\(\int \frac {\text {csch}^2(x)}{a+b \sinh (x)} \, dx\) [77]

Optimal result
Mathematica [A] (verified)
Rubi [C] (verified)
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 = 13, antiderivative size = 59 \[ \int \frac {\text {csch}^2(x)}{a+b \sinh (x)} \, dx=\frac {b \text {arctanh}(\cosh (x))}{a^2}-\frac {2 b^2 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \sqrt {a^2+b^2}}-\frac {\coth (x)}{a} \] Output:

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

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.53 \[ \int \frac {\text {csch}^2(x)}{a+b \sinh (x)} \, dx=-\frac {a \coth \left (\frac {x}{2}\right )+2 b \left (-\frac {2 b \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-\log \left (\cosh \left (\frac {x}{2}\right )\right )+\log \left (\sinh \left (\frac {x}{2}\right )\right )\right )+a \tanh \left (\frac {x}{2}\right )}{2 a^2} \] Input:

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

Output:

-1/2*(a*Coth[x/2] + 2*b*((-2*b*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]]) 
/Sqrt[-a^2 - b^2] - Log[Cosh[x/2]] + Log[Sinh[x/2]]) + a*Tanh[x/2])/a^2
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.077, Rules used = {3042, 25, 3281, 27, 3042, 26, 3226, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {1}{\sin (i x)^2 (a-i b \sin (i x))}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {1}{\sin (i x)^2 (a-i b \sin (i x))}dx\)

\(\Big \downarrow \) 3281

\(\displaystyle -\frac {\int \frac {b \text {csch}(x)}{a+b \sinh (x)}dx}{a}-\frac {\coth (x)}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {b \int \frac {\text {csch}(x)}{a+b \sinh (x)}dx}{a}-\frac {\coth (x)}{a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3226

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

\(\displaystyle -\frac {\coth (x)}{a}-\frac {i b \left (\frac {i b \int \frac {1}{a-i b \sin (i x)}dx}{a}+\frac {\int \csc (i x)dx}{a}\right )}{a}\)

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

((-I)*b*((I*ArcTanh[Cosh[x]])/a - ((2*I)*b*ArcTanh[(2*b - 2*a*Tanh[x/2])/( 
2*Sqrt[a^2 + b^2])])/(a*Sqrt[a^2 + b^2])))/a - Coth[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 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 3226
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), 
 x], x] - Simp[d/(b*c - a*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 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 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.24

method result size
default \(-\frac {\tanh \left (\frac {x}{2}\right )}{2 a}-\frac {1}{2 a \tanh \left (\frac {x}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{2}}+\frac {2 b^{2} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} \sqrt {a^{2}+b^{2}}}\) \(73\)
risch \(-\frac {2}{a \left ({\mathrm e}^{2 x}-1\right )}+\frac {b \ln \left ({\mathrm e}^{x}+1\right )}{a^{2}}+\frac {b^{2} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, a^{2}}-\frac {b^{2} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, a^{2}}-\frac {b \ln \left ({\mathrm e}^{x}-1\right )}{a^{2}}\) \(143\)

Input:

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

Output:

-1/2/a*tanh(1/2*x)-1/2/a/tanh(1/2*x)-1/a^2*b*ln(tanh(1/2*x))+2*b^2/a^2/(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. 345 vs. \(2 (55) = 110\).

Time = 0.11 (sec) , antiderivative size = 345, normalized size of antiderivative = 5.85 \[ \int \frac {\text {csch}^2(x)}{a+b \sinh (x)} \, dx=\frac {2 \, a^{3} + 2 \, a b^{2} - {\left (b^{2} \cosh \left (x\right )^{2} + 2 \, b^{2} \cosh \left (x\right ) \sinh \left (x\right ) + b^{2} \sinh \left (x\right )^{2} - b^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) + {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{4} + a^{2} b^{2} - {\left (a^{4} + a^{2} b^{2}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (a^{4} + a^{2} b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (a^{4} + a^{2} b^{2}\right )} \sinh \left (x\right )^{2}} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\text {csch}^2(x)}{a+b \sinh (x)} \, dx=\int \frac {\operatorname {csch}^{2}{\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \] Input:

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

Output:

Integral(csch(x)**2/(a + b*sinh(x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.69 \[ \int \frac {\text {csch}^2(x)}{a+b \sinh (x)} \, dx=\frac {b^{2} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} + \frac {b \log \left (e^{\left (-x\right )} + 1\right )}{a^{2}} - \frac {b \log \left (e^{\left (-x\right )} - 1\right )}{a^{2}} + \frac {2}{a e^{\left (-2 \, x\right )} - a} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.66 \[ \int \frac {\text {csch}^2(x)}{a+b \sinh (x)} \, dx=\frac {b^{2} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} + \frac {b \log \left (e^{x} + 1\right )}{a^{2}} - \frac {b \log \left ({\left | e^{x} - 1 \right |}\right )}{a^{2}} - \frac {2}{a {\left (e^{\left (2 \, x\right )} - 1\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 2.00 (sec) , antiderivative size = 292, normalized size of antiderivative = 4.95 \[ \int \frac {\text {csch}^2(x)}{a+b \sinh (x)} \, dx=\frac {2}{a-a\,{\mathrm {e}}^{2\,x}}-\frac {b\,\ln \left (32\,{\mathrm {e}}^x-32\right )}{a^2}+\frac {b\,\ln \left (32\,{\mathrm {e}}^x+32\right )}{a^2}+\frac {b^2\,\ln \left (128\,a^4\,{\mathrm {e}}^x-64\,a\,b^3-64\,a^3\,b-32\,b^3\,\sqrt {a^2+b^2}+32\,b^4\,{\mathrm {e}}^x+128\,a^3\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^x-64\,a^2\,b\,\sqrt {a^2+b^2}+96\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2}-\frac {b^2\,\ln \left (32\,b^3\,\sqrt {a^2+b^2}-64\,a\,b^3-64\,a^3\,b+128\,a^4\,{\mathrm {e}}^x+32\,b^4\,{\mathrm {e}}^x-128\,a^3\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^x+64\,a^2\,b\,\sqrt {a^2+b^2}-96\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^4+a^2\,b^2} \] Input:

int(1/(sinh(x)^2*(a + b*sinh(x))),x)
 

Output:

2/(a - a*exp(2*x)) - (b*log(32*exp(x) - 32))/a^2 + (b*log(32*exp(x) + 32)) 
/a^2 + (b^2*log(128*a^4*exp(x) - 64*a*b^3 - 64*a^3*b - 32*b^3*(a^2 + b^2)^ 
(1/2) + 32*b^4*exp(x) + 128*a^3*exp(x)*(a^2 + b^2)^(1/2) + 160*a^2*b^2*exp 
(x) - 64*a^2*b*(a^2 + b^2)^(1/2) + 96*a*b^2*exp(x)*(a^2 + b^2)^(1/2))*(a^2 
 + b^2)^(1/2))/(a^4 + a^2*b^2) - (b^2*log(32*b^3*(a^2 + b^2)^(1/2) - 64*a* 
b^3 - 64*a^3*b + 128*a^4*exp(x) + 32*b^4*exp(x) - 128*a^3*exp(x)*(a^2 + b^ 
2)^(1/2) + 160*a^2*b^2*exp(x) + 64*a^2*b*(a^2 + b^2)^(1/2) - 96*a*b^2*exp( 
x)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^4 + a^2*b^2)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 242, normalized size of antiderivative = 4.10 \[ \int \frac {\text {csch}^2(x)}{a+b \sinh (x)} \, dx=\frac {2 e^{2 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{2} i -2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{2} i -e^{2 x} \mathrm {log}\left (e^{x}-1\right ) a^{2} b -e^{2 x} \mathrm {log}\left (e^{x}-1\right ) b^{3}+e^{2 x} \mathrm {log}\left (e^{x}+1\right ) a^{2} b +e^{2 x} \mathrm {log}\left (e^{x}+1\right ) b^{3}-2 e^{2 x} a^{3}-2 e^{2 x} a \,b^{2}+\mathrm {log}\left (e^{x}-1\right ) a^{2} b +\mathrm {log}\left (e^{x}-1\right ) b^{3}-\mathrm {log}\left (e^{x}+1\right ) a^{2} b -\mathrm {log}\left (e^{x}+1\right ) b^{3}}{a^{2} \left (e^{2 x} a^{2}+e^{2 x} b^{2}-a^{2}-b^{2}\right )} \] Input:

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

Output:

(2*e**(2*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*b** 
2*i - 2*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*b**2*i 
- e**(2*x)*log(e**x - 1)*a**2*b - e**(2*x)*log(e**x - 1)*b**3 + e**(2*x)*l 
og(e**x + 1)*a**2*b + e**(2*x)*log(e**x + 1)*b**3 - 2*e**(2*x)*a**3 - 2*e* 
*(2*x)*a*b**2 + log(e**x - 1)*a**2*b + log(e**x - 1)*b**3 - log(e**x + 1)* 
a**2*b - log(e**x + 1)*b**3)/(a**2*(e**(2*x)*a**2 + e**(2*x)*b**2 - a**2 - 
 b**2))