\(\int \frac {\text {csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx\) [251]

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

Optimal result

Integrand size = 21, antiderivative size = 113 \[ \int \frac {\text {csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\left (a^2-2 b^2\right ) \text {arctanh}(\cosh (c+d x))}{2 a^3 d}+\frac {2 b^3 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^3 \sqrt {a^2+b^2} d}+\frac {b \coth (c+d x)}{a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d} \] Output:

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

Mathematica [A] (verified)

Time = 2.14 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.48 \[ \int \frac {\text {csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {16 b^3 \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-4 a b \coth \left (\frac {1}{2} (c+d x)\right )+a^2 \text {csch}^2\left (\frac {1}{2} (c+d x)\right )-4 \left (a^2-2 b^2\right ) \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )+4 \left (a^2-2 b^2\right ) \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )+a^2 \text {sech}^2\left (\frac {1}{2} (c+d x)\right )-4 a b \tanh \left (\frac {1}{2} (c+d x)\right )}{8 a^3 d} \] Input:

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

Output:

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

Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.91 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.22, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3042, 26, 3281, 26, 3042, 25, 3534, 25, 3042, 26, 3480, 26, 3042, 26, 3139, 1083, 217, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

\(\displaystyle -i \left (\frac {i \left (\frac {2 b \coth (c+d x)}{a d}-\frac {i \left (\frac {i \left (a^2-2 b^2\right ) \text {arctanh}(\cosh (c+d x))}{a d}-\frac {4 i b^3 \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{a d \sqrt {a^2+b^2}}\right )}{a}\right )}{2 a}-\frac {i \coth (c+d x) \text {csch}(c+d x)}{2 a d}\right )\)

Input:

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

Output:

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

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 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[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 = 0.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.26

method result size
derivativedivides \(\frac {\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}+2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {2 b^{3} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3} \sqrt {a^{2}+b^{2}}}-\frac {1}{8 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-2 a^{2}+4 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) \(142\)
default \(\frac {\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}+2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {2 b^{3} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3} \sqrt {a^{2}+b^{2}}}-\frac {1}{8 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-2 a^{2}+4 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) \(142\)
risch \(-\frac {{\mathrm e}^{3 d x +3 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a \,{\mathrm e}^{d x +c}+2 b}{d \,a^{2} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d \,a^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{d x +c}+1\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right ) b^{2}}{d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{d x +c}-1\right )}{2 d a}+\frac {\ln \left ({\mathrm e}^{d x +c}-1\right ) b^{2}}{d \,a^{3}}\) \(252\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1203 vs. \(2 (106) = 212\).

Time = 0.16 (sec) , antiderivative size = 1203, normalized size of antiderivative = 10.65 \[ \int \frac {\text {csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(4*a^3*b + 4*a*b^3 + 2*(a^4 + a^2*b^2)*cosh(d*x + c)^3 + 2*(a^4 + a^2 
*b^2)*sinh(d*x + c)^3 - 4*(a^3*b + a*b^3)*cosh(d*x + c)^2 - 2*(2*a^3*b + 2 
*a*b^3 - 3*(a^4 + a^2*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 - 2*(b^3*cosh(d* 
x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 - 2*b 
^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)^2 - b^3)*sinh(d*x + c)^2 
 + 4*(b^3*cosh(d*x + c)^3 - b^3*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 + b 
^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 
 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^ 
2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d* 
x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) 
 + 2*(a^4 + a^2*b^2)*cosh(d*x + c) - ((a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c 
)^4 + 4*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4 - a^2 
*b^2 - 2*b^4)*sinh(d*x + c)^4 + a^4 - a^2*b^2 - 2*b^4 - 2*(a^4 - a^2*b^2 - 
 2*b^4)*cosh(d*x + c)^2 - 2*(a^4 - a^2*b^2 - 2*b^4 - 3*(a^4 - a^2*b^2 - 2* 
b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^4 - a^2*b^2 - 2*b^4)*cosh(d* 
x + c)^3 - (a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c))*sinh(d*x + c))*log(cosh( 
d*x + c) + sinh(d*x + c) + 1) + ((a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^4 + 
 4*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4 - a^2*b^2 
- 2*b^4)*sinh(d*x + c)^4 + a^4 - a^2*b^2 - 2*b^4 - 2*(a^4 - a^2*b^2 - 2*b^ 
4)*cosh(d*x + c)^2 - 2*(a^4 - a^2*b^2 - 2*b^4 - 3*(a^4 - a^2*b^2 - 2*b^...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.87 \[ \int \frac {\text {csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {b^{3} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3} d} + \frac {a e^{\left (-d x - c\right )} + 2 \, b e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-3 \, d x - 3 \, c\right )} - 2 \, b}{{\left (2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )} - a^{2} e^{\left (-4 \, d x - 4 \, c\right )} - a^{2}\right )} d} + \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{2 \, a^{3} d} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{2 \, a^{3} d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.56 \[ \int \frac {\text {csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {2 \, b^{3} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{3}} + \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{3}} + \frac {2 \, {\left (a e^{\left (3 \, d x + 3 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a e^{\left (d x + c\right )} + 2 \, b\right )}}{a^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}}{2 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.64 (sec) , antiderivative size = 776, normalized size of antiderivative = 6.87 \[ \int \frac {\text {csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 724, normalized size of antiderivative = 6.41 \[ \int \frac {\text {csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{4}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2} b^{2}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{4}-e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2} b^{2}+2 e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) b^{4}-2 e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) b^{4}+2 e^{4 d x +4 c} a^{3} b +2 e^{4 d x +4 c} a \,b^{3}-2 e^{3 d x +3 c} a^{2} b^{2}+2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{4}-4 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) b^{4}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{4}+4 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) b^{4}-2 e^{d x +c} a^{2} b^{2}+\mathrm {log}\left (e^{d x +c}-1\right ) a^{2} b^{2}-\mathrm {log}\left (e^{d x +c}+1\right ) a^{2} b^{2}-4 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{3} i -2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2} b^{2}+2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2} b^{2}-\mathrm {log}\left (e^{d x +c}-1\right ) a^{4}+\mathrm {log}\left (e^{d x +c}+1\right ) a^{4}-2 e^{3 d x +3 c} a^{4}-2 e^{d x +c} a^{4}+2 \,\mathrm {log}\left (e^{d x +c}-1\right ) b^{4}-2 \,\mathrm {log}\left (e^{d x +c}+1\right ) b^{4}-2 a^{3} b -2 a \,b^{3}-4 e^{4 d x +4 c} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{3} i +8 e^{2 d x +2 c} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{3} i}{2 a^{3} d \left (e^{4 d x +4 c} a^{2}+e^{4 d x +4 c} b^{2}-2 e^{2 d x +2 c} a^{2}-2 e^{2 d x +2 c} b^{2}+a^{2}+b^{2}\right )} \] Input:

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

Output:

( - 4*e**(4*c + 4*d*x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqr 
t(a**2 + b**2))*b**3*i + 8*e**(2*c + 2*d*x)*sqrt(a**2 + b**2)*atan((e**(c 
+ d*x)*b*i + a*i)/sqrt(a**2 + b**2))*b**3*i - 4*sqrt(a**2 + b**2)*atan((e* 
*(c + d*x)*b*i + a*i)/sqrt(a**2 + b**2))*b**3*i - e**(4*c + 4*d*x)*log(e** 
(c + d*x) - 1)*a**4 + e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*a**2*b**2 + 2 
*e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*b**4 + e**(4*c + 4*d*x)*log(e**(c 
+ d*x) + 1)*a**4 - e**(4*c + 4*d*x)*log(e**(c + d*x) + 1)*a**2*b**2 - 2*e* 
*(4*c + 4*d*x)*log(e**(c + d*x) + 1)*b**4 + 2*e**(4*c + 4*d*x)*a**3*b + 2* 
e**(4*c + 4*d*x)*a*b**3 - 2*e**(3*c + 3*d*x)*a**4 - 2*e**(3*c + 3*d*x)*a** 
2*b**2 + 2*e**(2*c + 2*d*x)*log(e**(c + d*x) - 1)*a**4 - 2*e**(2*c + 2*d*x 
)*log(e**(c + d*x) - 1)*a**2*b**2 - 4*e**(2*c + 2*d*x)*log(e**(c + d*x) - 
1)*b**4 - 2*e**(2*c + 2*d*x)*log(e**(c + d*x) + 1)*a**4 + 2*e**(2*c + 2*d* 
x)*log(e**(c + d*x) + 1)*a**2*b**2 + 4*e**(2*c + 2*d*x)*log(e**(c + d*x) + 
 1)*b**4 - 2*e**(c + d*x)*a**4 - 2*e**(c + d*x)*a**2*b**2 - log(e**(c + d* 
x) - 1)*a**4 + log(e**(c + d*x) - 1)*a**2*b**2 + 2*log(e**(c + d*x) - 1)*b 
**4 + log(e**(c + d*x) + 1)*a**4 - log(e**(c + d*x) + 1)*a**2*b**2 - 2*log 
(e**(c + d*x) + 1)*b**4 - 2*a**3*b - 2*a*b**3)/(2*a**3*d*(e**(4*c + 4*d*x) 
*a**2 + e**(4*c + 4*d*x)*b**2 - 2*e**(2*c + 2*d*x)*a**2 - 2*e**(2*c + 2*d* 
x)*b**2 + a**2 + b**2))