\(\int \frac {\cosh ^4(x)}{a+b \sinh (x)} \, dx\) [190]

Optimal result

Integrand size = 13, antiderivative size = 97 \[ \int \frac {\cosh ^4(x)}{a+b \sinh (x)} \, dx=-\frac {a \left (2 a^2+3 b^2\right ) x}{2 b^4}-\frac {2 \left (a^2+b^2\right )^{3/2} \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4}+\frac {\cosh ^3(x)}{3 b}+\frac {\cosh (x) \left (2 \left (a^2+b^2\right )-a b \sinh (x)\right )}{2 b^3} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.66 (sec) , antiderivative size = 553, normalized size of antiderivative = 5.70 \[ \int \frac {\cosh ^4(x)}{a+b \sinh (x)} \, dx=\frac {\cosh ^3(x) \left (-12 \sqrt {a-i b} \sqrt {a+i b} \left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {-\frac {b (-i+\sinh (x))}{a+i b}}}\right ) \sqrt {1+i \sinh (x)}+12 (a-i b)^2 (a+i b) \text {arctanh}\left (\frac {\sqrt {a-i b} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {a+i b} \sqrt {-\frac {b (-i+\sinh (x))}{a+i b}}}\right ) \sqrt {1+i \sinh (x)}+\sqrt {a+i b} \sqrt {-\frac {b (-i+\sinh (x))}{a+i b}} \left ((3-3 i) \sqrt {2} \sqrt {b} \left (2 a^2-i a b+2 b^2\right ) \arcsin \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a-i b} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {b}}\right )+2 \sqrt {a-i b} \left (3 a^2+4 b^2\right ) \sqrt {1+i \sinh (x)} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}-3 a \sqrt {a-i b} b \sqrt {1+i \sinh (x)} \sinh (x) \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}+2 \sqrt {a-i b} b^2 \sqrt {1+i \sinh (x)} \sinh ^2(x) \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}\right )\right )}{6 (a-i b)^{3/2} (a+i b)^{3/2} b \sqrt {1+i \sinh (x)} \left (-\frac {b (-i+\sinh (x))}{a+i b}\right )^{3/2} \left (-\frac {b (i+\sinh (x))}{a-i b}\right )^{3/2}} \] Input:

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

Output:

(Cosh[x]^3*(-12*Sqrt[a - I*b]*Sqrt[a + I*b]*(a^2 + b^2)*ArcTanh[Sqrt[-((b* 
(I + Sinh[x]))/(a - I*b))]/Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))]]*Sqrt[1 + 
 I*Sinh[x]] + 12*(a - I*b)^2*(a + I*b)*ArcTanh[(Sqrt[a - I*b]*Sqrt[-((b*(I 
 + Sinh[x]))/(a - I*b))])/(Sqrt[a + I*b]*Sqrt[-((b*(-I + Sinh[x]))/(a + I* 
b))])]*Sqrt[1 + I*Sinh[x]] + Sqrt[a + I*b]*Sqrt[-((b*(-I + Sinh[x]))/(a + 
I*b))]*((3 - 3*I)*Sqrt[2]*Sqrt[b]*(2*a^2 - I*a*b + 2*b^2)*ArcSin[((1/2 + I 
/2)*Sqrt[a - I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/Sqrt[b]] + 2*Sqrt[ 
a - I*b]*(3*a^2 + 4*b^2)*Sqrt[1 + I*Sinh[x]]*Sqrt[-((b*(I + Sinh[x]))/(a - 
 I*b))] - 3*a*Sqrt[a - I*b]*b*Sqrt[1 + I*Sinh[x]]*Sinh[x]*Sqrt[-((b*(I + S 
inh[x]))/(a - I*b))] + 2*Sqrt[a - I*b]*b^2*Sqrt[1 + I*Sinh[x]]*Sinh[x]^2*S 
qrt[-((b*(I + Sinh[x]))/(a - I*b))])))/(6*(a - I*b)^(3/2)*(a + I*b)^(3/2)* 
b*Sqrt[1 + I*Sinh[x]]*(-((b*(-I + Sinh[x]))/(a + I*b)))^(3/2)*(-((b*(I + S 
inh[x]))/(a - I*b)))^(3/2))
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.16, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {3042, 3174, 26, 3042, 3344, 25, 3042, 3214, 3042, 3139, 1083, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (i x)^4}{a-i b \sin (i x)}dx\)

\(\Big \downarrow \) 3174

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3344

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {-\frac {8 \left (a^2+b^2\right )^2 \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{b}-\frac {a x \left (2 a^2+3 b^2\right )}{b}}{2 b^2}+\frac {\cosh (x) \left (2 \left (a^2+b^2\right )-a b \sinh (x)\right )}{2 b^2}}{b}+\frac {\cosh ^3(x)}{3 b}\)

\(\Big \downarrow \) 219

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

Input:

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

Output:

Cosh[x]^3/(3*b) + ((-((a*(2*a^2 + 3*b^2)*x)/b) - (4*(a^2 + b^2)^(3/2)*ArcT 
anh[(2*b - 2*a*Tanh[x/2])/(2*Sqrt[a^2 + b^2])])/b)/(2*b^2) + (Cosh[x]*(2*( 
a^2 + b^2) - a*b*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 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*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 [A] (verified)

Time = 8.22 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.67

method	result	size
risch	\(-\frac {a^{3} x}{b^{4}}-\frac {3 a x}{2 b^{2}}+\frac {{\mathrm e}^{3 x}}{24 b}-\frac {a \,{\mathrm e}^{2 x}}{8 b^{2}}+\frac {{\mathrm e}^{x} a^{2}}{2 b^{3}}+\frac {5 \,{\mathrm e}^{x}}{8 b}+\frac {{\mathrm e}^{-x} a^{2}}{2 b^{3}}+\frac {5 \,{\mathrm e}^{-x}}{8 b}+\frac {a \,{\mathrm e}^{-2 x}}{8 b^{2}}+\frac {{\mathrm e}^{-3 x}}{24 b}+\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} \ln \left ({\mathrm e}^{x}-\frac {-a +\sqrt {a^{2}+b^{2}}}{b}\right )}{b^{4}}-\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} \ln \left ({\mathrm e}^{x}+\frac {a +\sqrt {a^{2}+b^{2}}}{b}\right )}{b^{4}}\)	\(162\)
default	\(\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {-a +b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-2 a^{2}+a b -3 b^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \left (2 a^{2}+3 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{4}}-\frac {2 \left (-a^{4}-2 a^{2} b^{2}-b^{4}\right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{4} \sqrt {a^{2}+b^{2}}}-\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a +b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +3 b^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \left (2 a^{2}+3 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{4}}\)	\(220\)

Input:

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

Output:

-a^3*x/b^4-3/2*a*x/b^2+1/24/b*exp(x)^3-1/8*a/b^2*exp(x)^2+1/2/b^3*exp(x)*a 
^2+5/8/b*exp(x)+1/2/b^3/exp(x)*a^2+5/8/b/exp(x)+1/8*a/b^2/exp(x)^2+1/24/b/ 
exp(x)^3+(a^2+b^2)^(3/2)/b^4*ln(exp(x)-(-a+(a^2+b^2)^(1/2))/b)-(a^2+b^2)^( 
3/2)/b^4*ln(exp(x)+(a+(a^2+b^2)^(1/2))/b)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 569 vs. \(2 (87) = 174\).

Time = 0.10 (sec) , antiderivative size = 569, normalized size of antiderivative = 5.87 \[ \int \frac {\cosh ^4(x)}{a+b \sinh (x)} \, dx =\text {Too large to display} \] Input:

integrate(cosh(x)^4/(a+b*sinh(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)*sin 
h(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))*si 
nh(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)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.75 \[ \int \frac {\cosh ^4(x)}{a+b \sinh (x)} \, dx=-\frac {{\left (3 \, a b e^{\left (-x\right )} - b^{2} - 3 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} e^{\left (-2 \, x\right )}\right )} e^{\left (3 \, x\right )}}{24 \, b^{3}} + \frac {3 \, a b e^{\left (-2 \, x\right )} + b^{2} e^{\left (-3 \, x\right )} + 3 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} e^{\left (-x\right )}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} + 3 \, a b^{2}\right )} x}{2 \, b^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.73 \[ \int \frac {\cosh ^4(x)}{a+b \sinh (x)} \, dx=\frac {b^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 12 \, a^{2} e^{x} + 15 \, b^{2} e^{x}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} + 3 \, a b^{2}\right )} x}{2 \, b^{4}} + \frac {{\left (3 \, a b^{2} e^{x} + b^{3} + 3 \, {\left (4 \, a^{2} b + 5 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, b^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} \] Input:

integrate(cosh(x)^4/(a+b*sinh(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 + (a^4 + 2*a^2*b^2 + b^4)*log(abs(2*b*e^x + 2*a - 2*sqrt( 
a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^4)
 

Mupad [B] (verification not implemented)

Time = 1.71 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.06 \[ \int \frac {\cosh ^4(x)}{a+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^{-3\,x}}{24\,b}+\frac {{\mathrm {e}}^{3\,x}}{24\,b}-\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x\,{\left (a^2+b^2\right )}^2}{b^5}-\frac {2\,\left (b-a\,{\mathrm {e}}^x\right )\,{\left (a^2+b^2\right )}^{3/2}}{b^5}\right )\,{\left (a^2+b^2\right )}^{3/2}}{b^4}+\frac {\ln \left (\frac {2\,\left (b-a\,{\mathrm {e}}^x\right )\,{\left (a^2+b^2\right )}^{3/2}}{b^5}-\frac {2\,{\mathrm {e}}^x\,{\left (a^2+b^2\right )}^2}{b^5}\right )\,{\left (a^2+b^2\right )}^{3/2}}{b^4}-\frac {x\,\left (2\,a^3+3\,a\,b^2\right )}{2\,b^4}+\frac {{\mathrm {e}}^x\,\left (4\,a^2+5\,b^2\right )}{8\,b^3}+\frac {a\,{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {a\,{\mathrm {e}}^{2\,x}}{8\,b^2}+\frac {{\mathrm {e}}^{-x}\,\left (4\,a^2+5\,b^2\right )}{8\,b^3} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.98 \[ \int \frac {\cosh ^4(x)}{a+b \sinh (x)} \, dx=\frac {48 e^{3 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{2} i +48 e^{3 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{2} i +e^{6 x} b^{3}-3 e^{5 x} a \,b^{2}+12 e^{4 x} a^{2} b +15 e^{4 x} b^{3}-24 e^{3 x} a^{3} x -36 e^{3 x} a \,b^{2} x +12 e^{2 x} a^{2} b +15 e^{2 x} b^{3}+3 e^{x} a \,b^{2}+b^{3}}{24 e^{3 x} b^{4}} \] Input:

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

Output:

(48*e**(3*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a* 
*2*i + 48*e**(3*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b** 
2))*b**2*i + 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)