\(\int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx\) [97]

Optimal result

Integrand size = 13, antiderivative size = 155 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (1+\coth (x))}{16 (a-b)^3}-\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}-\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )} \]

[Out]

-1/16*(3*a^2+9*a*b+8*b^2)*ln(1-coth(x))/(a+b)^3+1/16*(3*a^2-9*a*b+8*b^2)*ln(1+coth(x))/(a-b)^3-b^5*ln(a+b*coth
(x))/(a^2-b^2)^3-1/8*(4*b^3-a*(7-3*a^2/b^2)*b^2*coth(x))*sinh(x)^2/(a^2-b^2)^2-1/4*(b-a*coth(x))*sinh(x)^4/(a^
2-b^2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3587, 755, 837, 815} \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (\coth (x)+1)}{16 (a-b)^3}-\frac {\sinh ^4(x) (b-a \coth (x))}{4 \left (a^2-b^2\right )}-\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}-\frac {\sinh ^2(x) \left (4 b^3-a b^2 \left (7-\frac {3 a^2}{b^2}\right ) \coth (x)\right )}{8 \left (a^2-b^2\right )^2} \]

[In]

Int[Sinh[x]^4/(a + b*Coth[x]),x]

[Out]

-1/16*((3*a^2 + 9*a*b + 8*b^2)*Log[1 - Coth[x]])/(a + b)^3 + ((3*a^2 - 9*a*b + 8*b^2)*Log[1 + Coth[x]])/(16*(a
- b)^3) - (b^5*Log[a + b*Coth[x]])/(a^2 - b^2)^3 - ((4*b^3 - a*(7 - (3*a^2)/b^2)*b^2*Coth[x])*Sinh[x]^2)/(8*(
a^2 - b^2)^2) - ((b - a*Coth[x])*Sinh[x]^4)/(4*(a^2 - b^2))

Rule 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 3587

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{(a+x) \left (1-\frac {x^2}{b^2}\right )^3} \, dx,x,b \coth (x)\right )}{b} \\ & = -\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}+\frac {b \text {Subst}\left (\int \frac {-4+\frac {3 a^2}{b^2}+\frac {3 a x}{b^2}}{(a+x) \left (1-\frac {x^2}{b^2}\right )^2} \, dx,x,b \coth (x)\right )}{4 \left (a^2-b^2\right )} \\ & = -\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {b^5 \text {Subst}\left (\int \frac {-\frac {3 a^4-7 a^2 b^2+8 b^4}{b^6}+\frac {a \left (7-\frac {3 a^2}{b^2}\right ) x}{b^4}}{(a+x) \left (1-\frac {x^2}{b^2}\right )} \, dx,x,b \coth (x)\right )}{8 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {b^5 \text {Subst}\left (\int \left (-\frac {(a-b)^2 \left (3 a^2+9 a b+8 b^2\right )}{2 b^5 (a+b) (b-x)}+\frac {8}{(a-b) (a+b) (a+x)}-\frac {(a+b)^2 \left (3 a^2-9 a b+8 b^2\right )}{2 (a-b) b^5 (b+x)}\right ) \, dx,x,b \coth (x)\right )}{8 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (1+\coth (x))}{16 (a-b)^3}-\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}-\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.01 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=\frac {12 a^5 x-40 a^3 b^2 x+60 a b^4 x+4 b \left (a^4-4 a^2 b^2+3 b^4\right ) \cosh (2 x)-b \left (a^2-b^2\right )^2 \cosh (4 x)-32 b^5 \log (b \cosh (x)+a \sinh (x))-8 a^5 \sinh (2 x)+24 a^3 b^2 \sinh (2 x)-16 a b^4 \sinh (2 x)+a^5 \sinh (4 x)-2 a^3 b^2 \sinh (4 x)+a b^4 \sinh (4 x)}{32 (a-b)^3 (a+b)^3} \]

[In]

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

[Out]

(12*a^5*x - 40*a^3*b^2*x + 60*a*b^4*x + 4*b*(a^4 - 4*a^2*b^2 + 3*b^4)*Cosh[2*x] - b*(a^2 - b^2)^2*Cosh[4*x] -
32*b^5*Log[b*Cosh[x] + a*Sinh[x]] - 8*a^5*Sinh[2*x] + 24*a^3*b^2*Sinh[2*x] - 16*a*b^4*Sinh[2*x] + a^5*Sinh[4*x
] - 2*a^3*b^2*Sinh[4*x] + a*b^4*Sinh[4*x])/(32*(a - b)^3*(a + b)^3)

Maple [A] (verified)

Time = 5.86 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.24

method	result	size
risch	\(\frac {3 a^{2} x}{8 \left (a +b \right )^{3}}+\frac {9 a x b}{8 \left (a +b \right )^{3}}+\frac {x \,b^{2}}{\left (a +b \right )^{3}}+\frac {{\mathrm e}^{4 x}}{64 a +64 b}-\frac {{\mathrm e}^{2 x} a}{8 \left (a +b \right )^{2}}-\frac {3 \,{\mathrm e}^{2 x} b}{16 \left (a +b \right )^{2}}+\frac {{\mathrm e}^{-2 x} a}{8 \left (a -b \right )^{2}}-\frac {3 \,{\mathrm e}^{-2 x} b}{16 \left (a -b \right )^{2}}-\frac {{\mathrm e}^{-4 x}}{64 \left (a -b \right )}+\frac {2 b^{5} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {b^{5} \ln \left ({\mathrm e}^{2 x}-\frac {a -b}{a +b}\right )}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}\)	\(192\)
default	\(-\frac {16}{\left (64 a -64 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {64}{\left (128 a -128 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {-a +3 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 a -5 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (3 a^{2}-9 a b +8 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 \left (a -b \right )^{3}}-\frac {b^{5} \ln \left (\tanh \left (\frac {x}{2}\right )^{2} b +2 a \tanh \left (\frac {x}{2}\right )+b \right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}+\frac {16}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {64}{\left (128 a +128 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a +3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3 a +5 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-3 a^{2}-9 a b -8 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}\)	\(263\)

[In]

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

[Out]

3/8*a^2*x/(a+b)^3+9/8*a*x/(a+b)^3*b+x/(a+b)^3*b^2+1/64/(a+b)*exp(4*x)-1/8/(a+b)^2*exp(2*x)*a-3/16/(a+b)^2*exp(
2*x)*b+1/8/(a-b)^2*exp(-2*x)*a-3/16/(a-b)^2*exp(-2*x)*b-1/64/(a-b)*exp(-4*x)+2*b^5/(a^6-3*a^4*b^2+3*a^2*b^4-b^
6)*x-b^5/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*ln(exp(2*x)-(a-b)/(a+b))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1279 vs. \(2 (147) = 294\).

Time = 0.27 (sec) , antiderivative size = 1279, normalized size of antiderivative = 8.25 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/64*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^8 + 8*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 +
a*b^4 - b^5)*cosh(x)*sinh(x)^7 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^8 - 4*(2*a^5 - a
^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4*a*b^4 - 3*b^5)*cosh(x)^6 - 4*(2*a^5 - a^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4*a*b^4
- 3*b^5 - 7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^6 + 8*(3*a^5 - 10*a^3*b^2
+ 15*a*b^4 + 8*b^5)*x*cosh(x)^4 + 8*(7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(2*a^
5 - a^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4*a*b^4 - 3*b^5)*cosh(x))*sinh(x)^5 - a^5 - a^4*b + 2*a^3*b^2 + 2*a^2*b^3
- a*b^4 - b^5 + 2*(35*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 - 30*(2*a^5 - a^4*b - 6*a^
3*b^2 + 4*a^2*b^3 + 4*a*b^4 - 3*b^5)*cosh(x)^2 + 4*(3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^5)*x)*sinh(x)^4 + 8*(7
*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^5 - 10*(2*a^5 - a^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4
*a*b^4 - 3*b^5)*cosh(x)^3 + 4*(3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^5)*x*cosh(x))*sinh(x)^3 + 4*(2*a^5 + a^4*b
- 6*a^3*b^2 - 4*a^2*b^3 + 4*a*b^4 + 3*b^5)*cosh(x)^2 + 4*(7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5
)*cosh(x)^6 + 2*a^5 + a^4*b - 6*a^3*b^2 - 4*a^2*b^3 + 4*a*b^4 + 3*b^5 - 15*(2*a^5 - a^4*b - 6*a^3*b^2 + 4*a^2*
b^3 + 4*a*b^4 - 3*b^5)*cosh(x)^4 + 12*(3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^5)*x*cosh(x)^2)*sinh(x)^2 - 64*(b^5
*cosh(x)^4 + 4*b^5*cosh(x)^3*sinh(x) + 6*b^5*cosh(x)^2*sinh(x)^2 + 4*b^5*cosh(x)*sinh(x)^3 + b^5*sinh(x)^4)*lo
g(2*(b*cosh(x) + a*sinh(x))/(cosh(x) - sinh(x))) + 8*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh
(x)^7 - 3*(2*a^5 - a^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4*a*b^4 - 3*b^5)*cosh(x)^5 + 4*(3*a^5 - 10*a^3*b^2 + 15*a*b
^4 + 8*b^5)*x*cosh(x)^3 + (2*a^5 + a^4*b - 6*a^3*b^2 - 4*a^2*b^3 + 4*a*b^4 + 3*b^5)*cosh(x))*sinh(x))/((a^6 -
3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^4 + 4*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3*sinh(x) + 6*(a^6 - 3*
a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x)^2 + 4*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^3 + (a^
6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.07 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=-\frac {b^{5} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} + 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (4 \, {\left (2 \, a + 3 \, b\right )} e^{\left (-2 \, x\right )} - a - b\right )} e^{\left (4 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {4 \, {\left (2 \, a - 3 \, b\right )} e^{\left (-2 \, x\right )} - {\left (a - b\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} \]

[In]

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

[Out]

-b^5*log(-(a - b)*e^(-2*x) + a + b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + 1/8*(3*a^2 + 9*a*b + 8*b^2)*x/(a^3 +
3*a^2*b + 3*a*b^2 + b^3) - 1/64*(4*(2*a + 3*b)*e^(-2*x) - a - b)*e^(4*x)/(a^2 + 2*a*b + b^2) + 1/64*(4*(2*a -
3*b)*e^(-2*x) - (a - b)*e^(-4*x))/(a^2 - 2*a*b + b^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.48 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=-\frac {b^{5} \log \left ({\left | -a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} - 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {{\left (18 \, a^{2} e^{\left (4 \, x\right )} - 54 \, a b e^{\left (4 \, x\right )} + 48 \, b^{2} e^{\left (4 \, x\right )} - 8 \, a^{2} e^{\left (2 \, x\right )} + 20 \, a b e^{\left (2 \, x\right )} - 12 \, b^{2} e^{\left (2 \, x\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 8 \, a e^{\left (2 \, x\right )} - 12 \, b e^{\left (2 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \]

[In]

integrate(sinh(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 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + 1/8*(3*a^2 - 9*a*b + 8*b^2
)*x/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 1/64*(18*a^2*e^(4*x) - 54*a*b*e^(4*x) + 48*b^2*e^(4*x) - 8*a^2*e^(2*x) +
20*a*b*e^(2*x) - 12*b^2*e^(2*x) + a^2 - 2*a*b + b^2)*e^(-4*x)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 1/64*(a*e^(4*
x) + b*e^(4*x) - 8*a*e^(2*x) - 12*b*e^(2*x))/(a^2 + 2*a*b + b^2)

Mupad [B] (verification not implemented)

Time = 2.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.92 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=\frac {{\mathrm {e}}^{4\,x}}{64\,a+64\,b}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a-64\,b}+\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a-3\,b\right )}{16\,{\left (a-b\right )}^2}-\frac {b^5\,\ln \left (b-a+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {x\,\left (3\,a^2-9\,a\,b+8\,b^2\right )}{8\,{\left (a-b\right )}^3}-\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a+3\,b\right )}{16\,{\left (a+b\right )}^2} \]

[In]

int(sinh(x)^4/(a + b*coth(x)),x)

[Out]

exp(4*x)/(64*a + 64*b) - exp(-4*x)/(64*a - 64*b) + (exp(-2*x)*(2*a - 3*b))/(16*(a - b)^2) - (b^5*log(b - a + a
*exp(2*x) + b*exp(2*x)))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (x*(3*a^2 - 9*a*b + 8*b^2))/(8*(a - b)^3) - (ex
p(2*x)*(2*a + 3*b))/(16*(a + b)^2)