3.140 \(\int \frac{\tanh ^4(x)}{a+b \coth (x)} \, dx\)
Optimal. Leaf size=97 \[ \frac{a x}{a^2-b^2}-\frac{\left (a^2+b^2\right ) \tanh (x)}{a^3}-\frac{b \left (a^2+b^2\right ) \log (\cosh (x))}{a^4}-\frac{b^5 \log (a \sinh (x)+b \cosh (x))}{a^4 \left (a^2-b^2\right )}+\frac{b \tanh ^2(x)}{2 a^2}-\frac{\tanh ^3(x)}{3 a} \]
[Out]
(a*x)/(a^2 - b^2) - (b*(a^2 + b^2)*Log[Cosh[x]])/a^4 - (b^5*Log[b*Cosh[x] + a*Sinh[x]])/(a^4*(a^2 - b^2)) - ((
a^2 + b^2)*Tanh[x])/a^3 + (b*Tanh[x]^2)/(2*a^2) - Tanh[x]^3/(3*a)
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Rubi [A] time = 0.525196, antiderivative size = 97, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used =
{3569, 3649, 3650, 3651, 3530, 3475} \[ \frac{a x}{a^2-b^2}-\frac{\left (a^2+b^2\right ) \tanh (x)}{a^3}-\frac{b \left (a^2+b^2\right ) \log (\cosh (x))}{a^4}-\frac{b^5 \log (a \sinh (x)+b \cosh (x))}{a^4 \left (a^2-b^2\right )}+\frac{b \tanh ^2(x)}{2 a^2}-\frac{\tanh ^3(x)}{3 a} \]
Antiderivative was successfully verified.
[In]
Int[Tanh[x]^4/(a + b*Coth[x]),x]
[Out]
(a*x)/(a^2 - b^2) - (b*(a^2 + b^2)*Log[Cosh[x]])/a^4 - (b^5*Log[b*Cosh[x] + a*Sinh[x]])/(a^4*(a^2 - b^2)) - ((
a^2 + b^2)*Tanh[x])/a^3 + (b*Tanh[x]^2)/(2*a^2) - Tanh[x]^3/(3*a)
Rule 3569
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 3649
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*T
an[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 3650
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 3651
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]
Rule 3530
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), 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 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{\tanh ^4(x)}{a+b \coth (x)} \, dx &=-\frac{\tanh ^3(x)}{3 a}-\frac{i \int \frac{\left (-3 i b+3 i a \coth (x)+3 i b \coth ^2(x)\right ) \tanh ^3(x)}{a+b \coth (x)} \, dx}{3 a}\\ &=\frac{b \tanh ^2(x)}{2 a^2}-\frac{\tanh ^3(x)}{3 a}-\frac{\int \frac{\left (-6 \left (a^2+b^2\right )+6 b^2 \coth ^2(x)\right ) \tanh ^2(x)}{a+b \coth (x)} \, dx}{6 a^2}\\ &=-\frac{\left (a^2+b^2\right ) \tanh (x)}{a^3}+\frac{b \tanh ^2(x)}{2 a^2}-\frac{\tanh ^3(x)}{3 a}+\frac{i \int \frac{\left (6 i b \left (a^2+b^2\right )-6 i a^3 \coth (x)-6 i b \left (a^2+b^2\right ) \coth ^2(x)\right ) \tanh (x)}{a+b \coth (x)} \, dx}{6 a^3}\\ &=\frac{a x}{a^2-b^2}-\frac{\left (a^2+b^2\right ) \tanh (x)}{a^3}+\frac{b \tanh ^2(x)}{2 a^2}-\frac{\tanh ^3(x)}{3 a}-\frac{\left (i b^5\right ) \int \frac{-i b-i a \coth (x)}{a+b \coth (x)} \, dx}{a^4 \left (a^2-b^2\right )}-\frac{\left (b \left (a^2+b^2\right )\right ) \int \tanh (x) \, dx}{a^4}\\ &=\frac{a x}{a^2-b^2}-\frac{b \left (a^2+b^2\right ) \log (\cosh (x))}{a^4}-\frac{b^5 \log (b \cosh (x)+a \sinh (x))}{a^4 \left (a^2-b^2\right )}-\frac{\left (a^2+b^2\right ) \tanh (x)}{a^3}+\frac{b \tanh ^2(x)}{2 a^2}-\frac{\tanh ^3(x)}{3 a}\\ \end{align*}
Mathematica [A] time = 0.327879, size = 105, normalized size = 1.08 \[ \frac{\left (2 a^3 b^2-8 a^5+6 a b^4\right ) \tanh (x)+6 \left (b^5-a^4 b\right ) \log (\cosh (x))+a^2 \left (a^2-b^2\right ) \text{sech}^2(x) (2 a \tanh (x)-3 b)+6 a^5 x-6 b^5 \log (a \sinh (x)+b \cosh (x))}{6 a^4 (a-b) (a+b)} \]
Antiderivative was successfully verified.
[In]
Integrate[Tanh[x]^4/(a + b*Coth[x]),x]
[Out]
(6*a^5*x + 6*(-(a^4*b) + b^5)*Log[Cosh[x]] - 6*b^5*Log[b*Cosh[x] + a*Sinh[x]] + (-8*a^5 + 2*a^3*b^2 + 6*a*b^4)
*Tanh[x] + a^2*(a^2 - b^2)*Sech[x]^2*(-3*b + 2*a*Tanh[x]))/(6*a^4*(a - b)*(a + b))
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Maple [B] time = 0.056, size = 283, normalized size = 2.9 \begin{align*} 64\,{\frac{\ln \left ( \tanh \left ( x/2 \right ) +1 \right ) }{64\,a-64\,b}}-64\,{\frac{\ln \left ( \tanh \left ( x/2 \right ) -1 \right ) }{64\,a+64\,b}}-2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{5}}{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{5}{b}^{2}}{{a}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{b \left ( \tanh \left ( x/2 \right ) \right ) ^{4}}{{a}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{20}{3\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}-4\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{3}{b}^{2}}{{a}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b}{{a}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-2\,{\frac{\tanh \left ( x/2 \right ) }{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-2\,{\frac{\tanh \left ( x/2 \right ){b}^{2}}{{a}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{b}{{a}^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }-{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }-{\frac{{b}^{5}}{ \left ( a+b \right ) \left ( a-b \right ){a}^{4}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b+2\,a\tanh \left ( x/2 \right ) +b \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(x)^4/(a+b*coth(x)),x)
[Out]
64/(64*a-64*b)*ln(tanh(1/2*x)+1)-64/(64*a+64*b)*ln(tanh(1/2*x)-1)-2/a/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^5-2/a^3/
(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^5*b^2+2/a^2/(tanh(1/2*x)^2+1)^3*b*tanh(1/2*x)^4-20/3/a/(tanh(1/2*x)^2+1)^3*tan
h(1/2*x)^3-4/a^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^3*b^2+2/a^2/(tanh(1/2*x)^2+1)^3*b*tanh(1/2*x)^2-2/a/(tanh(1/2
*x)^2+1)^3*tanh(1/2*x)-2/a^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)*b^2-1/a^2*b*ln(tanh(1/2*x)^2+1)-1/a^4*ln(tanh(1/2
*x)^2+1)*b^3-b^5/(a+b)/(a-b)/a^4*ln(tanh(1/2*x)^2*b+2*a*tanh(1/2*x)+b)
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Maxima [A] time = 1.75174, size = 197, normalized size = 2.03 \begin{align*} -\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 (-2 \, x\right )} + 1\right )}{a^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)^4/(a+b*coth(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^(-2*x) + 1)/a^4
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Fricas [B] time = 3.18172, size = 3087, normalized size = 31.82 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)^4/(a+b*coth(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*(b*cosh(x) + a*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*cosh(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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{4}{\left (x \right )}}{a + b \coth{\left (x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)**4/(a+b*coth(x)),x)
[Out]
Integral(tanh(x)**4/(a + b*coth(x)), x)
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Giac [A] time = 1.13076, size = 190, normalized size = 1.96 \begin{align*} -\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 (e^{\left (2 \, x\right )} + 1\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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)^4/(a+b*coth(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(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)