\(\int \frac {\coth ^4(x)}{a+b \tanh (x)} \, dx\) [142]

Optimal result

Integrand size = 13, antiderivative size = 97 \[ \int \frac {\coth ^4(x)}{a+b \tanh (x)} \, dx=\frac {a x}{a^2-b^2}-\frac {\left (a^2+b^2\right ) \coth (x)}{a^3}+\frac {b \coth ^2(x)}{2 a^2}-\frac {\coth ^3(x)}{3 a}-\frac {b \left (a^2+b^2\right ) \log (\sinh (x))}{a^4}-\frac {b^5 \log (a \cosh (x)+b \sinh (x))}{a^4 \left (a^2-b^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3650, 3730, 3731, 3732, 3611, 3556} \[ \int \frac {\coth ^4(x)}{a+b \tanh (x)} \, dx=\frac {a x}{a^2-b^2}+\frac {b \coth ^2(x)}{2 a^2}-\frac {b \left (a^2+b^2\right ) \log (\sinh (x))}{a^4}-\frac {b^5 \log (a \cosh (x)+b \sinh (x))}{a^4 \left (a^2-b^2\right )}-\frac {\left (a^2+b^2\right ) \coth (x)}{a^3}-\frac {\coth ^3(x)}{3 a} \]

[In]

Int[Coth[x]^4/(a + b*Tanh[x]),x]

[Out]

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

Rule 3556

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

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3650

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 + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*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] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3730

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3731

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

Rule 3732

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\coth ^3(x)}{3 a}-\frac {i \int \frac {\coth ^3(x) \left (-3 i b+3 i a \tanh (x)+3 i b \tanh ^2(x)\right )}{a+b \tanh (x)} \, dx}{3 a} \\ & = \frac {b \coth ^2(x)}{2 a^2}-\frac {\coth ^3(x)}{3 a}-\frac {\int \frac {\coth ^2(x) \left (-6 \left (a^2+b^2\right )+6 b^2 \tanh ^2(x)\right )}{a+b \tanh (x)} \, dx}{6 a^2} \\ & = -\frac {\left (a^2+b^2\right ) \coth (x)}{a^3}+\frac {b \coth ^2(x)}{2 a^2}-\frac {\coth ^3(x)}{3 a}+\frac {i \int \frac {\coth (x) \left (6 i b \left (a^2+b^2\right )-6 i a^3 \tanh (x)-6 i b \left (a^2+b^2\right ) \tanh ^2(x)\right )}{a+b \tanh (x)} \, dx}{6 a^3} \\ & = \frac {a x}{a^2-b^2}-\frac {\left (a^2+b^2\right ) \coth (x)}{a^3}+\frac {b \coth ^2(x)}{2 a^2}-\frac {\coth ^3(x)}{3 a}-\frac {\left (i b^5\right ) \int \frac {-i b-i a \tanh (x)}{a+b \tanh (x)} \, dx}{a^4 \left (a^2-b^2\right )}-\frac {\left (b \left (a^2+b^2\right )\right ) \int \coth (x) \, dx}{a^4} \\ & = \frac {a x}{a^2-b^2}-\frac {\left (a^2+b^2\right ) \coth (x)}{a^3}+\frac {b \coth ^2(x)}{2 a^2}-\frac {\coth ^3(x)}{3 a}-\frac {b \left (a^2+b^2\right ) \log (\sinh (x))}{a^4}-\frac {b^5 \log (a \cosh (x)+b \sinh (x))}{a^4 \left (a^2-b^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.95 \[ \int \frac {\coth ^4(x)}{a+b \tanh (x)} \, dx=\frac {1}{6} \left (-\frac {6 \left (a^2+b^2\right ) \coth (x)}{a^3}+\frac {3 b \coth ^2(x)}{a^2}-\frac {2 \coth ^3(x)}{a}-\frac {3 \log (1-\coth (x))}{a+b}+\frac {3 \log (1+\coth (x))}{a-b}+\frac {6 b^5 \log (b+a \coth (x))}{a^4 \left (-a^2+b^2\right )}\right ) \]

[In]

Integrate[Coth[x]^4/(a + b*Tanh[x]),x]

[Out]

((-6*(a^2 + b^2)*Coth[x])/a^3 + (3*b*Coth[x]^2)/a^2 - (2*Coth[x]^3)/a - (3*Log[1 - Coth[x]])/(a + b) + (3*Log[
1 + Coth[x]])/(a - b) + (6*b^5*Log[b + a*Coth[x]])/(a^4*(-a^2 + b^2)))/6

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}-\frac {b^{5} \ln \left (a +b \tanh \left (x \right )\right )}{a^{4} \left (a +b \right ) \left (a -b \right )}+\frac {b}{2 a^{2} \tanh \left (x \right )^{2}}+\frac {-a^{2}-b^{2}}{a^{3} \tanh \left (x \right )}-\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tanh \left (x \right )\right )}{a^{4}}-\frac {1}{3 a \tanh \left (x \right )^{3}}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 a +2 b}\)	\(114\)
default	\(\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}-\frac {b^{5} \ln \left (a +b \tanh \left (x \right )\right )}{a^{4} \left (a +b \right ) \left (a -b \right )}+\frac {b}{2 a^{2} \tanh \left (x \right )^{2}}+\frac {-a^{2}-b^{2}}{a^{3} \tanh \left (x \right )}-\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tanh \left (x \right )\right )}{a^{4}}-\frac {1}{3 a \tanh \left (x \right )^{3}}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 a +2 b}\)	\(114\)
parallelrisch	\(\frac {-6 \ln \left (a +b \tanh \left (x \right )\right ) b^{5}+6 \ln \left (1-\tanh \left (x \right )\right ) a^{4} b +\left (-6 a^{4} b +6 b^{5}\right ) \ln \left (\tanh \left (x \right )\right )+\left (-2 a^{5}+2 a^{3} b^{2}\right ) \coth \left (x \right )^{3}+\left (3 a^{4} b -3 a^{2} b^{3}\right ) \coth \left (x \right )^{2}+\left (-6 a^{5}+6 a \,b^{4}\right ) \coth \left (x \right )+6 a^{4} x \left (a +b \right )}{6 a^{6}-6 a^{4} b^{2}}\)	\(123\)
risch	\(\frac {x}{a +b}+\frac {2 x b}{a^{2}}+\frac {2 x \,b^{3}}{a^{4}}+\frac {2 x \,b^{5}}{a^{4} \left (a^{2}-b^{2}\right )}-\frac {2 \left (6 a^{2} {\mathrm e}^{4 x}-3 a b \,{\mathrm e}^{4 x}+3 b^{2} {\mathrm e}^{4 x}-6 a^{2} {\mathrm e}^{2 x}+3 b \,{\mathrm e}^{2 x} a -6 b^{2} {\mathrm e}^{2 x}+4 a^{2}+3 b^{2}\right )}{3 a^{3} \left ({\mathrm e}^{2 x}-1\right )^{3}}-\frac {b \ln \left ({\mathrm e}^{2 x}-1\right )}{a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 x}-1\right )}{a^{4}}-\frac {b^{5} \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{4} \left (a^{2}-b^{2}\right )}\)	\(185\)

[In]

int(coth(x)^4/(a+b*tanh(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1299 vs. \(2 (93) = 186\).

Time = 0.28 (sec) , antiderivative size = 1299, normalized size of antiderivative = 13.39 \[ \int \frac {\coth ^4(x)}{a+b \tanh (x)} \, dx=\text {Too large to display} \]

[In]

integrate(coth(x)^4/(a+b*tanh(x)),x, algorithm="fricas")

[Out]

1/3*(3*(a^5 + a^4*b)*x*cosh(x)^6 + 18*(a^5 + a^4*b)*x*cosh(x)*sinh(x)^5 + 3*(a^5 + a^4*b)*x*sinh(x)^6 - 8*a^5
+ 2*a^3*b^2 + 6*a*b^4 - 3*(4*a^5 - 2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 + 3*(a^5 + a^4*b)*x)*cosh(x)^4 -
3*(4*a^5 - 2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 - 15*(a^5 + a^4*b)*x*cosh(x)^2 + 3*(a^5 + a^4*b)*x)*sinh(
x)^4 + 12*(5*(a^5 + a^4*b)*x*cosh(x)^3 - (4*a^5 - 2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 + 3*(a^5 + a^4*b)*
x)*cosh(x))*sinh(x)^3 + 3*(4*a^5 - 2*a^4*b + 2*a^2*b^3 - 4*a*b^4 + 3*(a^5 + a^4*b)*x)*cosh(x)^2 + 3*(15*(a^5 +
a^4*b)*x*cosh(x)^4 + 4*a^5 - 2*a^4*b + 2*a^2*b^3 - 4*a*b^4 - 6*(4*a^5 - 2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 - 2*a
*b^4 + 3*(a^5 + a^4*b)*x)*cosh(x)^2 + 3*(a^5 + a^4*b)*x)*sinh(x)^2 - 3*(a^5 + a^4*b)*x - 3*(b^5*cosh(x)^6 + 6*
b^5*cosh(x)*sinh(x)^5 + b^5*sinh(x)^6 - 3*b^5*cosh(x)^4 + 3*b^5*cosh(x)^2 - b^5 + 3*(5*b^5*cosh(x)^2 - b^5)*si
nh(x)^4 + 4*(5*b^5*cosh(x)^3 - 3*b^5*cosh(x))*sinh(x)^3 + 3*(5*b^5*cosh(x)^4 - 6*b^5*cosh(x)^2 + b^5)*sinh(x)^
2 + 6*(b^5*cosh(x)^5 - 2*b^5*cosh(x)^3 + b^5*cosh(x))*sinh(x))*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x
))) - 3*((a^4*b - b^5)*cosh(x)^6 + 6*(a^4*b - b^5)*cosh(x)*sinh(x)^5 + (a^4*b - b^5)*sinh(x)^6 - a^4*b + b^5 -
3*(a^4*b - b^5)*cosh(x)^4 - 3*(a^4*b - b^5 - 5*(a^4*b - b^5)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^4*b - b^5)*cosh(x
)^3 - 3*(a^4*b - b^5)*cosh(x))*sinh(x)^3 + 3*(a^4*b - b^5)*cosh(x)^2 + 3*(a^4*b - b^5 + 5*(a^4*b - b^5)*cosh(x
)^4 - 6*(a^4*b - b^5)*cosh(x)^2)*sinh(x)^2 + 6*((a^4*b - b^5)*cosh(x)^5 - 2*(a^4*b - b^5)*cosh(x)^3 + (a^4*b -
b^5)*cosh(x))*sinh(x))*log(2*sinh(x)/(cosh(x) - sinh(x))) + 6*(3*(a^5 + a^4*b)*x*cosh(x)^5 - 2*(4*a^5 - 2*a^4
*b - 2*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 + 3*(a^5 + a^4*b)*x)*cosh(x)^3 + (4*a^5 - 2*a^4*b + 2*a^2*b^3 - 4*a*b^4 +
3*(a^5 + a^4*b)*x)*cosh(x))*sinh(x))/((a^6 - a^4*b^2)*cosh(x)^6 + 6*(a^6 - a^4*b^2)*cosh(x)*sinh(x)^5 + (a^6
- a^4*b^2)*sinh(x)^6 - a^6 + a^4*b^2 - 3*(a^6 - a^4*b^2)*cosh(x)^4 - 3*(a^6 - a^4*b^2 - 5*(a^6 - a^4*b^2)*cosh
(x)^2)*sinh(x)^4 + 4*(5*(a^6 - a^4*b^2)*cosh(x)^3 - 3*(a^6 - a^4*b^2)*cosh(x))*sinh(x)^3 + 3*(a^6 - a^4*b^2)*c
osh(x)^2 + 3*(a^6 - a^4*b^2 + 5*(a^6 - a^4*b^2)*cosh(x)^4 - 6*(a^6 - a^4*b^2)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 -
a^4*b^2)*cosh(x)^5 - 2*(a^6 - a^4*b^2)*cosh(x)^3 + (a^6 - a^4*b^2)*cosh(x))*sinh(x))

Sympy [F]

\[ \int \frac {\coth ^4(x)}{a+b \tanh (x)} \, dx=\int \frac {\coth ^{4}{\left (x \right )}}{a + b \tanh {\left (x \right )}}\, dx \]

[In]

integrate(coth(x)**4/(a+b*tanh(x)),x)

[Out]

Integral(coth(x)**4/(a + b*tanh(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.78 \[ \int \frac {\coth ^4(x)}{a+b \tanh (x)} \, dx=-\frac {b^{5} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{6} - a^{4} b^{2}} + \frac {2 \, {\left (4 \, a^{2} + 3 \, b^{2} - 3 \, {\left (2 \, a^{2} + a b + 2 \, b^{2}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (2 \, a^{2} + a b + b^{2}\right )} e^{\left (-4 \, x\right )}\right )}}{3 \, {\left (3 \, a^{3} e^{\left (-2 \, x\right )} - 3 \, a^{3} e^{\left (-4 \, x\right )} + a^{3} e^{\left (-6 \, x\right )} - a^{3}\right )}} + \frac {x}{a + b} - \frac {{\left (a^{2} b + b^{3}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{a^{4}} - \frac {{\left (a^{2} b + b^{3}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{a^{4}} \]

[In]

integrate(coth(x)^4/(a+b*tanh(x)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.46 \[ \int \frac {\coth ^4(x)}{a+b \tanh (x)} \, dx=-\frac {b^{5} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - a^{4} b^{2}} + \frac {x}{a - b} - \frac {{\left (a^{2} b + b^{3}\right )} \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right )}{a^{4}} - \frac {2 \, {\left (4 \, a^{3} + 3 \, a b^{2} + 3 \, {\left (2 \, a^{3} - a^{2} b + a b^{2}\right )} e^{\left (4 \, x\right )} - 3 \, {\left (2 \, a^{3} - a^{2} b + 2 \, a b^{2}\right )} e^{\left (2 \, x\right )}\right )}}{3 \, a^{4} {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \]

[In]

integrate(coth(x)^4/(a+b*tanh(x)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.10 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.68 \[ \int \frac {\coth ^4(x)}{a+b \tanh (x)} \, dx=\frac {x}{a-b}-\frac {8}{3\,a\,\left (3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1\right )}-\frac {b^5\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^6-a^4\,b^2}-\frac {\ln \left ({\mathrm {e}}^{2\,x}-1\right )\,\left (a^2\,b+b^3\right )}{a^4}-\frac {2\,\left (2\,a^3+a^2\,b+b^3\right )}{a^3\,\left (a+b\right )\,\left ({\mathrm {e}}^{2\,x}-1\right )}-\frac {2\,\left (2\,a^2+a\,b-b^2\right )}{a^2\,\left (a+b\right )\,\left ({\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1\right )} \]

[In]

int(coth(x)^4/(a + b*tanh(x)),x)

[Out]

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