\(\int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx\) [80]

Optimal result

Integrand size = 13, antiderivative size = 162 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {\left (6 a^2-b^2\right ) x}{2 b^4}+\frac {2 a^3 \left (3 a^2+4 b^2\right ) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4 \left (a^2+b^2\right )^{3/2}}-\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {\left (3 a^2+b^2\right ) \cosh (x) \sinh (x)}{2 b^2 \left (a^2+b^2\right )}-\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.73 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {-2 \left (-6 a^2+b^2\right ) x+\frac {8 a^3 \left (3 a^2+4 b^2\right ) \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}-8 a b \cosh (x)-\frac {4 a^4 b \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+b^2 \sinh (2 x)}{4 b^4} \] Input:

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

Output:

(-2*(-6*a^2 + b^2)*x + (8*a^3*(3*a^2 + 4*b^2)*ArcTan[(b - a*Tanh[x/2])/Sqr 
t[-a^2 - b^2]])/(-a^2 - b^2)^(3/2) - 8*a*b*Cosh[x] - (4*a^4*b*Cosh[x])/((a 
^2 + b^2)*(a + b*Sinh[x])) + b^2*Sinh[2*x])/(4*b^4)
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 1.00 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.16, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.231, Rules used = {3042, 3271, 26, 3042, 26, 3528, 25, 3042, 3502, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {i \left (\frac {i \left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {2 a \left (3 a^2+2 b^2\right ) \cosh (x)}{b}+\frac {i \int -\frac {i \left (a b \left (3 a^2+b^2\right )-\left (6 a^2-b^2\right ) \left (a^2+b^2\right ) \sinh (x)\right )}{a+b \sinh (x)}dx}{b}\right )}{2 b}\right )}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {i \left (\frac {i \left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {\int \frac {a b \left (3 a^2+b^2\right )-\left (6 a^2-b^2\right ) \left (a^2+b^2\right ) \sinh (x)}{a+b \sinh (x)}dx}{b}+\frac {2 a \left (3 a^2+2 b^2\right ) \cosh (x)}{b}\right )}{2 b}\right )}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {i \left (\frac {i \left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {2 a \left (3 a^2+2 b^2\right ) \cosh (x)}{b}+\frac {\int \frac {a b \left (3 a^2+b^2\right )+i \left (6 a^2-b^2\right ) \left (a^2+b^2\right ) \sin (i x)}{a-i b \sin (i x)}dx}{b}\right )}{2 b}\right )}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {i \left (\frac {i \left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {\frac {2 a^3 \left (3 a^2+4 b^2\right ) \int \frac {1}{a+b \sinh (x)}dx}{b}-\frac {x \left (6 a^2-b^2\right ) \left (a^2+b^2\right )}{b}}{b}+\frac {2 a \left (3 a^2+2 b^2\right ) \cosh (x)}{b}\right )}{2 b}\right )}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {i \left (\frac {i \left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {2 a \left (3 a^2+2 b^2\right ) \cosh (x)}{b}+\frac {-\frac {x \left (6 a^2-b^2\right ) \left (a^2+b^2\right )}{b}+\frac {2 a^3 \left (3 a^2+4 b^2\right ) \int \frac {1}{a-i b \sin (i x)}dx}{b}}{b}\right )}{2 b}\right )}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3139

\(\displaystyle -\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {i \left (\frac {i \left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {\frac {4 a^3 \left (3 a^2+4 b^2\right ) \int \frac {1}{-a \tanh ^2\left (\frac {x}{2}\right )+2 b \tanh \left (\frac {x}{2}\right )+a}d\tanh \left (\frac {x}{2}\right )}{b}-\frac {x \left (6 a^2-b^2\right ) \left (a^2+b^2\right )}{b}}{b}+\frac {2 a \left (3 a^2+2 b^2\right ) \cosh (x)}{b}\right )}{2 b}\right )}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {i \left (\frac {i \left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {-\frac {8 a^3 \left (3 a^2+4 b^2\right ) \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{b}-\frac {x \left (6 a^2-b^2\right ) \left (a^2+b^2\right )}{b}}{b}+\frac {2 a \left (3 a^2+2 b^2\right ) \cosh (x)}{b}\right )}{2 b}\right )}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {i \left (\frac {i \left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {2 a \left (3 a^2+2 b^2\right ) \cosh (x)}{b}+\frac {-\frac {x \left (6 a^2-b^2\right ) \left (a^2+b^2\right )}{b}-\frac {4 a^3 \left (3 a^2+4 b^2\right ) \text {arctanh}\left (\frac {2 b-2 a \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2}}}{b}\right )}{2 b}\right )}{b \left (a^2+b^2\right )}\)

Input:

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

Output:

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

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 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 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f* 
(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin 
[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 
2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 
+ b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - 
 d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x] 
, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || 
IntegersQ[2*m, 2*n])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3528

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ 
.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[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] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.35

method	result	size
default	\(-\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-b +4 a}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (6 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{4}}+\frac {2 a^{3} \left (\frac {\frac {b^{2} \tanh \left (\frac {x}{2}\right )}{a^{2}+b^{2}}+\frac {a b}{a^{2}+b^{2}}}{\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a}-\frac {\left (3 a^{2}+4 b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{b^{4}}+\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-b -4 a}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-6 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{4}}\)	\(218\)
risch	\(\frac {3 x \,a^{2}}{b^{4}}-\frac {x}{2 b^{2}}+\frac {{\mathrm e}^{2 x}}{8 b^{2}}-\frac {a \,{\mathrm e}^{x}}{b^{3}}-\frac {a \,{\mathrm e}^{-x}}{b^{3}}-\frac {{\mathrm e}^{-2 x}}{8 b^{2}}+\frac {2 a^{4} \left ({\mathrm e}^{x} a -b \right )}{b^{4} \left (a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right )}+\frac {3 a^{5} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{4}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{2}}-\frac {3 a^{5} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{4}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{2}}\)	\(343\)

Input:

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

Output:

-1/2/b^2/(tanh(1/2*x)+1)^2-1/2*(-b+4*a)/b^3/(tanh(1/2*x)+1)+1/2*(6*a^2-b^2 
)/b^4*ln(tanh(1/2*x)+1)+2*a^3/b^4*((b^2/(a^2+b^2)*tanh(1/2*x)+a*b/(a^2+b^2 
))/(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a)-(3*a^2+4*b^2)/(a^2+b^2)^(3/2)*arcta 
nh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2)))+1/2/b^2/(tanh(1/2*x)-1)^2-1 
/2*(-b-4*a)/b^3/(tanh(1/2*x)-1)+1/2/b^4*(-6*a^2+b^2)*ln(tanh(1/2*x)-1)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1769 vs. \(2 (154) = 308\).

Time = 0.11 (sec) , antiderivative size = 1769, normalized size of antiderivative = 10.92 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*(a^4*b^3 + 2*a^2*b^5 + b^7 + (a^4*b^3 + 2*a^2*b^5 + b^7)*cosh(x)^6 + ( 
a^4*b^3 + 2*a^2*b^5 + b^7)*sinh(x)^6 - 6*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos 
h(x)^5 - 6*(a^5*b^2 + 2*a^3*b^4 + a*b^6 - (a^4*b^3 + 2*a^2*b^5 + b^7)*cosh 
(x))*sinh(x)^5 - (16*a^6*b + 33*a^4*b^3 + 18*a^2*b^5 + b^7 - 4*(6*a^6*b + 
11*a^4*b^3 + 4*a^2*b^5 - b^7)*x)*cosh(x)^4 - (16*a^6*b + 33*a^4*b^3 + 18*a 
^2*b^5 + b^7 - 15*(a^4*b^3 + 2*a^2*b^5 + b^7)*cosh(x)^2 - 4*(6*a^6*b + 11* 
a^4*b^3 + 4*a^2*b^5 - b^7)*x + 30*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cosh(x))*s 
inh(x)^4 + 8*(2*a^7 + 2*a^5*b^2 + (6*a^7 + 11*a^5*b^2 + 4*a^3*b^4 - a*b^6) 
*x)*cosh(x)^3 + 4*(4*a^7 + 4*a^5*b^2 + 5*(a^4*b^3 + 2*a^2*b^5 + b^7)*cosh( 
x)^3 - 15*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cosh(x)^2 + 2*(6*a^7 + 11*a^5*b^2 
+ 4*a^3*b^4 - a*b^6)*x - (16*a^6*b + 33*a^4*b^3 + 18*a^2*b^5 + b^7 - 4*(6* 
a^6*b + 11*a^4*b^3 + 4*a^2*b^5 - b^7)*x)*cosh(x))*sinh(x)^3 - (32*a^6*b + 
49*a^4*b^3 + 18*a^2*b^5 + b^7 + 4*(6*a^6*b + 11*a^4*b^3 + 4*a^2*b^5 - b^7) 
*x)*cosh(x)^2 - (32*a^6*b + 49*a^4*b^3 + 18*a^2*b^5 + b^7 - 15*(a^4*b^3 + 
2*a^2*b^5 + b^7)*cosh(x)^4 + 60*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cosh(x)^3 + 
6*(16*a^6*b + 33*a^4*b^3 + 18*a^2*b^5 + b^7 - 4*(6*a^6*b + 11*a^4*b^3 + 4* 
a^2*b^5 - b^7)*x)*cosh(x)^2 + 4*(6*a^6*b + 11*a^4*b^3 + 4*a^2*b^5 - b^7)*x 
 - 24*(2*a^7 + 2*a^5*b^2 + (6*a^7 + 11*a^5*b^2 + 4*a^3*b^4 - a*b^6)*x)*cos 
h(x))*sinh(x)^2 + 8*((3*a^5*b + 4*a^3*b^3)*cosh(x)^4 + (3*a^5*b + 4*a^3*b^ 
3)*sinh(x)^4 + 2*(3*a^6 + 4*a^4*b^2)*cosh(x)^3 + 2*(3*a^6 + 4*a^4*b^2 +...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.58 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=-\frac {{\left (3 \, a^{2} + 4 \, b^{2}\right )} a^{3} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {a^{2} b^{3} + b^{5} - 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} e^{\left (-x\right )} - {\left (32 \, a^{4} b + 17 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-2 \, x\right )} - 8 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} e^{\left (-3 \, x\right )}}{8 \, {\left ({\left (a^{2} b^{5} + b^{7}\right )} e^{\left (-2 \, x\right )} + 2 \, {\left (a^{3} b^{4} + a b^{6}\right )} e^{\left (-3 \, x\right )} - {\left (a^{2} b^{5} + b^{7}\right )} e^{\left (-4 \, x\right )}\right )}} - \frac {8 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )}}{8 \, b^{3}} + \frac {{\left (6 \, a^{2} - b^{2}\right )} x}{2 \, b^{4}} \] Input:

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

Output:

-(3*a^2 + 4*b^2)*a^3*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + 
sqrt(a^2 + b^2)))/((a^2*b^4 + b^6)*sqrt(a^2 + b^2)) + 1/8*(a^2*b^3 + b^5 - 
 6*(a^3*b^2 + a*b^4)*e^(-x) - (32*a^4*b + 17*a^2*b^3 + b^5)*e^(-2*x) - 8*( 
2*a^5 - a^3*b^2 - a*b^4)*e^(-3*x))/((a^2*b^5 + b^7)*e^(-2*x) + 2*(a^3*b^4 
+ a*b^6)*e^(-3*x) - (a^2*b^5 + b^7)*e^(-4*x)) - 1/8*(8*a*e^(-x) + b*e^(-2* 
x))/b^3 + 1/2*(6*a^2 - b^2)*x/b^4
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.45 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=-\frac {{\left (3 \, a^{5} + 4 \, a^{3} b^{2}\right )} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {{\left (6 \, a^{2} - b^{2}\right )} x}{2 \, b^{4}} + \frac {b^{2} e^{\left (2 \, x\right )} - 8 \, a b e^{x}}{8 \, b^{4}} + \frac {{\left (a^{2} b^{3} + b^{5} + 8 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} e^{\left (3 \, x\right )} - {\left (32 \, a^{4} b + 17 \, a^{2} b^{3} + b^{5}\right )} e^{\left (2 \, x\right )} + 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} e^{x}\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (a^{2} + b^{2}\right )} {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )} b^{4}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 2.10 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.88 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,b^2}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {\frac {2\,a^4}{b^2\,\left (a^2\,b+b^3\right )}-\frac {2\,a^5\,{\mathrm {e}}^x}{b^3\,\left (a^2\,b+b^3\right )}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}+\frac {x\,\left (6\,a^2-b^2\right )}{2\,b^4}-\frac {a\,{\mathrm {e}}^x}{b^3}-\frac {a\,{\mathrm {e}}^{-x}}{b^3}-\frac {a^3\,\ln \left (\frac {2\,{\mathrm {e}}^x\,\left (3\,a^5+4\,a^3\,b^2\right )}{a^2\,b^5+b^7}-\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^x\right )\,\left (3\,a^2+4\,b^2\right )}{b^5\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (3\,a^2+4\,b^2\right )}{b^4\,{\left (a^2+b^2\right )}^{3/2}}+\frac {a^3\,\ln \left (\frac {2\,{\mathrm {e}}^x\,\left (3\,a^5+4\,a^3\,b^2\right )}{a^2\,b^5+b^7}+\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^x\right )\,\left (3\,a^2+4\,b^2\right )}{b^5\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (3\,a^2+4\,b^2\right )}{b^4\,{\left (a^2+b^2\right )}^{3/2}} \] Input:

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

Output:

exp(2*x)/(8*b^2) - exp(-2*x)/(8*b^2) - ((2*a^4)/(b^2*(a^2*b + b^3)) - (2*a 
^5*exp(x))/(b^3*(a^2*b + b^3)))/(2*a*exp(x) - b + b*exp(2*x)) + (x*(6*a^2 
- b^2))/(2*b^4) - (a*exp(x))/b^3 - (a*exp(-x))/b^3 - (a^3*log((2*exp(x)*(3 
*a^5 + 4*a^3*b^2))/(b^7 + a^2*b^5) - (2*a^3*(b - a*exp(x))*(3*a^2 + 4*b^2) 
)/(b^5*(a^2 + b^2)^(3/2)))*(3*a^2 + 4*b^2))/(b^4*(a^2 + b^2)^(3/2)) + (a^3 
*log((2*exp(x)*(3*a^5 + 4*a^3*b^2))/(b^7 + a^2*b^5) + (2*a^3*(b - a*exp(x) 
)*(3*a^2 + 4*b^2))/(b^5*(a^2 + b^2)^(3/2)))*(3*a^2 + 4*b^2))/(b^4*(a^2 + b 
^2)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 719, normalized size of antiderivative = 4.44 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {24 e^{4 x} a^{6} b x +44 e^{4 x} a^{4} b^{3} x +16 e^{4 x} a^{2} b^{5} x +88 e^{3 x} a^{5} b^{2} x +32 e^{3 x} a^{3} b^{4} x -8 e^{3 x} a \,b^{6} x -24 e^{2 x} a^{6} b x -44 e^{2 x} a^{4} b^{3} x -16 e^{2 x} a^{2} b^{5} x -48 e^{4 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{5} b i -64 e^{4 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{3} b^{3} i -128 e^{3 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{4} b^{2} i +48 e^{2 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{5} b i +64 e^{2 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{3} b^{3} i +2 e^{6 x} a^{2} b^{5}-6 e^{5 x} a^{5} b^{2}-12 e^{5 x} a^{3} b^{4}-6 e^{5 x} a \,b^{6}-24 e^{4 x} a^{6} b -41 e^{4 x} a^{4} b^{3}-18 e^{4 x} a^{2} b^{5}-4 e^{4 x} b^{7} x +48 e^{3 x} a^{7} x -24 e^{2 x} a^{6} b -41 e^{2 x} a^{4} b^{3}-18 e^{2 x} a^{2} b^{5}+4 e^{2 x} b^{7} x +6 e^{x} a^{5} b^{2}+12 e^{x} a^{3} b^{4}+6 e^{x} a \,b^{6}+e^{6 x} b^{7}-e^{4 x} b^{7}-e^{2 x} b^{7}+a^{4} b^{3}+2 a^{2} b^{5}-96 e^{3 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{6} i +e^{6 x} a^{4} b^{3}+b^{7}}{8 e^{2 x} b^{4} \left (e^{2 x} a^{4} b +2 e^{2 x} a^{2} b^{3}+e^{2 x} b^{5}+2 e^{x} a^{5}+4 e^{x} a^{3} b^{2}+2 e^{x} a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}\right )} \] Input:

int(sinh(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**5*b*i - 64*e**(4*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 
+ b**2))*a**3*b**3*i - 96*e**(3*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i) 
/sqrt(a**2 + b**2))*a**6*i - 128*e**(3*x)*sqrt(a**2 + b**2)*atan((e**x*b*i 
 + a*i)/sqrt(a**2 + b**2))*a**4*b**2*i + 48*e**(2*x)*sqrt(a**2 + b**2)*ata 
n((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a**5*b*i + 64*e**(2*x)*sqrt(a**2 + b 
**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a**3*b**3*i + e**(6*x)*a**4* 
b**3 + 2*e**(6*x)*a**2*b**5 + e**(6*x)*b**7 - 6*e**(5*x)*a**5*b**2 - 12*e* 
*(5*x)*a**3*b**4 - 6*e**(5*x)*a*b**6 + 24*e**(4*x)*a**6*b*x - 24*e**(4*x)* 
a**6*b + 44*e**(4*x)*a**4*b**3*x - 41*e**(4*x)*a**4*b**3 + 16*e**(4*x)*a** 
2*b**5*x - 18*e**(4*x)*a**2*b**5 - 4*e**(4*x)*b**7*x - e**(4*x)*b**7 + 48* 
e**(3*x)*a**7*x + 88*e**(3*x)*a**5*b**2*x + 32*e**(3*x)*a**3*b**4*x - 8*e* 
*(3*x)*a*b**6*x - 24*e**(2*x)*a**6*b*x - 24*e**(2*x)*a**6*b - 44*e**(2*x)* 
a**4*b**3*x - 41*e**(2*x)*a**4*b**3 - 16*e**(2*x)*a**2*b**5*x - 18*e**(2*x 
)*a**2*b**5 + 4*e**(2*x)*b**7*x - e**(2*x)*b**7 + 6*e**x*a**5*b**2 + 12*e* 
*x*a**3*b**4 + 6*e**x*a*b**6 + a**4*b**3 + 2*a**2*b**5 + b**7)/(8*e**(2*x) 
*b**4*(e**(2*x)*a**4*b + 2*e**(2*x)*a**2*b**3 + e**(2*x)*b**5 + 2*e**x*a** 
5 + 4*e**x*a**3*b**2 + 2*e**x*a*b**4 - a**4*b - 2*a**2*b**3 - b**5))