\(\int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx\) [168]

Optimal result

Integrand size = 13, antiderivative size = 104 \[ \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}+\frac {\left (2 \left (a^2-b^2\right )-a b \cosh (x)\right ) \sinh (x)}{2 b^3}+\frac {\sinh ^3(x)}{3 b} \] Output:

-1/2*a*(2*a^2-3*b^2)*x/b^4+2*(a-b)^(3/2)*(a+b)^(3/2)*arctanh((a-b)^(1/2)*t 
anh(1/2*x)/(a+b)^(1/2))/b^4+1/2*(2*a^2-2*b^2-a*b*cosh(x))*sinh(x)/b^3+1/3* 
sinh(x)^3/b
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.91 \[ \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx=\frac {-12 a^3 x+18 a b^2 x-24 \left (-a^2+b^2\right )^{3/2} \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )+12 a^2 b \sinh (x)-15 b^3 \sinh (x)-3 a b^2 \sinh (2 x)+b^3 \sinh (3 x)}{12 b^4} \] Input:

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

Output:

(-12*a^3*x + 18*a*b^2*x - 24*(-a^2 + b^2)^(3/2)*ArcTan[((a - b)*Tanh[x/2]) 
/Sqrt[-a^2 + b^2]] + 12*a^2*b*Sinh[x] - 15*b^3*Sinh[x] - 3*a*b^2*Sinh[2*x] 
 + b^3*Sinh[3*x])/(12*b^4)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.20, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {3042, 3174, 3042, 25, 3344, 25, 3042, 3214, 3042, 3138, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos \left (-\frac {\pi }{2}+i x\right )^4}{a-b \sin \left (-\frac {\pi }{2}+i x\right )}dx\)

\(\Big \downarrow \) 3174

\(\displaystyle \frac {\sinh ^3(x)}{3 b}-\frac {\int \frac {(b+a \cosh (x)) \sinh ^2(x)}{a+b \cosh (x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sinh ^3(x)}{3 b}-\frac {\int -\frac {\cos \left (i x+\frac {\pi }{2}\right )^2 \left (b+a \sin \left (i x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sinh ^3(x)}{3 b}+\frac {\int \frac {\cos \left (i x+\frac {\pi }{2}\right )^2 \left (b+a \sin \left (i x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{b}\)

\(\Big \downarrow \) 3344

\(\displaystyle \frac {\frac {\int -\frac {b \left (a^2-2 b^2\right )+a \left (2 a^2-3 b^2\right ) \cosh (x)}{a+b \cosh (x)}dx}{2 b^2}+\frac {\sinh (x) \left (2 \left (a^2-b^2\right )-a b \cosh (x)\right )}{2 b^2}}{b}+\frac {\sinh ^3(x)}{3 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\sinh (x) \left (2 \left (a^2-b^2\right )-a b \cosh (x)\right )}{2 b^2}-\frac {\int \frac {b \left (a^2-2 b^2\right )+a \left (2 a^2-3 b^2\right ) \cosh (x)}{a+b \cosh (x)}dx}{2 b^2}}{b}+\frac {\sinh ^3(x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sinh ^3(x)}{3 b}+\frac {\frac {\sinh (x) \left (2 \left (a^2-b^2\right )-a b \cosh (x)\right )}{2 b^2}-\frac {\int \frac {b \left (a^2-2 b^2\right )+a \left (2 a^2-3 b^2\right ) \sin \left (i x+\frac {\pi }{2}\right )}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{2 b^2}}{b}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {\sinh (x) \left (2 \left (a^2-b^2\right )-a b \cosh (x)\right )}{2 b^2}-\frac {\frac {a x \left (2 a^2-3 b^2\right )}{b}-\frac {2 \left (a^2-b^2\right )^2 \int \frac {1}{a+b \cosh (x)}dx}{b}}{2 b^2}}{b}+\frac {\sinh ^3(x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sinh ^3(x)}{3 b}+\frac {\frac {\sinh (x) \left (2 \left (a^2-b^2\right )-a b \cosh (x)\right )}{2 b^2}-\frac {\frac {a x \left (2 a^2-3 b^2\right )}{b}-\frac {2 \left (a^2-b^2\right )^2 \int \frac {1}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{b}}{2 b^2}}{b}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\sinh (x) \left (2 \left (a^2-b^2\right )-a b \cosh (x)\right )}{2 b^2}-\frac {\frac {a x \left (2 a^2-3 b^2\right )}{b}-\frac {4 \left (a^2-b^2\right )^2 \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{b}}{2 b^2}}{b}+\frac {\sinh ^3(x)}{3 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\sinh (x) \left (2 \left (a^2-b^2\right )-a b \cosh (x)\right )}{2 b^2}-\frac {\frac {a x \left (2 a^2-3 b^2\right )}{b}-\frac {4 \left (a^2-b^2\right )^2 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}}}{2 b^2}}{b}+\frac {\sinh ^3(x)}{3 b}\)

Input:

Int[Sinh[x]^4/(a + b*Cosh[x]),x]
 

Output:

Sinh[x]^3/(3*b) + (-1/2*((a*(2*a^2 - 3*b^2)*x)/b - (4*(a^2 - b^2)^2*ArcTan 
h[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]))/b^2 + 
 ((2*(a^2 - b^2) - a*b*Cosh[x])*Sinh[x])/(2*b^2))/b
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3174

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x 
])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(b*(m + p)))   Int[(g*Cos 
[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; F 
reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 
 0] && IntegersQ[2*m, 2*p]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3344

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g* 
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* 
p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( 
(p - 1)/(b^2*(m + p)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si 
n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 
2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 
, 0] && IntegerQ[2*m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(222\) vs. \(2(87)=174\).

Time = 30.00 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.14

method	result	size
default	\(-\frac {2 \left (-a^{4}+2 a^{2} b^{2}-b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {1}{3 b \left (1+\tanh \left (\frac {x}{2}\right )\right )^{3}}-\frac {-a -b}{2 b^{2} \left (1+\tanh \left (\frac {x}{2}\right )\right )^{2}}-\frac {2 a^{2}+a b -2 b^{2}}{2 b^{3} \left (1+\tanh \left (\frac {x}{2}\right )\right )}-\frac {a \left (2 a^{2}-3 b^{2}\right ) \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2 b^{4}}-\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a +b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b -2 b^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \left (2 a^{2}-3 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{4}}\)	\(223\)
risch	\(-\frac {a^{3} x}{b^{4}}+\frac {3 a x}{2 b^{2}}+\frac {{\mathrm e}^{3 x}}{24 b}-\frac {a \,{\mathrm e}^{2 x}}{8 b^{2}}+\frac {{\mathrm e}^{x} a^{2}}{2 b^{3}}-\frac {5 \,{\mathrm e}^{x}}{8 b}-\frac {{\mathrm e}^{-x} a^{2}}{2 b^{3}}+\frac {5 \,{\mathrm e}^{-x}}{8 b}+\frac {a \,{\mathrm e}^{-2 x}}{8 b^{2}}-\frac {{\mathrm e}^{-3 x}}{24 b}+\frac {\sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{b^{4}}-\frac {\sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{b^{2}}-\frac {\sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{b^{4}}+\frac {\sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{b^{2}}\)	\(326\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (86) = 172\).

Time = 0.10 (sec) , antiderivative size = 1099, normalized size of antiderivative = 10.57 \[ \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/24*(b^3*cosh(x)^6 + b^3*sinh(x)^6 - 3*a*b^2*cosh(x)^5 + 3*(2*b^3*cosh(x 
) - a*b^2)*sinh(x)^5 - 12*(2*a^3 - 3*a*b^2)*x*cosh(x)^3 + 3*(4*a^2*b - 5*b 
^3)*cosh(x)^4 + 3*(5*b^3*cosh(x)^2 - 5*a*b^2*cosh(x) + 4*a^2*b - 5*b^3)*si 
nh(x)^4 + 3*a*b^2*cosh(x) + 2*(10*b^3*cosh(x)^3 - 15*a*b^2*cosh(x)^2 - 6*( 
2*a^3 - 3*a*b^2)*x + 6*(4*a^2*b - 5*b^3)*cosh(x))*sinh(x)^3 - b^3 - 3*(4*a 
^2*b - 5*b^3)*cosh(x)^2 + 3*(5*b^3*cosh(x)^4 - 10*a*b^2*cosh(x)^3 - 4*a^2* 
b + 5*b^3 - 12*(2*a^3 - 3*a*b^2)*x*cosh(x) + 6*(4*a^2*b - 5*b^3)*cosh(x)^2 
)*sinh(x)^2 - 24*((a^2 - b^2)*cosh(x)^3 + 3*(a^2 - b^2)*cosh(x)^2*sinh(x) 
+ 3*(a^2 - b^2)*cosh(x)*sinh(x)^2 + (a^2 - b^2)*sinh(x)^3)*sqrt(a^2 - b^2) 
*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2 
*cosh(x) + a*b)*sinh(x) + 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/( 
b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)) 
+ 3*(2*b^3*cosh(x)^5 - 5*a*b^2*cosh(x)^4 - 12*(2*a^3 - 3*a*b^2)*x*cosh(x)^ 
2 + 4*(4*a^2*b - 5*b^3)*cosh(x)^3 + a*b^2 - 2*(4*a^2*b - 5*b^3)*cosh(x))*s 
inh(x))/(b^4*cosh(x)^3 + 3*b^4*cosh(x)^2*sinh(x) + 3*b^4*cosh(x)*sinh(x)^2 
 + b^4*sinh(x)^3), 1/24*(b^3*cosh(x)^6 + b^3*sinh(x)^6 - 3*a*b^2*cosh(x)^5 
 + 3*(2*b^3*cosh(x) - a*b^2)*sinh(x)^5 - 12*(2*a^3 - 3*a*b^2)*x*cosh(x)^3 
+ 3*(4*a^2*b - 5*b^3)*cosh(x)^4 + 3*(5*b^3*cosh(x)^2 - 5*a*b^2*cosh(x) + 4 
*a^2*b - 5*b^3)*sinh(x)^4 + 3*a*b^2*cosh(x) + 2*(10*b^3*cosh(x)^3 - 15*a*b 
^2*cosh(x)^2 - 6*(2*a^3 - 3*a*b^2)*x + 6*(4*a^2*b - 5*b^3)*cosh(x))*sin...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.40 \[ \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx=\frac {b^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 12 \, a^{2} e^{x} - 15 \, b^{2} e^{x}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} x}{2 \, b^{4}} + \frac {{\left (3 \, a b^{2} e^{x} - b^{3} - 3 \, {\left (4 \, a^{2} b - 5 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, b^{4}} + \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{4}} \] Input:

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

Output:

1/24*(b^2*e^(3*x) - 3*a*b*e^(2*x) + 12*a^2*e^x - 15*b^2*e^x)/b^3 - 1/2*(2* 
a^3 - 3*a*b^2)*x/b^4 + 1/24*(3*a*b^2*e^x - b^3 - 3*(4*a^2*b - 5*b^3)*e^(2* 
x))*e^(-3*x)/b^4 + 2*(a^4 - 2*a^2*b^2 + b^4)*arctan((b*e^x + a)/sqrt(-a^2 
+ b^2))/(sqrt(-a^2 + b^2)*b^4)
 

Mupad [B] (verification not implemented)

Time = 2.36 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.13 \[ \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx=\frac {{\mathrm {e}}^{3\,x}}{24\,b}-\frac {{\mathrm {e}}^{-3\,x}}{24\,b}+\frac {x\,\left (3\,a\,b^2-2\,a^3\right )}{2\,b^4}+\frac {{\mathrm {e}}^x\,\left (4\,a^2-5\,b^2\right )}{8\,b^3}+\frac {a\,{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {a\,{\mathrm {e}}^{2\,x}}{8\,b^2}-\frac {{\mathrm {e}}^{-x}\,\left (4\,a^2-5\,b^2\right )}{8\,b^3}+\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x\,\left (a^4-2\,a^2\,b^2+b^4\right )}{b^5}-\frac {2\,{\left (a+b\right )}^{3/2}\,\left (b+a\,{\mathrm {e}}^x\right )\,{\left (a-b\right )}^{3/2}}{b^5}\right )\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}{b^4}-\frac {\ln \left (\frac {2\,{\left (a+b\right )}^{3/2}\,\left (b+a\,{\mathrm {e}}^x\right )\,{\left (a-b\right )}^{3/2}}{b^5}-\frac {2\,{\mathrm {e}}^x\,\left (a^4-2\,a^2\,b^2+b^4\right )}{b^5}\right )\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}{b^4} \] Input:

int(sinh(x)^4/(a + b*cosh(x)),x)
 

Output:

exp(3*x)/(24*b) - exp(-3*x)/(24*b) + (x*(3*a*b^2 - 2*a^3))/(2*b^4) + (exp( 
x)*(4*a^2 - 5*b^2))/(8*b^3) + (a*exp(-2*x))/(8*b^2) - (a*exp(2*x))/(8*b^2) 
 - (exp(-x)*(4*a^2 - 5*b^2))/(8*b^3) + (log(- (2*exp(x)*(a^4 + b^4 - 2*a^2 
*b^2))/b^5 - (2*(a + b)^(3/2)*(b + a*exp(x))*(a - b)^(3/2))/b^5)*(a + b)^( 
3/2)*(a - b)^(3/2))/b^4 - (log((2*(a + b)^(3/2)*(b + a*exp(x))*(a - b)^(3/ 
2))/b^5 - (2*exp(x)*(a^4 + b^4 - 2*a^2*b^2))/b^5)*(a + b)^(3/2)*(a - b)^(3 
/2))/b^4
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.87 \[ \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx=\frac {-48 e^{3 x} \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2}+48 e^{3 x} \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) b^{2}+e^{6 x} b^{3}-3 e^{5 x} a \,b^{2}+12 e^{4 x} a^{2} b -15 e^{4 x} b^{3}-24 e^{3 x} a^{3} x +36 e^{3 x} a \,b^{2} x -12 e^{2 x} a^{2} b +15 e^{2 x} b^{3}+3 e^{x} a \,b^{2}-b^{3}}{24 e^{3 x} b^{4}} \] Input:

int(sinh(x)^4/(a+b*cosh(x)),x)
 

Output:

( - 48*e**(3*x)*sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqrt( - a**2 + b**2 
))*a**2 + 48*e**(3*x)*sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqrt( - a**2 
+ b**2))*b**2 + e**(6*x)*b**3 - 3*e**(5*x)*a*b**2 + 12*e**(4*x)*a**2*b - 1 
5*e**(4*x)*b**3 - 24*e**(3*x)*a**3*x + 36*e**(3*x)*a*b**2*x - 12*e**(2*x)* 
a**2*b + 15*e**(2*x)*b**3 + 3*e**x*a*b**2 - b**3)/(24*e**(3*x)*b**4)