\(\int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx\) [78]

Optimal result
Mathematica [A] (verified)
Rubi [C] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 81 \[ \int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx=\frac {\left (a^2-2 b^2\right ) \text {arctanh}(\cosh (x))}{2 a^3}+\frac {2 b^3 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^3 \sqrt {a^2+b^2}}+\frac {b \coth (x)}{a^2}-\frac {\coth (x) \text {csch}(x)}{2 a} \] Output:

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

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.68 \[ \int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx=-\frac {\frac {16 b^3 \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-4 a b \coth \left (\frac {x}{2}\right )+a^2 \text {csch}^2\left (\frac {x}{2}\right )-4 \left (a^2-2 b^2\right ) \log \left (\cosh \left (\frac {x}{2}\right )\right )+4 \left (a^2-2 b^2\right ) \log \left (\sinh \left (\frac {x}{2}\right )\right )+a^2 \text {sech}^2\left (\frac {x}{2}\right )-4 a b \tanh \left (\frac {x}{2}\right )}{8 a^3} \] Input:

Integrate[Csch[x]^3/(a + b*Sinh[x]),x]
 

Output:

-1/8*((16*b^3*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] 
 - 4*a*b*Coth[x/2] + a^2*Csch[x/2]^2 - 4*(a^2 - 2*b^2)*Log[Cosh[x/2]] + 4* 
(a^2 - 2*b^2)*Log[Sinh[x/2]] + a^2*Sech[x/2]^2 - 4*a*b*Tanh[x/2])/a^3
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.84 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.40, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.385, Rules used = {3042, 26, 3281, 26, 3042, 25, 3534, 25, 3042, 26, 3480, 26, 3042, 26, 3139, 1083, 219, 4257}

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.

\(\displaystyle \int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {1}{\sin (i x)^3 (a-i b \sin (i x))}dx\)

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (-\frac {i \int -\frac {-b \sin (i x)^2-i a \sin (i x)+2 b}{\sin (i x)^2 (a-i b \sin (i x))}dx}{2 a}-\frac {i \coth (x) \text {csch}(x)}{2 a}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -i \left (\frac {i \int \frac {-b \sin (i x)^2-i a \sin (i x)+2 b}{\sin (i x)^2 (a-i b \sin (i x))}dx}{2 a}-\frac {i \coth (x) \text {csch}(x)}{2 a}\right )\)

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (\frac {i \left (\frac {2 b \coth (x)}{a}-\frac {\int \frac {i \left (a^2-i b \sin (i x) a-2 b^2\right )}{\sin (i x) (a-i b \sin (i x))}dx}{a}\right )}{2 a}-\frac {i \coth (x) \text {csch}(x)}{2 a}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle -i \left (\frac {i \left (\frac {2 b \coth (x)}{a}-\frac {i \int \frac {a^2-i b \sin (i x) a-2 b^2}{\sin (i x) (a-i b \sin (i x))}dx}{a}\right )}{2 a}-\frac {i \coth (x) \text {csch}(x)}{2 a}\right )\)

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (\frac {i \left (\frac {2 b \coth (x)}{a}-\frac {i \left (-\frac {i \left (a^2-2 b^2\right ) \int i \csc (i x)dx}{a}-\frac {2 i b^3 \int \frac {1}{a-i b \sin (i x)}dx}{a}\right )}{a}\right )}{2 a}-\frac {i \coth (x) \text {csch}(x)}{2 a}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle -i \left (\frac {i \left (\frac {2 b \coth (x)}{a}-\frac {i \left (\frac {\left (a^2-2 b^2\right ) \int \csc (i x)dx}{a}-\frac {2 i b^3 \int \frac {1}{a-i b \sin (i x)}dx}{a}\right )}{a}\right )}{2 a}-\frac {i \coth (x) \text {csch}(x)}{2 a}\right )\)

\(\Big \downarrow \) 3139

\(\displaystyle -i \left (\frac {i \left (\frac {2 b \coth (x)}{a}-\frac {i \left (\frac {\left (a^2-2 b^2\right ) \int \csc (i x)dx}{a}-\frac {4 i b^3 \int \frac {1}{-a \tanh ^2\left (\frac {x}{2}\right )+2 b \tanh \left (\frac {x}{2}\right )+a}d\tanh \left (\frac {x}{2}\right )}{a}\right )}{a}\right )}{2 a}-\frac {i \coth (x) \text {csch}(x)}{2 a}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle -i \left (\frac {i \left (\frac {2 b \coth (x)}{a}-\frac {i \left (\frac {\left (a^2-2 b^2\right ) \int \csc (i x)dx}{a}+\frac {8 i b^3 \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{a}\right )}{a}\right )}{2 a}-\frac {i \coth (x) \text {csch}(x)}{2 a}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle -i \left (\frac {i \left (\frac {2 b \coth (x)}{a}-\frac {i \left (\frac {\left (a^2-2 b^2\right ) \int \csc (i x)dx}{a}+\frac {4 i b^3 \text {arctanh}\left (\frac {2 b-2 a \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2}}\right )}{a}\right )}{2 a}-\frac {i \coth (x) \text {csch}(x)}{2 a}\right )\)

\(\Big \downarrow \) 4257

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

Input:

Int[Csch[x]^3/(a + b*Sinh[x]),x]
 

Output:

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

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 3281
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 
))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))
 

rule 3480
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b 
- a*B)/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ 
(b*c - a*d)   Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3534
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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.33

method result size
default \(\frac {\frac {\tanh \left (\frac {x}{2}\right )^{2} a}{2}+2 b \tanh \left (\frac {x}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tanh \left (\frac {x}{2}\right )^{2}}+\frac {\left (-2 a^{2}+4 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tanh \left (\frac {x}{2}\right )}-\frac {2 b^{3} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3} \sqrt {a^{2}+b^{2}}}\) \(108\)
risch \(-\frac {a \,{\mathrm e}^{3 x}-2 b \,{\mathrm e}^{2 x}+{\mathrm e}^{x} a +2 b}{\left ({\mathrm e}^{2 x}-1\right )^{2} a^{2}}-\frac {\ln \left ({\mathrm e}^{x}-1\right )}{2 a}+\frac {\ln \left ({\mathrm e}^{x}-1\right ) b^{2}}{a^{3}}+\frac {\ln \left ({\mathrm e}^{x}+1\right )}{2 a}-\frac {\ln \left ({\mathrm e}^{x}+1\right ) b^{2}}{a^{3}}+\frac {b^{3} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, a^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, a^{3}}\) \(188\)

Input:

int(csch(x)^3/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
 

Output:

1/4/a^2*(1/2*tanh(1/2*x)^2*a+2*b*tanh(1/2*x))-1/8/a/tanh(1/2*x)^2+1/4/a^3* 
(-2*a^2+4*b^2)*ln(tanh(1/2*x))+1/2/a^2*b/tanh(1/2*x)-2*b^3/a^3/(a^2+b^2)^( 
1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 929 vs. \(2 (73) = 146\).

Time = 0.19 (sec) , antiderivative size = 929, normalized size of antiderivative = 11.47 \[ \int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx=\text {Too large to display} \] Input:

integrate(csch(x)^3/(a+b*sinh(x)),x, algorithm="fricas")
 

Output:

-1/2*(4*a^3*b + 4*a*b^3 + 2*(a^4 + a^2*b^2)*cosh(x)^3 + 2*(a^4 + a^2*b^2)* 
sinh(x)^3 - 4*(a^3*b + a*b^3)*cosh(x)^2 - 2*(2*a^3*b + 2*a*b^3 - 3*(a^4 + 
a^2*b^2)*cosh(x))*sinh(x)^2 - 2*(b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + 
 b^3*sinh(x)^4 - 2*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 - b^3)*sinh(x) 
^2 + 4*(b^3*cosh(x)^3 - b^3*cosh(x))*sinh(x))*sqrt(a^2 + b^2)*log((b^2*cos 
h(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*(a^4 + a^ 
2*b^2)*cosh(x) - ((a^4 - a^2*b^2 - 2*b^4)*cosh(x)^4 + 4*(a^4 - a^2*b^2 - 2 
*b^4)*cosh(x)*sinh(x)^3 + (a^4 - a^2*b^2 - 2*b^4)*sinh(x)^4 + a^4 - a^2*b^ 
2 - 2*b^4 - 2*(a^4 - a^2*b^2 - 2*b^4)*cosh(x)^2 - 2*(a^4 - a^2*b^2 - 2*b^4 
 - 3*(a^4 - a^2*b^2 - 2*b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 - a^2*b^2 - 2* 
b^4)*cosh(x)^3 - (a^4 - a^2*b^2 - 2*b^4)*cosh(x))*sinh(x))*log(cosh(x) + s 
inh(x) + 1) + ((a^4 - a^2*b^2 - 2*b^4)*cosh(x)^4 + 4*(a^4 - a^2*b^2 - 2*b^ 
4)*cosh(x)*sinh(x)^3 + (a^4 - a^2*b^2 - 2*b^4)*sinh(x)^4 + a^4 - a^2*b^2 - 
 2*b^4 - 2*(a^4 - a^2*b^2 - 2*b^4)*cosh(x)^2 - 2*(a^4 - a^2*b^2 - 2*b^4 - 
3*(a^4 - a^2*b^2 - 2*b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 - a^2*b^2 - 2*b^4 
)*cosh(x)^3 - (a^4 - a^2*b^2 - 2*b^4)*cosh(x))*sinh(x))*log(cosh(x) + sinh 
(x) - 1) + 2*(a^4 + a^2*b^2 + 3*(a^4 + a^2*b^2)*cosh(x)^2 - 4*(a^3*b + a*b 
^3)*cosh(x))*sinh(x))/(a^5 + a^3*b^2 + (a^5 + a^3*b^2)*cosh(x)^4 + 4*(a...
 

Sympy [F]

\[ \int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx=\int \frac {\operatorname {csch}^{3}{\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \] Input:

integrate(csch(x)**3/(a+b*sinh(x)),x)
 

Output:

Integral(csch(x)**3/(a + b*sinh(x)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (73) = 146\).

Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.90 \[ \int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx=-\frac {b^{3} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} + \frac {a e^{\left (-x\right )} + 2 \, b e^{\left (-2 \, x\right )} + a e^{\left (-3 \, x\right )} - 2 \, b}{2 \, a^{2} e^{\left (-2 \, x\right )} - a^{2} e^{\left (-4 \, x\right )} - a^{2}} + \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \, a^{3}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \, a^{3}} \] Input:

integrate(csch(x)^3/(a+b*sinh(x)),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.69 \[ \int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx=-\frac {b^{3} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} + \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{x} + 1\right )}{2 \, a^{3}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | e^{x} - 1 \right |}\right )}{2 \, a^{3}} - \frac {a e^{\left (3 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a e^{x} + 2 \, b}{a^{2} {\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} \] Input:

integrate(csch(x)^3/(a+b*sinh(x)),x, algorithm="giac")
 

Output:

-b^3*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)*a^3) + 1/2*(a^2 - 2*b^2)*log(e^x + 1)/a^3 - 
 1/2*(a^2 - 2*b^2)*log(abs(e^x - 1))/a^3 - (a*e^(3*x) - 2*b*e^(2*x) + a*e^ 
x + 2*b)/(a^2*(e^(2*x) - 1)^2)
 

Mupad [B] (verification not implemented)

Time = 2.34 (sec) , antiderivative size = 617, normalized size of antiderivative = 7.62 \[ \int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^x}{a-a\,{\mathrm {e}}^{2\,x}}-\frac {2\,{\mathrm {e}}^x}{a-2\,a\,{\mathrm {e}}^{2\,x}+a\,{\mathrm {e}}^{4\,x}}-\frac {\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^x-24\,b^4\,{\mathrm {e}}^x+20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{2\,a}+\frac {\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2+4\,a^4\,{\mathrm {e}}^x+24\,b^4\,{\mathrm {e}}^x-20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{2\,a}+\frac {2\,b}{a^2\,{\mathrm {e}}^{2\,x}-a^2}+\frac {b^2\,\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^x-24\,b^4\,{\mathrm {e}}^x+20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{a^3}-\frac {b^2\,\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2+4\,a^4\,{\mathrm {e}}^x+24\,b^4\,{\mathrm {e}}^x-20\,a^2\,b^2\,{\mathrm {e}}^x\right )}{a^3}-\frac {b^3\,\ln \left (16\,a^5\,b-48\,a\,b^5-24\,b^5\,\sqrt {a^2+b^2}-32\,a^3\,b^3-32\,a^6\,{\mathrm {e}}^x+24\,b^6\,{\mathrm {e}}^x-40\,a^2\,b^3\,\sqrt {a^2+b^2}-32\,a^5\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+112\,a^2\,b^4\,{\mathrm {e}}^x+56\,a^4\,b^2\,{\mathrm {e}}^x+16\,a^4\,b\,\sqrt {a^2+b^2}+72\,a\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+72\,a^3\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^5+a^3\,b^2}+\frac {b^3\,\ln \left (24\,b^5\,\sqrt {a^2+b^2}-48\,a\,b^5+16\,a^5\,b-32\,a^3\,b^3-32\,a^6\,{\mathrm {e}}^x+24\,b^6\,{\mathrm {e}}^x+40\,a^2\,b^3\,\sqrt {a^2+b^2}+32\,a^5\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+112\,a^2\,b^4\,{\mathrm {e}}^x+56\,a^4\,b^2\,{\mathrm {e}}^x-16\,a^4\,b\,\sqrt {a^2+b^2}-72\,a\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}-72\,a^3\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^5+a^3\,b^2} \] Input:

int(1/(sinh(x)^3*(a + b*sinh(x))),x)
 

Output:

exp(x)/(a - a*exp(2*x)) - (2*exp(x))/(a - 2*a*exp(2*x) + a*exp(4*x)) - log 
(4*a^4 + 24*b^4 - 20*a^2*b^2 - 4*a^4*exp(x) - 24*b^4*exp(x) + 20*a^2*b^2*e 
xp(x))/(2*a) + log(4*a^4 + 24*b^4 - 20*a^2*b^2 + 4*a^4*exp(x) + 24*b^4*exp 
(x) - 20*a^2*b^2*exp(x))/(2*a) + (2*b)/(a^2*exp(2*x) - a^2) + (b^2*log(4*a 
^4 + 24*b^4 - 20*a^2*b^2 - 4*a^4*exp(x) - 24*b^4*exp(x) + 20*a^2*b^2*exp(x 
)))/a^3 - (b^2*log(4*a^4 + 24*b^4 - 20*a^2*b^2 + 4*a^4*exp(x) + 24*b^4*exp 
(x) - 20*a^2*b^2*exp(x)))/a^3 - (b^3*log(16*a^5*b - 48*a*b^5 - 24*b^5*(a^2 
 + b^2)^(1/2) - 32*a^3*b^3 - 32*a^6*exp(x) + 24*b^6*exp(x) - 40*a^2*b^3*(a 
^2 + b^2)^(1/2) - 32*a^5*exp(x)*(a^2 + b^2)^(1/2) + 112*a^2*b^4*exp(x) + 5 
6*a^4*b^2*exp(x) + 16*a^4*b*(a^2 + b^2)^(1/2) + 72*a*b^4*exp(x)*(a^2 + b^2 
)^(1/2) + 72*a^3*b^2*exp(x)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^5 + a 
^3*b^2) + (b^3*log(24*b^5*(a^2 + b^2)^(1/2) - 48*a*b^5 + 16*a^5*b - 32*a^3 
*b^3 - 32*a^6*exp(x) + 24*b^6*exp(x) + 40*a^2*b^3*(a^2 + b^2)^(1/2) + 32*a 
^5*exp(x)*(a^2 + b^2)^(1/2) + 112*a^2*b^4*exp(x) + 56*a^4*b^2*exp(x) - 16* 
a^4*b*(a^2 + b^2)^(1/2) - 72*a*b^4*exp(x)*(a^2 + b^2)^(1/2) - 72*a^3*b^2*e 
xp(x)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^5 + a^3*b^2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 519, normalized size of antiderivative = 6.41 \[ \int \frac {\text {csch}^3(x)}{a+b \sinh (x)} \, dx=\frac {-\mathrm {log}\left (e^{x}-1\right ) a^{4}+\mathrm {log}\left (e^{x}+1\right ) a^{4}-2 e^{3 x} a^{4}-2 e^{x} a^{4}+2 \,\mathrm {log}\left (e^{x}-1\right ) b^{4}-2 \,\mathrm {log}\left (e^{x}+1\right ) b^{4}-4 e^{4 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{3} i +8 e^{2 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{3} i -2 a^{3} b -2 a \,b^{3}+e^{4 x} \mathrm {log}\left (e^{x}-1\right ) a^{2} b^{2}+e^{4 x} \mathrm {log}\left (e^{x}+1\right ) a^{4}-e^{4 x} \mathrm {log}\left (e^{x}+1\right ) a^{2} b^{2}+\mathrm {log}\left (e^{x}-1\right ) a^{2} b^{2}-\mathrm {log}\left (e^{x}+1\right ) a^{2} b^{2}-4 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{3} i -2 e^{2 x} \mathrm {log}\left (e^{x}-1\right ) a^{2} b^{2}+2 e^{2 x} \mathrm {log}\left (e^{x}+1\right ) a^{2} b^{2}-e^{4 x} \mathrm {log}\left (e^{x}-1\right ) a^{4}+2 e^{4 x} \mathrm {log}\left (e^{x}-1\right ) b^{4}-2 e^{4 x} \mathrm {log}\left (e^{x}+1\right ) b^{4}+2 e^{4 x} a^{3} b +2 e^{4 x} a \,b^{3}-2 e^{3 x} a^{2} b^{2}+2 e^{2 x} \mathrm {log}\left (e^{x}-1\right ) a^{4}-4 e^{2 x} \mathrm {log}\left (e^{x}-1\right ) b^{4}-2 e^{2 x} \mathrm {log}\left (e^{x}+1\right ) a^{4}+4 e^{2 x} \mathrm {log}\left (e^{x}+1\right ) b^{4}-2 e^{x} a^{2} b^{2}}{2 a^{3} \left (e^{4 x} a^{2}+e^{4 x} b^{2}-2 e^{2 x} a^{2}-2 e^{2 x} b^{2}+a^{2}+b^{2}\right )} \] Input:

int(csch(x)^3/(a+b*sinh(x)),x)
 

Output:

( - 4*e**(4*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))* 
b**3*i + 8*e**(2*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b* 
*2))*b**3*i - 4*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2)) 
*b**3*i - e**(4*x)*log(e**x - 1)*a**4 + e**(4*x)*log(e**x - 1)*a**2*b**2 + 
 2*e**(4*x)*log(e**x - 1)*b**4 + e**(4*x)*log(e**x + 1)*a**4 - e**(4*x)*lo 
g(e**x + 1)*a**2*b**2 - 2*e**(4*x)*log(e**x + 1)*b**4 + 2*e**(4*x)*a**3*b 
+ 2*e**(4*x)*a*b**3 - 2*e**(3*x)*a**4 - 2*e**(3*x)*a**2*b**2 + 2*e**(2*x)* 
log(e**x - 1)*a**4 - 2*e**(2*x)*log(e**x - 1)*a**2*b**2 - 4*e**(2*x)*log(e 
**x - 1)*b**4 - 2*e**(2*x)*log(e**x + 1)*a**4 + 2*e**(2*x)*log(e**x + 1)*a 
**2*b**2 + 4*e**(2*x)*log(e**x + 1)*b**4 - 2*e**x*a**4 - 2*e**x*a**2*b**2 
- log(e**x - 1)*a**4 + log(e**x - 1)*a**2*b**2 + 2*log(e**x - 1)*b**4 + lo 
g(e**x + 1)*a**4 - log(e**x + 1)*a**2*b**2 - 2*log(e**x + 1)*b**4 - 2*a**3 
*b - 2*a*b**3)/(2*a**3*(e**(4*x)*a**2 + e**(4*x)*b**2 - 2*e**(2*x)*a**2 - 
2*e**(2*x)*b**2 + a**2 + b**2))