\(\int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx\) [200]

Optimal result

Integrand size = 13, antiderivative size = 94 \[ \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {3 \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {6 a \sqrt {a^2+b^2} \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4}-\frac {3 \cosh (x) (2 a-b \sinh (x))}{2 b^3}-\frac {\cosh ^3(x)}{b (a+b \sinh (x))} \] Output:

3/2*(2*a^2+b^2)*x/b^4+6*a*(a^2+b^2)^(1/2)*arctanh((b-a*tanh(1/2*x))/(a^2+b 
^2)^(1/2))/b^4-3/2*cosh(x)*(2*a-b*sinh(x))/b^3-cosh(x)^3/b/(a+b*sinh(x))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.37 (sec) , antiderivative size = 660, normalized size of antiderivative = 7.02 \[ \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {\cosh ^3(x) \left (12 a \sqrt {a-i b} (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {-\frac {b (-i+\sinh (x))}{a+i b}}}\right ) \sqrt {1+i \sinh (x)} (a+b \sinh (x))-12 a \left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-i b} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {a+i b} \sqrt {-\frac {b (-i+\sinh (x))}{a+i b}}}\right ) \sqrt {1+i \sinh (x)} (a+b \sinh (x))+\sqrt {a+i b} \sqrt {-\frac {b (-i+\sinh (x))}{a+i b}} \left (6 (-1)^{3/4} a \sqrt {b} \left (2 a^2+i a b+b^2\right ) \arcsin \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a-i b} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {b}}\right )+6 (-1)^{3/4} b^{3/2} \left (2 a^2+i a b+b^2\right ) \arcsin \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a-i b} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {b}}\right ) \sinh (x)-2 \sqrt {a-i b} \left (3 a^3+3 i a^2 b+a b^2+i b^3\right ) \sqrt {1+i \sinh (x)} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}-3 a \sqrt {a-i b} (a+i b) b \sqrt {1+i \sinh (x)} \sinh (x) \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}+\sqrt {a-i b} (a+i b) b^2 \sqrt {1+i \sinh (x)} \sinh ^2(x) \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}\right )\right )}{2 (a-i b)^{3/2} (a+i b)^{5/2} b \sqrt {1+i \sinh (x)} \left (-\frac {b (-i+\sinh (x))}{a+i b}\right )^{3/2} \left (-\frac {b (i+\sinh (x))}{a-i b}\right )^{3/2} (a+b \sinh (x))} \] Input:

Integrate[Cosh[x]^4/(a + b*Sinh[x])^2,x]
 

Output:

(Cosh[x]^3*(12*a*Sqrt[a - I*b]*(a + I*b)^(3/2)*ArcTanh[Sqrt[-((b*(I + Sinh 
[x]))/(a - I*b))]/Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))]]*Sqrt[1 + I*Sinh[x 
]]*(a + b*Sinh[x]) - 12*a*(a^2 + b^2)*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]]*(a + b*Sinh[x]) + Sqrt[a + I*b]*Sqrt[-((b*(-I + 
Sinh[x]))/(a + I*b))]*(6*(-1)^(3/4)*a*Sqrt[b]*(2*a^2 + I*a*b + b^2)*ArcSin 
[((1/2 + I/2)*Sqrt[a - I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/Sqrt[b]] 
 + 6*(-1)^(3/4)*b^(3/2)*(2*a^2 + I*a*b + b^2)*ArcSin[((1/2 + I/2)*Sqrt[a - 
 I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/Sqrt[b]]*Sinh[x] - 2*Sqrt[a - 
I*b]*(3*a^3 + (3*I)*a^2*b + a*b^2 + I*b^3)*Sqrt[1 + I*Sinh[x]]*Sqrt[-((b*( 
I + Sinh[x]))/(a - I*b))] - 3*a*Sqrt[a - I*b]*(a + I*b)*b*Sqrt[1 + I*Sinh[ 
x]]*Sinh[x]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))] + Sqrt[a - I*b]*(a + I*b) 
*b^2*Sqrt[1 + I*Sinh[x]]*Sinh[x]^2*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))]))) 
/(2*(a - I*b)^(3/2)*(a + I*b)^(5/2)*b*Sqrt[1 + I*Sinh[x]]*(-((b*(-I + Sinh 
[x]))/(a + I*b)))^(3/2)*(-((b*(I + Sinh[x]))/(a - I*b)))^(3/2)*(a + b*Sinh 
[x]))
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.24, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 3172, 26, 3042, 26, 3344, 26, 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.

\(\displaystyle \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3172

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3344

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

Input:

Int[Cosh[x]^4/(a + b*Sinh[x])^2,x]
 

Output:

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

Defintions of rubi rules used

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 3172

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 + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1)))   Int[(g*Cos 
[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre 
eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I 
ntegersQ[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 = 16.70 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.71

method	result	size
risch	\(\frac {3 x \,a^{2}}{b^{4}}+\frac {3 x}{2 b^{2}}+\frac {{\mathrm e}^{2 x}}{8 b^{2}}-\frac {a \,{\mathrm e}^{x}}{b^{3}}-\frac {a \,{\mathrm e}^{-x}}{b^{3}}-\frac {{\mathrm e}^{-2 x}}{8 b^{2}}+\frac {2 \left (a^{2}+b^{2}\right ) \left ({\mathrm e}^{x} a -b \right )}{b^{4} \left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right )}+\frac {3 \sqrt {a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{x}+\frac {a +\sqrt {a^{2}+b^{2}}}{b}\right )}{b^{4}}-\frac {3 \sqrt {a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{x}-\frac {-a +\sqrt {a^{2}+b^{2}}}{b}\right )}{b^{4}}\)	\(161\)
default	\(-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2} b^{2}}-\frac {4 a -b}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (6 a^{2}+3 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{4}}+\frac {\frac {2 \left (\frac {b^{2} \left (a^{2}+b^{2}\right ) \tanh \left (\frac {x}{2}\right )}{a}+b \left (a^{2}+b^{2}\right )\right )}{\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a}-6 a \sqrt {a^{2}+b^{2}}\, \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{4}}+\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-4 a -b}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-6 a^{2}-3 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{4}}\)	\(205\)

Input:

int(cosh(x)^4/(a+b*sinh(x))^2,x,method=_RETURNVERBOSE)
 

Output:

3*x/b^4*a^2+3/2*x/b^2+1/8/b^2*exp(x)^2-1/b^3*a*exp(x)-1/b^3*a/exp(x)-1/8/b 
^2/exp(x)^2+2*(a^2+b^2)*(exp(x)*a-b)/b^4/(b*exp(x)^2+2*exp(x)*a-b)+3*(a^2+ 
b^2)^(1/2)*a/b^4*ln(exp(x)+(a+(a^2+b^2)^(1/2))/b)-3*(a^2+b^2)^(1/2)*a/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. 833 vs. \(2 (85) = 170\).

Time = 0.10 (sec) , antiderivative size = 833, normalized size of antiderivative = 8.86 \[ \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx=\text {Too large to display} \] Input:

integrate(cosh(x)^4/(a+b*sinh(x))^2,x, algorithm="fricas")
 

Output:

1/8*(b^3*cosh(x)^6 + b^3*sinh(x)^6 - 6*a*b^2*cosh(x)^5 + 6*(b^3*cosh(x) - 
a*b^2)*sinh(x)^5 - (16*a^2*b + b^3 - 12*(2*a^2*b + b^3)*x)*cosh(x)^4 + (15 
*b^3*cosh(x)^2 - 30*a*b^2*cosh(x) - 16*a^2*b - b^3 + 12*(2*a^2*b + b^3)*x) 
*sinh(x)^4 + 6*a*b^2*cosh(x) + 8*(2*a^3 + 2*a*b^2 + 3*(2*a^3 + a*b^2)*x)*c 
osh(x)^3 + 4*(5*b^3*cosh(x)^3 - 15*a*b^2*cosh(x)^2 + 4*a^3 + 4*a*b^2 + 6*( 
2*a^3 + a*b^2)*x - (16*a^2*b + b^3 - 12*(2*a^2*b + b^3)*x)*cosh(x))*sinh(x 
)^3 + b^3 - (32*a^2*b + 17*b^3 + 12*(2*a^2*b + b^3)*x)*cosh(x)^2 + (15*b^3 
*cosh(x)^4 - 60*a*b^2*cosh(x)^3 - 32*a^2*b - 17*b^3 - 6*(16*a^2*b + b^3 - 
12*(2*a^2*b + b^3)*x)*cosh(x)^2 - 12*(2*a^2*b + b^3)*x + 24*(2*a^3 + 2*a*b 
^2 + 3*(2*a^3 + a*b^2)*x)*cosh(x))*sinh(x)^2 + 24*(a*b*cosh(x)^4 + a*b*sin 
h(x)^4 + 2*a^2*cosh(x)^3 - a*b*cosh(x)^2 + 2*(2*a*b*cosh(x) + a^2)*sinh(x) 
^3 + (6*a*b*cosh(x)^2 + 6*a^2*cosh(x) - a*b)*sinh(x)^2 + 2*(2*a*b*cosh(x)^ 
3 + 3*a^2*cosh(x)^2 - a*b*cosh(x))*sinh(x))*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)) + 2*(3*b^3*cosh 
(x)^5 - 15*a*b^2*cosh(x)^4 - 2*(16*a^2*b + b^3 - 12*(2*a^2*b + b^3)*x)*cos 
h(x)^3 + 3*a*b^2 + 12*(2*a^3 + 2*a*b^2 + 3*(2*a^3 + a*b^2)*x)*cosh(x)^2 - 
(32*a^2*b + 17*b^3 + 12*(2*a^2*b + b^3)*x)*cosh(x))*sinh(x))/(b^5*cosh(x)^ 
4 + b^5*sinh(x)^4 + 2*a*b^4*cosh(x)^3 - b^5*cosh(x)^2 + 2*(2*b^5*cosh(x...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx=\text {Timed out} \] Input:

integrate(cosh(x)**4/(a+b*sinh(x))**2,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 176 vs. \(2 (85) = 170\).

Time = 0.12 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.87 \[ \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx=-\frac {6 \, a b^{2} e^{\left (-x\right )} - b^{3} + {\left (32 \, a^{2} b + 17 \, b^{3}\right )} e^{\left (-2 \, x\right )} + 8 \, {\left (2 \, a^{3} + a b^{2}\right )} e^{\left (-3 \, x\right )}}{8 \, {\left (b^{5} e^{\left (-2 \, x\right )} + 2 \, a b^{4} e^{\left (-3 \, x\right )} - b^{5} e^{\left (-4 \, x\right )}\right )}} - \frac {3 \, \sqrt {a^{2} + b^{2}} a \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{b^{4}} - \frac {8 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )}}{8 \, b^{3}} + \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} x}{2 \, b^{4}} \] Input:

integrate(cosh(x)^4/(a+b*sinh(x))^2,x, algorithm="maxima")
 

Output:

-1/8*(6*a*b^2*e^(-x) - b^3 + (32*a^2*b + 17*b^3)*e^(-2*x) + 8*(2*a^3 + a*b 
^2)*e^(-3*x))/(b^5*e^(-2*x) + 2*a*b^4*e^(-3*x) - b^5*e^(-4*x)) - 3*sqrt(a^ 
2 + b^2)*a*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2 + 
 b^2)))/b^4 - 1/8*(8*a*e^(-x) + b*e^(-2*x))/b^3 + 3/2*(2*a^2 + b^2)*x/b^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (85) = 170\).

Time = 0.14 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.89 \[ \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} x}{2 \, b^{4}} - \frac {3 \, {\left (a^{3} + a b^{2}\right )} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} + \frac {b^{2} e^{\left (2 \, x\right )} - 8 \, a b e^{x}}{8 \, b^{4}} + \frac {{\left (6 \, a b^{2} e^{x} + b^{3} + 8 \, {\left (2 \, a^{3} + a b^{2}\right )} e^{\left (3 \, x\right )} - {\left (32 \, a^{2} b + 17 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )} b^{4}} \] Input:

integrate(cosh(x)^4/(a+b*sinh(x))^2,x, algorithm="giac")
 

Output:

3/2*(2*a^2 + b^2)*x/b^4 - 3*(a^3 + a*b^2)*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) + 
 1/8*(b^2*e^(2*x) - 8*a*b*e^x)/b^4 + 1/8*(6*a*b^2*e^x + b^3 + 8*(2*a^3 + a 
*b^2)*e^(3*x) - (32*a^2*b + 17*b^3)*e^(2*x))*e^(-2*x)/((b*e^(2*x) + 2*a*e^ 
x - b)*b^4)
 

Mupad [B] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.72 \[ \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,b^2}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {\frac {2\,\left (a^4\,b^2+2\,a^2\,b^4+b^6\right )}{b^4\,\left (a^2\,b+b^3\right )}-\frac {2\,{\mathrm {e}}^x\,\left (a^5\,b^2+2\,a^3\,b^4+a\,b^6\right )}{b^5\,\left (a^2\,b+b^3\right )}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}+\frac {x\,\left (6\,a^2+3\,b^2\right )}{2\,b^4}-\frac {a\,{\mathrm {e}}^x}{b^3}-\frac {a\,{\mathrm {e}}^{-x}}{b^3}-\frac {3\,a\,\ln \left (\frac {6\,a\,{\mathrm {e}}^x\,\left (a^2+b^2\right )}{b^5}-\frac {6\,a\,\left (b-a\,{\mathrm {e}}^x\right )\,\sqrt {a^2+b^2}}{b^5}\right )\,\sqrt {a^2+b^2}}{b^4}+\frac {3\,a\,\ln \left (\frac {6\,a\,\left (b-a\,{\mathrm {e}}^x\right )\,\sqrt {a^2+b^2}}{b^5}+\frac {6\,a\,{\mathrm {e}}^x\,\left (a^2+b^2\right )}{b^5}\right )\,\sqrt {a^2+b^2}}{b^4} \] Input:

int(cosh(x)^4/(a + b*sinh(x))^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 296, normalized size of antiderivative = 3.15 \[ \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {-48 e^{4 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a b i -96 e^{3 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{2} i +48 e^{2 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a b i +e^{6 x} b^{3}-6 e^{5 x} a \,b^{2}+24 e^{4 x} a^{2} b x -24 e^{4 x} a^{2} b +12 e^{4 x} b^{3} x -9 e^{4 x} b^{3}+48 e^{3 x} a^{3} x +24 e^{3 x} a \,b^{2} x -24 e^{2 x} a^{2} b x -24 e^{2 x} a^{2} b -12 e^{2 x} b^{3} x -9 e^{2 x} b^{3}+6 e^{x} a \,b^{2}+b^{3}}{8 e^{2 x} b^{4} \left (e^{2 x} b +2 e^{x} a -b \right )} \] Input:

int(cosh(x)^4/(a+b*sinh(x))^2,x)
 

Output:

( - 48*e**(4*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2)) 
*a*b*i - 96*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**(2*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a* 
*2 + b**2))*a*b*i + e**(6*x)*b**3 - 6*e**(5*x)*a*b**2 + 24*e**(4*x)*a**2*b 
*x - 24*e**(4*x)*a**2*b + 12*e**(4*x)*b**3*x - 9*e**(4*x)*b**3 + 48*e**(3* 
x)*a**3*x + 24*e**(3*x)*a*b**2*x - 24*e**(2*x)*a**2*b*x - 24*e**(2*x)*a**2 
*b - 12*e**(2*x)*b**3*x - 9*e**(2*x)*b**3 + 6*e**x*a*b**2 + b**3)/(8*e**(2 
*x)*b**4*(e**(2*x)*b + 2*e**x*a - b))