosh(x)^4 + 4*(5*cosh(x)^3 + 5*cos
h(x)^2 - cosh(x) - 1)*sinh(x)^3 - 4*cosh(x)^3 + (15*cosh(x)^4 + 20*cosh(x)^3 - 6*cosh(x)^2 - 12*cosh(x) - 1)*s
inh(x)^2 - cosh(x)^2 + 2*(3*cosh(x)^5 + 5*cosh(x)^4 - 2*cosh(x)^3 - 6*cosh(x)^2 - cosh(x) + 1)*sinh(x) + 2*cos
h(x) + 1)*log(cosh(x) + sinh(x) - 1) + 2*(25*cosh(x)^4 + 8*cosh(x)^3 + 6*cosh(x)^2 + 4*cosh(x) + 5)*sinh(x) +
10*cosh(x))/(a*cosh(x)^6 + a*sinh(x)^6 + 2*a*cosh(x)^5 + 2*(3*a*cosh(x) + a)*sinh(x)^5 - a*cosh(x)^4 + (15*a*c
osh(x)^2 + 10*a*cosh(x) - a)*sinh(x)^4 - 4*a*cosh(x)^3 + 4*(5*a*cosh(x)^3 + 5*a*cosh(x)^2 - a*cosh(x) - a)*sin
h(x)^3 - a*cosh(x)^2 + (15*a*cosh(x)^4 + 20*a*cosh(x)^3 - 6*a*cosh(x)^2 - 12*a*cosh(x) - a)*sinh(x)^2 + 2*a*co
sh(x) + 2*(3*a*cosh(x)^5 + 5*a*cosh(x)^4 - 2*a*cosh(x)^3 - 6*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 ^{3}{\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)**3/(a+a*cosh(x)),x)

[Out]

Integral(coth(x)**3/(cosh(x) + 1), x)/a

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Giac [B]  time = 1.17454, size = 127, normalized size = 2.76 \begin{align*} -\frac{3 \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{16 \, a} + \frac{3 \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{16 \, a} - \frac{3 \, e^{\left (-x\right )} + 3 \, e^{x} - 2}{16 \, a{\left (e^{\left (-x\right )} + e^{x} - 2\right )}} + \frac{9 \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 4 \, e^{\left (-x\right )} + 4 \, e^{x} - 12}{32 \, a{\left (e^{\left (-x\right )} + e^{x} + 2\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^3/(a+a*cosh(x)),x, algorithm="giac")

[Out]

-3/16*log(e^(-x) + e^x + 2)/a + 3/16*log(e^(-x) + e^x - 2)/a - 1/16*(3*e^(-x) + 3*e^x - 2)/(a*(e^(-x) + e^x -
2)) + 1/32*(9*(e^(-x) + e^x)^2 + 4*e^(-x) + 4*e^x - 12)/(a*(e^(-x) + e^x + 2)^2)