\(\int \frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)} \, dx\) [246]

Optimal result

Integrand size = 21, antiderivative size = 80 \[ \int \frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b \text {arctanh}(\cosh (c+d x))}{a^2 d}-\frac {2 b^2 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \sqrt {a^2+b^2} d}-\frac {\coth (c+d x)}{a d} \] Output:

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

Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.41 \[ \int \frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a \coth \left (\frac {1}{2} (c+d x)\right )+2 b \left (-\frac {2 b \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-\log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )\right )+a \tanh \left (\frac {1}{2} (c+d x)\right )}{2 a^2 d} \] Input:

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

Output:

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

Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.11, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 25, 3281, 27, 3042, 26, 3226, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3226

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

((-I)*b*((I*ArcTanh[Cosh[c + d*x]])/(a*d) + ((2*I)*b*ArcTanh[Tanh[(c + d*x 
)/2]/(2*Sqrt[a^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d)))/a - Coth[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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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 = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b^{2} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} \sqrt {a^{2}+b^{2}}}}{d}\)	\(97\)
default	\(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b^{2} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} \sqrt {a^{2}+b^{2}}}}{d}\)	\(97\)
risch	\(-\frac {2}{d a \left ({\mathrm e}^{2 d x +2 c}-1\right )}+\frac {b^{2} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d \,a^{2}}-\frac {b^{2} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{d x +c}-1\right )}{d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{d x +c}+1\right )}{d \,a^{2}}\)	\(179\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 479 vs. \(2 (77) = 154\).

Time = 0.13 (sec) , antiderivative size = 479, normalized size of antiderivative = 5.99 \[ \int \frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 \, a^{3} + 2 \, a b^{2} - {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) + {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) - {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) - {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) - {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right )}{{\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{4} + a^{2} b^{2}\right )} d \sinh \left (d x + c\right )^{2} - {\left (a^{4} + a^{2} b^{2}\right )} d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.71 \[ \int \frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b^{2} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{2} d} + \frac {b \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} - \frac {b \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} + \frac {2}{{\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.54 \[ \int \frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {b^{2} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} + \frac {b \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{2}} - \frac {b \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{2}} - \frac {2}{a {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}}{d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 360, normalized size of antiderivative = 4.50 \[ \int \frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2}{a\,d-a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}}+\frac {b^2\,\ln \left (128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a\,b^3-64\,a^3\,b-32\,b^3\,\sqrt {a^2+b^2}+32\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a^2\,b\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+128\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+96\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^4+d\,a^2\,b^2}-\frac {b^2\,\ln \left (32\,b^3\,\sqrt {a^2+b^2}-64\,a\,b^3-64\,a^3\,b+128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+64\,a^2\,b\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-128\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}-96\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^4+d\,a^2\,b^2}-\frac {b\,\ln \left (32\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-32\right )}{a^2\,d}+\frac {b\,\ln \left (32\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\right )}{a^2\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 330, normalized size of antiderivative = 4.12 \[ \int \frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 e^{2 d x +2 c} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{2} i -2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{2} i -e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2} b -e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) b^{3}+e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2} b +e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) b^{3}-2 e^{2 d x +2 c} a^{3}-2 e^{2 d x +2 c} a \,b^{2}+\mathrm {log}\left (e^{d x +c}-1\right ) a^{2} b +\mathrm {log}\left (e^{d x +c}-1\right ) b^{3}-\mathrm {log}\left (e^{d x +c}+1\right ) a^{2} b -\mathrm {log}\left (e^{d x +c}+1\right ) b^{3}}{a^{2} d \left (e^{2 d x +2 c} a^{2}+e^{2 d x +2 c} b^{2}-a^{2}-b^{2}\right )} \] Input:

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

Output:

(2*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**2*i - 2*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sq 
rt(a**2 + b**2))*b**2*i - e**(2*c + 2*d*x)*log(e**(c + d*x) - 1)*a**2*b - 
e**(2*c + 2*d*x)*log(e**(c + d*x) - 1)*b**3 + e**(2*c + 2*d*x)*log(e**(c + 
 d*x) + 1)*a**2*b + e**(2*c + 2*d*x)*log(e**(c + d*x) + 1)*b**3 - 2*e**(2* 
c + 2*d*x)*a**3 - 2*e**(2*c + 2*d*x)*a*b**2 + log(e**(c + d*x) - 1)*a**2*b 
 + log(e**(c + d*x) - 1)*b**3 - log(e**(c + d*x) + 1)*a**2*b - log(e**(c + 
 d*x) + 1)*b**3)/(a**2*d*(e**(2*c + 2*d*x)*a**2 + e**(2*c + 2*d*x)*b**2 - 
a**2 - b**2))