3.149 \(\int \frac{\coth ^5(x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=94 \[ -\frac{b x}{a^2-b^2}-\frac{\left (a^2+b^2\right ) \coth (x)}{b^3}+\frac{a \log (\sinh (x))}{a^2-b^2}+\frac{a^5 \log (a+b \coth (x))}{b^4 \left (a^2-b^2\right )}+\frac{a \coth ^2(x)}{2 b^2}-\frac{\coth ^3(x)}{3 b} \]

[Out]

-((b*x)/(a^2 - b^2)) - ((a^2 + b^2)*Coth[x])/b^3 + (a*Coth[x]^2)/(2*b^2) - Coth[x]^3/(3*b) + (a^5*Log[a + b*Co
th[x]])/(b^4*(a^2 - b^2)) + (a*Log[Sinh[x]])/(a^2 - b^2)

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Rubi [A]  time = 0.391848, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3566, 3647, 3648, 3626, 3617, 31, 3475} \[ -\frac{b x}{a^2-b^2}-\frac{\left (a^2+b^2\right ) \coth (x)}{b^3}+\frac{a \log (\sinh (x))}{a^2-b^2}+\frac{a^5 \log (a+b \coth (x))}{b^4 \left (a^2-b^2\right )}+\frac{a \coth ^2(x)}{2 b^2}-\frac{\coth ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^5/(a + b*Coth[x]),x]

[Out]

-((b*x)/(a^2 - b^2)) - ((a^2 + b^2)*Coth[x])/b^3 + (a*Coth[x]^2)/(2*b^2) - Coth[x]^3/(3*b) + (a^5*Log[a + b*Co
th[x]])/(b^4*(a^2 - b^2)) + (a*Log[Sinh[x]])/(a^2 - b^2)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3648

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m
+ n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m
+ n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b - b*C)*(m + n + 1)*Tan[e + f*x] - C*m*(b*c - a*d)*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3475

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

Rubi steps

\begin{align*} \int \frac{\coth ^5(x)}{a+b \coth (x)} \, dx &=-\frac{\coth ^3(x)}{3 b}-\frac{\int \frac{\coth ^2(x) \left (-3 a-3 b \coth (x)+3 a \coth ^2(x)\right )}{a+b \coth (x)} \, dx}{3 b}\\ &=\frac{a \coth ^2(x)}{2 b^2}-\frac{\coth ^3(x)}{3 b}-\frac{\int \frac{\coth (x) \left (6 a^2-6 \left (a^2+b^2\right ) \coth ^2(x)\right )}{a+b \coth (x)} \, dx}{6 b^2}\\ &=-\frac{\left (a^2+b^2\right ) \coth (x)}{b^3}+\frac{a \coth ^2(x)}{2 b^2}-\frac{\coth ^3(x)}{3 b}-\frac{\int \frac{-6 a \left (a^2+b^2\right )-6 b^3 \coth (x)+6 a \left (a^2+b^2\right ) \coth ^2(x)}{a+b \coth (x)} \, dx}{6 b^3}\\ &=-\frac{b x}{a^2-b^2}-\frac{\left (a^2+b^2\right ) \coth (x)}{b^3}+\frac{a \coth ^2(x)}{2 b^2}-\frac{\coth ^3(x)}{3 b}+\frac{a \int \coth (x) \, dx}{a^2-b^2}+\frac{a^5 \int \frac{1-\coth ^2(x)}{a+b \coth (x)} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=-\frac{b x}{a^2-b^2}-\frac{\left (a^2+b^2\right ) \coth (x)}{b^3}+\frac{a \coth ^2(x)}{2 b^2}-\frac{\coth ^3(x)}{3 b}+\frac{a \log (\sinh (x))}{a^2-b^2}+\frac{a^5 \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \coth (x)\right )}{b^4 \left (a^2-b^2\right )}\\ &=-\frac{b x}{a^2-b^2}-\frac{\left (a^2+b^2\right ) \coth (x)}{b^3}+\frac{a \coth ^2(x)}{2 b^2}-\frac{\coth ^3(x)}{3 b}+\frac{a^5 \log (a+b \coth (x))}{b^4 \left (a^2-b^2\right )}+\frac{a \log (\sinh (x))}{a^2-b^2}\\ \end{align*}

Mathematica [A]  time = 0.286693, size = 108, normalized size = 1.15 \[ \frac{-3 a b^2 \left (a^2-b^2\right ) \text{csch}^2(x)+6 a \left (a^4-b^4\right ) \log (\sinh (x))+2 b \left (a^2-b^2\right ) \coth (x) \left (3 a^2+b^2 \text{csch}^2(x)+4 b^2\right )-6 a^5 \log (a \sinh (x)+b \cosh (x))+6 b^5 x}{6 b^4 (b-a) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^5/(a + b*Coth[x]),x]

[Out]

(6*b^5*x - 3*a*b^2*(a^2 - b^2)*Csch[x]^2 + 2*b*(a^2 - b^2)*Coth[x]*(3*a^2 + 4*b^2 + b^2*Csch[x]^2) + 6*a*(a^4
- b^4)*Log[Sinh[x]] - 6*a^5*Log[b*Cosh[x] + a*Sinh[x]])/(6*b^4*(-a + b)*(a + b))

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Maple [A]  time = 0.023, size = 96, normalized size = 1. \begin{align*} -{\frac{ \left ({\rm coth} \left (x\right ) \right ) ^{3}}{3\,b}}+{\frac{a \left ({\rm coth} \left (x\right ) \right ) ^{2}}{2\,{b}^{2}}}-{\frac{{a}^{2}{\rm coth} \left (x\right )}{{b}^{3}}}-{\frac{{\rm coth} \left (x\right )}{b}}-{\frac{\ln \left ( 1+{\rm coth} \left (x\right ) \right ) }{2\,a-2\,b}}-{\frac{\ln \left ({\rm coth} \left (x\right )-1 \right ) }{2\,b+2\,a}}+{\frac{{a}^{5}\ln \left ( a+b{\rm coth} \left (x\right ) \right ) }{{b}^{4} \left ( a+b \right ) \left ( a-b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^5/(a+b*coth(x)),x)

[Out]

-1/3*coth(x)^3/b+1/2*a*coth(x)^2/b^2-1/b^3*a^2*coth(x)-coth(x)/b-1/(2*a-2*b)*ln(1+coth(x))-1/(2*b+2*a)*ln(coth
(x)-1)+1/b^4*a^5/(a+b)/(a-b)*ln(a+b*coth(x))

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Maxima [A]  time = 1.26407, size = 228, normalized size = 2.43 \begin{align*} \frac{a^{5} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{2} b^{4} - b^{6}} + \frac{2 \,{\left (3 \, a^{2} + 4 \, b^{2} - 3 \,{\left (2 \, a^{2} + a b + 2 \, b^{2}\right )} e^{\left (-2 \, x\right )} + 3 \,{\left (a^{2} + a b + 2 \, b^{2}\right )} e^{\left (-4 \, x\right )}\right )}}{3 \,{\left (3 \, b^{3} e^{\left (-2 \, x\right )} - 3 \, b^{3} e^{\left (-4 \, x\right )} + b^{3} e^{\left (-6 \, x\right )} - b^{3}\right )}} + \frac{x}{a + b} - \frac{{\left (a^{3} + a b^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{b^{4}} - \frac{{\left (a^{3} + a b^{2}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5/(a+b*coth(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.09601, size = 3089, normalized size = 32.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5/(a+b*coth(x)),x, algorithm="fricas")

[Out]

-1/3*(3*(a*b^4 + b^5)*x*cosh(x)^6 + 18*(a*b^4 + b^5)*x*cosh(x)*sinh(x)^5 + 3*(a*b^4 + b^5)*x*sinh(x)^6 + 6*a^4
*b + 2*a^2*b^3 - 8*b^5 + 3*(2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 - 4*b^5 - 3*(a*b^4 + b^5)*x)*cosh(x)^4 +
3*(2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 - 4*b^5 + 15*(a*b^4 + b^5)*x*cosh(x)^2 - 3*(a*b^4 + b^5)*x)*sinh
(x)^4 + 12*(5*(a*b^4 + b^5)*x*cosh(x)^3 + (2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 - 4*b^5 - 3*(a*b^4 + b^5)
*x)*cosh(x))*sinh(x)^3 - 3*(4*a^4*b - 2*a^3*b^2 + 2*a*b^4 - 4*b^5 - 3*(a*b^4 + b^5)*x)*cosh(x)^2 + 3*(15*(a*b^
4 + b^5)*x*cosh(x)^4 - 4*a^4*b + 2*a^3*b^2 - 2*a*b^4 + 4*b^5 + 6*(2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 -
4*b^5 - 3*(a*b^4 + b^5)*x)*cosh(x)^2 + 3*(a*b^4 + b^5)*x)*sinh(x)^2 - 3*(a*b^4 + b^5)*x - 3*(a^5*cosh(x)^6 + 6
*a^5*cosh(x)*sinh(x)^5 + a^5*sinh(x)^6 - 3*a^5*cosh(x)^4 + 3*a^5*cosh(x)^2 - a^5 + 3*(5*a^5*cosh(x)^2 - a^5)*s
inh(x)^4 + 4*(5*a^5*cosh(x)^3 - 3*a^5*cosh(x))*sinh(x)^3 + 3*(5*a^5*cosh(x)^4 - 6*a^5*cosh(x)^2 + a^5)*sinh(x)
^2 + 6*(a^5*cosh(x)^5 - 2*a^5*cosh(x)^3 + a^5*cosh(x))*sinh(x))*log(2*(b*cosh(x) + a*sinh(x))/(cosh(x) - sinh(
x))) + 3*((a^5 - a*b^4)*cosh(x)^6 + 6*(a^5 - a*b^4)*cosh(x)*sinh(x)^5 + (a^5 - a*b^4)*sinh(x)^6 - a^5 + a*b^4
- 3*(a^5 - a*b^4)*cosh(x)^4 - 3*(a^5 - a*b^4 - 5*(a^5 - a*b^4)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^5 - a*b^4)*cosh(
x)^3 - 3*(a^5 - a*b^4)*cosh(x))*sinh(x)^3 + 3*(a^5 - a*b^4)*cosh(x)^2 + 3*(a^5 - a*b^4 + 5*(a^5 - a*b^4)*cosh(
x)^4 - 6*(a^5 - a*b^4)*cosh(x)^2)*sinh(x)^2 + 6*((a^5 - a*b^4)*cosh(x)^5 - 2*(a^5 - a*b^4)*cosh(x)^3 + (a^5 -
a*b^4)*cosh(x))*sinh(x))*log(2*sinh(x)/(cosh(x) - sinh(x))) + 6*(3*(a*b^4 + b^5)*x*cosh(x)^5 + 2*(2*a^4*b - 2*
a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 - 4*b^5 - 3*(a*b^4 + b^5)*x)*cosh(x)^3 - (4*a^4*b - 2*a^3*b^2 + 2*a*b^4 - 4*b^5
- 3*(a*b^4 + b^5)*x)*cosh(x))*sinh(x))/((a^2*b^4 - b^6)*cosh(x)^6 + 6*(a^2*b^4 - b^6)*cosh(x)*sinh(x)^5 + (a^2
*b^4 - b^6)*sinh(x)^6 - a^2*b^4 + b^6 - 3*(a^2*b^4 - b^6)*cosh(x)^4 - 3*(a^2*b^4 - b^6 - 5*(a^2*b^4 - b^6)*cos
h(x)^2)*sinh(x)^4 + 4*(5*(a^2*b^4 - b^6)*cosh(x)^3 - 3*(a^2*b^4 - b^6)*cosh(x))*sinh(x)^3 + 3*(a^2*b^4 - b^6)*
cosh(x)^2 + 3*(a^2*b^4 - b^6 + 5*(a^2*b^4 - b^6)*cosh(x)^4 - 6*(a^2*b^4 - b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^2*
b^4 - b^6)*cosh(x)^5 - 2*(a^2*b^4 - b^6)*cosh(x)^3 + (a^2*b^4 - b^6)*cosh(x))*sinh(x))

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Sympy [A]  time = 13.0737, size = 1013, normalized size = 10.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**5/(a+b*coth(x)),x)

[Out]

Piecewise((zoo*(x - 1/tanh(x) - 1/(3*tanh(x)**3)), Eq(a, 0) & Eq(b, 0)), ((x - 1/tanh(x) - 1/(3*tanh(x)**3))/b
, Eq(a, 0)), (27*x*tanh(x)**4/(6*b*tanh(x)**4 - 6*b*tanh(x)**3) - 27*x*tanh(x)**3/(6*b*tanh(x)**4 - 6*b*tanh(x
)**3) - 12*log(tanh(x) + 1)*tanh(x)**4/(6*b*tanh(x)**4 - 6*b*tanh(x)**3) + 12*log(tanh(x) + 1)*tanh(x)**3/(6*b
*tanh(x)**4 - 6*b*tanh(x)**3) + 12*log(tanh(x))*tanh(x)**4/(6*b*tanh(x)**4 - 6*b*tanh(x)**3) - 12*log(tanh(x))
*tanh(x)**3/(6*b*tanh(x)**4 - 6*b*tanh(x)**3) - 15*tanh(x)**4/(6*b*tanh(x)**4 - 6*b*tanh(x)**3) + 9*tanh(x)**2
/(6*b*tanh(x)**4 - 6*b*tanh(x)**3) + tanh(x)/(6*b*tanh(x)**4 - 6*b*tanh(x)**3) + 2/(6*b*tanh(x)**4 - 6*b*tanh(
x)**3), Eq(a, -b)), (3*x*tanh(x)**4/(6*b*tanh(x)**4 + 6*b*tanh(x)**3) + 3*x*tanh(x)**3/(6*b*tanh(x)**4 + 6*b*t
anh(x)**3) + 12*log(tanh(x) + 1)*tanh(x)**4/(6*b*tanh(x)**4 + 6*b*tanh(x)**3) + 12*log(tanh(x) + 1)*tanh(x)**3
/(6*b*tanh(x)**4 + 6*b*tanh(x)**3) - 12*log(tanh(x))*tanh(x)**4/(6*b*tanh(x)**4 + 6*b*tanh(x)**3) - 12*log(tan
h(x))*tanh(x)**3/(6*b*tanh(x)**4 + 6*b*tanh(x)**3) + 15*tanh(x)**4/(6*b*tanh(x)**4 + 6*b*tanh(x)**3) - 9*tanh(
x)**2/(6*b*tanh(x)**4 + 6*b*tanh(x)**3) + tanh(x)/(6*b*tanh(x)**4 + 6*b*tanh(x)**3) - 2/(6*b*tanh(x)**4 + 6*b*
tanh(x)**3), Eq(a, b)), ((x - log(tanh(x) + 1) + log(tanh(x)) - 1/(2*tanh(x)**2) - 1/(4*tanh(x)**4))/a, Eq(b,
0)), (6*a**5*log(tanh(x) + b/a)*tanh(x)**3/(6*a**2*b**4*tanh(x)**3 - 6*b**6*tanh(x)**3) - 6*a**5*log(tanh(x))*
tanh(x)**3/(6*a**2*b**4*tanh(x)**3 - 6*b**6*tanh(x)**3) - 6*a**4*b*tanh(x)**2/(6*a**2*b**4*tanh(x)**3 - 6*b**6
*tanh(x)**3) + 3*a**3*b**2*tanh(x)/(6*a**2*b**4*tanh(x)**3 - 6*b**6*tanh(x)**3) - 2*a**2*b**3/(6*a**2*b**4*tan
h(x)**3 - 6*b**6*tanh(x)**3) + 6*a*b**4*x*tanh(x)**3/(6*a**2*b**4*tanh(x)**3 - 6*b**6*tanh(x)**3) - 6*a*b**4*l
og(tanh(x) + 1)*tanh(x)**3/(6*a**2*b**4*tanh(x)**3 - 6*b**6*tanh(x)**3) + 6*a*b**4*log(tanh(x))*tanh(x)**3/(6*
a**2*b**4*tanh(x)**3 - 6*b**6*tanh(x)**3) - 3*a*b**4*tanh(x)/(6*a**2*b**4*tanh(x)**3 - 6*b**6*tanh(x)**3) - 6*
b**5*x*tanh(x)**3/(6*a**2*b**4*tanh(x)**3 - 6*b**6*tanh(x)**3) + 6*b**5*tanh(x)**2/(6*a**2*b**4*tanh(x)**3 - 6
*b**6*tanh(x)**3) + 2*b**5/(6*a**2*b**4*tanh(x)**3 - 6*b**6*tanh(x)**3), True))

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Giac [A]  time = 1.17641, size = 193, normalized size = 2.05 \begin{align*} \frac{a^{5} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} - a + b \right |}\right )}{a^{2} b^{4} - b^{6}} - \frac{x}{a - b} - \frac{{\left (a^{3} + a b^{2}\right )} \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right )}{b^{4}} - \frac{2 \,{\left (3 \, a^{2} b + 4 \, b^{3} + 3 \,{\left (a^{2} b - a b^{2} + 2 \, b^{3}\right )} e^{\left (4 \, x\right )} - 3 \,{\left (2 \, a^{2} b - a b^{2} + 2 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )}}{3 \, b^{4}{\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5/(a+b*coth(x)),x, algorithm="giac")

[Out]

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