\(\int \frac {\text {csch}^2(c+d x) \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx\) [471]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 29, antiderivative size = 144 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b \text {arctanh}(\cosh (c+d x))}{a^2 d}-\frac {2 b^4 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{3/2} d}-\frac {\coth (c+d x)}{a d}-\frac {b \text {sech}(c+d x)}{a^2 d}+\frac {b^2 \text {sech}(c+d x) (b+a \sinh (c+d x))}{a^2 \left (a^2+b^2\right ) d}-\frac {\tanh (c+d x)}{a d} \] Output:

b*arctanh(cosh(d*x+c))/a^2/d-2*b^4*arctanh((b-a*tanh(1/2*d*x+1/2*c))/(a^2+ 
b^2)^(1/2))/a^2/(a^2+b^2)^(3/2)/d-coth(d*x+c)/a/d-b*sech(d*x+c)/a^2/d+b^2* 
sech(d*x+c)*(b+a*sinh(d*x+c))/a^2/(a^2+b^2)/d-tanh(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 1.61 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.06 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {4 b^4 \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{a^2 \left (-a^2-b^2\right )^{3/2}}+\frac {\coth \left (\frac {1}{2} (c+d x)\right )}{a}-\frac {2 b \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}+\frac {2 b \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}+\frac {2 \text {sech}(c+d x) (b+a \sinh (c+d x))}{a^2+b^2}+\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{a}}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3042, 25, 3377, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3377
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a 
_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + 
 f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, 
 g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 
2, 0])
 
Maple [A] (verified)

Time = 8.79 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.97

method result size
derivativedivides \(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b^{4} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}+\frac {-2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{\left (a^{2}+b^{2}\right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}}{d}\) \(139\)
default \(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b^{4} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}+\frac {-2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{\left (a^{2}+b^{2}\right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}}{d}\) \(139\)
risch \(-\frac {2 \left ({\mathrm e}^{3 d x +3 c} a b +{\mathrm e}^{2 d x +2 c} b^{2}-b \,{\mathrm e}^{d x +c} a +2 a^{2}+b^{2}\right )}{d \left (a^{2}+b^{2}\right ) \left (1+{\mathrm e}^{2 d x +2 c}\right ) a \left ({\mathrm e}^{2 d x +2 c}-1\right )}+\frac {b^{4} \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d \,a^{2}}-\frac {b^{4} \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{d x +c}+1\right )}{d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{d x +c}-1\right )}{d \,a^{2}}\) \(261\)

Input:

int(csch(d*x+c)^2*sech(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/a^2*b^4/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/ 
2*c)-2*b)/(a^2+b^2)^(1/2))+2/(a^2+b^2)*(-a*tanh(1/2*d*x+1/2*c)-b)/(1+tanh( 
1/2*d*x+1/2*c)^2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1040 vs. \(2 (141) = 282\).

Time = 0.17 (sec) , antiderivative size = 1040, normalized size of antiderivative = 7.22 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-(4*a^5 + 6*a^3*b^2 + 2*a*b^4 + 2*(a^4*b + a^2*b^3)*cosh(d*x + c)^3 + 2*(a 
^4*b + a^2*b^3)*sinh(d*x + c)^3 + 2*(a^3*b^2 + a*b^4)*cosh(d*x + c)^2 + 2* 
(a^3*b^2 + a*b^4 + 3*(a^4*b + a^2*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - (b 
^4*cosh(d*x + c)^4 + 4*b^4*cosh(d*x + c)^3*sinh(d*x + c) + 6*b^4*cosh(d*x 
+ c)^2*sinh(d*x + c)^2 + 4*b^4*cosh(d*x + c)*sinh(d*x + c)^3 + b^4*sinh(d* 
x + c)^4 - b^4)*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)*sin 
h(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*b + a^2*b^3)*cosh(d*x + c) + (a^4*b 
+ 2*a^2*b^3 + b^5 - (a^4*b + 2*a^2*b^3 + b^5)*cosh(d*x + c)^4 - 4*(a^4*b + 
 2*a^2*b^3 + b^5)*cosh(d*x + c)^3*sinh(d*x + c) - 6*(a^4*b + 2*a^2*b^3 + b 
^5)*cosh(d*x + c)^2*sinh(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3 + b^5)*cosh(d*x 
 + c)*sinh(d*x + c)^3 - (a^4*b + 2*a^2*b^3 + b^5)*sinh(d*x + c)^4)*log(cos 
h(d*x + c) + sinh(d*x + c) + 1) - (a^4*b + 2*a^2*b^3 + b^5 - (a^4*b + 2*a^ 
2*b^3 + b^5)*cosh(d*x + c)^4 - 4*(a^4*b + 2*a^2*b^3 + b^5)*cosh(d*x + c)^3 
*sinh(d*x + c) - 6*(a^4*b + 2*a^2*b^3 + b^5)*cosh(d*x + c)^2*sinh(d*x + c) 
^2 - 4*(a^4*b + 2*a^2*b^3 + b^5)*cosh(d*x + c)*sinh(d*x + c)^3 - (a^4*b + 
2*a^2*b^3 + b^5)*sinh(d*x + c)^4)*log(cosh(d*x + c) + sinh(d*x + c) - 1) - 
 2*(a^4*b + a^2*b^3 - 3*(a^4*b + a^2*b^3)*cosh(d*x + c)^2 - 2*(a^3*b^2 ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.44 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b^{4} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + a^{2} b^{2}\right )} \sqrt {a^{2} + b^{2}} d} - \frac {2 \, {\left (a b e^{\left (-d x - c\right )} + b^{2} e^{\left (-2 \, d x - 2 \, c\right )} - a b e^{\left (-3 \, d x - 3 \, c\right )} + 2 \, a^{2} + b^{2}\right )}}{{\left (a^{3} + a b^{2} - {\left (a^{3} + a b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {b \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} - \frac {b \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.28 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {b^{4} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + a^{2} b^{2}\right )} \sqrt {a^{2} + b^{2}}} + \frac {b \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{2}} - \frac {b \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{2}} - \frac {2 \, {\left (a b e^{\left (3 \, d x + 3 \, c\right )} + b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a b e^{\left (d x + c\right )} + 2 \, a^{2} + b^{2}\right )}}{{\left (a^{3} + a b^{2}\right )} {\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}}}{d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 7.23 (sec) , antiderivative size = 768, normalized size of antiderivative = 5.33 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b^4\,\ln \left (\frac {64\,b^8\,\sqrt {{\left (a^2+b^2\right )}^3}-96\,a\,b^{10}-384\,a^3\,b^8-512\,a^5\,b^6-288\,a^7\,b^4-64\,a^9\,b^2+288\,a^2\,b^9\,{\mathrm {e}}^{c+d\,x}+960\,a^4\,b^7\,{\mathrm {e}}^{c+d\,x}+1152\,a^6\,b^5\,{\mathrm {e}}^{c+d\,x}+608\,a^8\,b^3\,{\mathrm {e}}^{c+d\,x}+128\,a^{10}\,b\,{\mathrm {e}}^{c+d\,x}-64\,a\,b^7\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}+32\,a^3\,b^5\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}}{a^3\,{\left ({\left (a^2+b^2\right )}^3\right )}^{3/2}\,\left (a^2+b^2\right )}-\frac {32\,b\,\left (-4\,{\mathrm {e}}^{c+d\,x}\,a^3+2\,a^2\,b-5\,{\mathrm {e}}^{c+d\,x}\,a\,b^2+2\,b^3\right )}{a^3\,{\left (a^2+b^2\right )}^2}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{d\,a^8+3\,d\,a^6\,b^2+3\,d\,a^4\,b^4+d\,a^2\,b^6}-\frac {\frac {2\,b^4\,{\mathrm {e}}^{3\,c+3\,d\,x}}{d\,\left (a^2\,b^3+b^5\right )}-\frac {2\,b^4\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2\,b^3+b^5\right )}+\frac {2\,b^3\,\left (2\,a^2+b^2\right )}{a\,d\,\left (a^2\,b^3+b^5\right )}+\frac {2\,b^5\,{\mathrm {e}}^{2\,c+2\,d\,x}}{a\,d\,\left (a^2\,b^3+b^5\right )}}{{\mathrm {e}}^{4\,c+4\,d\,x}-1}-\frac {b^4\,\ln \left (\frac {96\,a\,b^{10}+64\,b^8\,\sqrt {{\left (a^2+b^2\right )}^3}+384\,a^3\,b^8+512\,a^5\,b^6+288\,a^7\,b^4+64\,a^9\,b^2-288\,a^2\,b^9\,{\mathrm {e}}^{c+d\,x}-960\,a^4\,b^7\,{\mathrm {e}}^{c+d\,x}-1152\,a^6\,b^5\,{\mathrm {e}}^{c+d\,x}-608\,a^8\,b^3\,{\mathrm {e}}^{c+d\,x}-128\,a^{10}\,b\,{\mathrm {e}}^{c+d\,x}-64\,a\,b^7\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}+32\,a^3\,b^5\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}}{a^3\,{\left ({\left (a^2+b^2\right )}^3\right )}^{3/2}\,\left (a^2+b^2\right )}-\frac {32\,b\,\left (-4\,{\mathrm {e}}^{c+d\,x}\,a^3+2\,a^2\,b-5\,{\mathrm {e}}^{c+d\,x}\,a\,b^2+2\,b^3\right )}{a^3\,{\left (a^2+b^2\right )}^2}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{d\,a^8+3\,d\,a^6\,b^2+3\,d\,a^4\,b^4+d\,a^2\,b^6}-\frac {b\,\ln \left ({\mathrm {e}}^{c+d\,x}-1\right )}{a^2\,d}+\frac {b\,\ln \left ({\mathrm {e}}^{c+d\,x}+1\right )}{a^2\,d} \] Input:

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

Output:

(b^4*log((64*b^8*((a^2 + b^2)^3)^(1/2) - 96*a*b^10 - 384*a^3*b^8 - 512*a^5 
*b^6 - 288*a^7*b^4 - 64*a^9*b^2 + 288*a^2*b^9*exp(c + d*x) + 960*a^4*b^7*e 
xp(c + d*x) + 1152*a^6*b^5*exp(c + d*x) + 608*a^8*b^3*exp(c + d*x) + 128*a 
^10*b*exp(c + d*x) - 64*a*b^7*exp(c + d*x)*((a^2 + b^2)^3)^(1/2) + 32*a^3* 
b^5*exp(c + d*x)*((a^2 + b^2)^3)^(1/2))/(a^3*((a^2 + b^2)^3)^(3/2)*(a^2 + 
b^2)) - (32*b*(2*a^2*b - 4*a^3*exp(c + d*x) + 2*b^3 - 5*a*b^2*exp(c + d*x) 
))/(a^3*(a^2 + b^2)^2))*((a^2 + b^2)^3)^(1/2))/(a^8*d + a^2*b^6*d + 3*a^4* 
b^4*d + 3*a^6*b^2*d) - ((2*b^4*exp(3*c + 3*d*x))/(d*(b^5 + a^2*b^3)) - (2* 
b^4*exp(c + d*x))/(d*(b^5 + a^2*b^3)) + (2*b^3*(2*a^2 + b^2))/(a*d*(b^5 + 
a^2*b^3)) + (2*b^5*exp(2*c + 2*d*x))/(a*d*(b^5 + a^2*b^3)))/(exp(4*c + 4*d 
*x) - 1) - (b^4*log((96*a*b^10 + 64*b^8*((a^2 + b^2)^3)^(1/2) + 384*a^3*b^ 
8 + 512*a^5*b^6 + 288*a^7*b^4 + 64*a^9*b^2 - 288*a^2*b^9*exp(c + d*x) - 96 
0*a^4*b^7*exp(c + d*x) - 1152*a^6*b^5*exp(c + d*x) - 608*a^8*b^3*exp(c + d 
*x) - 128*a^10*b*exp(c + d*x) - 64*a*b^7*exp(c + d*x)*((a^2 + b^2)^3)^(1/2 
) + 32*a^3*b^5*exp(c + d*x)*((a^2 + b^2)^3)^(1/2))/(a^3*((a^2 + b^2)^3)^(3 
/2)*(a^2 + b^2)) - (32*b*(2*a^2*b - 4*a^3*exp(c + d*x) + 2*b^3 - 5*a*b^2*e 
xp(c + d*x)))/(a^3*(a^2 + b^2)^2))*((a^2 + b^2)^3)^(1/2))/(a^8*d + a^2*b^6 
*d + 3*a^4*b^4*d + 3*a^6*b^2*d) - (b*log(exp(c + d*x) - 1))/(a^2*d) + (b*l 
og(exp(c + d*x) + 1))/(a^2*d)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 562, normalized size of antiderivative = 3.90 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 e^{4 d x +4 c} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{4} i -2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{4} i -e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{4} b -2 e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2} b^{3}-e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) b^{5}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{4} b +2 e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2} b^{3}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) b^{5}-4 e^{4 d x +4 c} a^{5}-6 e^{4 d x +4 c} a^{3} b^{2}-2 e^{4 d x +4 c} a \,b^{4}-2 e^{3 d x +3 c} a^{4} b -2 e^{3 d x +3 c} a^{2} b^{3}-2 e^{2 d x +2 c} a^{3} b^{2}-2 e^{2 d x +2 c} a \,b^{4}+2 e^{d x +c} a^{4} b +2 e^{d x +c} a^{2} b^{3}+\mathrm {log}\left (e^{d x +c}-1\right ) a^{4} b +2 \,\mathrm {log}\left (e^{d x +c}-1\right ) a^{2} b^{3}+\mathrm {log}\left (e^{d x +c}-1\right ) b^{5}-\mathrm {log}\left (e^{d x +c}+1\right ) a^{4} b -2 \,\mathrm {log}\left (e^{d x +c}+1\right ) a^{2} b^{3}-\mathrm {log}\left (e^{d x +c}+1\right ) b^{5}}{a^{2} d \left (e^{4 d x +4 c} a^{4}+2 e^{4 d x +4 c} a^{2} b^{2}+e^{4 d x +4 c} b^{4}-a^{4}-2 a^{2} b^{2}-b^{4}\right )} \] Input:

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

Output:

(2*e**(4*c + 4*d*x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqrt(a 
**2 + b**2))*b**4*i - 2*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sq 
rt(a**2 + b**2))*b**4*i - e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*a**4*b - 
2*e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*a**2*b**3 - e**(4*c + 4*d*x)*log( 
e**(c + d*x) - 1)*b**5 + e**(4*c + 4*d*x)*log(e**(c + d*x) + 1)*a**4*b + 2 
*e**(4*c + 4*d*x)*log(e**(c + d*x) + 1)*a**2*b**3 + e**(4*c + 4*d*x)*log(e 
**(c + d*x) + 1)*b**5 - 4*e**(4*c + 4*d*x)*a**5 - 6*e**(4*c + 4*d*x)*a**3* 
b**2 - 2*e**(4*c + 4*d*x)*a*b**4 - 2*e**(3*c + 3*d*x)*a**4*b - 2*e**(3*c + 
 3*d*x)*a**2*b**3 - 2*e**(2*c + 2*d*x)*a**3*b**2 - 2*e**(2*c + 2*d*x)*a*b* 
*4 + 2*e**(c + d*x)*a**4*b + 2*e**(c + d*x)*a**2*b**3 + log(e**(c + d*x) - 
 1)*a**4*b + 2*log(e**(c + d*x) - 1)*a**2*b**3 + log(e**(c + d*x) - 1)*b** 
5 - log(e**(c + d*x) + 1)*a**4*b - 2*log(e**(c + d*x) + 1)*a**2*b**3 - log 
(e**(c + d*x) + 1)*b**5)/(a**2*d*(e**(4*c + 4*d*x)*a**4 + 2*e**(4*c + 4*d* 
x)*a**2*b**2 + e**(4*c + 4*d*x)*b**4 - a**4 - 2*a**2*b**2 - b**4))