\(\int \frac {a+b \text {csch}^2(x)}{c+d \sinh (x)} \, dx\) [337]

Optimal result

Integrand size = 17, antiderivative size = 69 \[ \int \frac {a+b \text {csch}^2(x)}{c+d \sinh (x)} \, dx=\frac {b d \text {arctanh}(\cosh (x))}{c^2}-\frac {2 \left (a c^2+b d^2\right ) \text {arctanh}\left (\frac {d-c \tanh \left (\frac {x}{2}\right )}{\sqrt {c^2+d^2}}\right )}{c^2 \sqrt {c^2+d^2}}-\frac {b \coth (x)}{c} \] Output:

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

Mathematica [A] (verified)

Time = 1.59 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.54 \[ \int \frac {a+b \text {csch}^2(x)}{c+d \sinh (x)} \, dx=\frac {\text {csch}\left (\frac {x}{2}\right ) \text {sech}\left (\frac {x}{2}\right ) \left (-b c \cosh (x)+\left (\frac {2 \left (a c^2+b d^2\right ) \arctan \left (\frac {d-c \tanh \left (\frac {x}{2}\right )}{\sqrt {-c^2-d^2}}\right )}{\sqrt {-c^2-d^2}}+b d \left (\log \left (\cosh \left (\frac {x}{2}\right )\right )-\log \left (\sinh \left (\frac {x}{2}\right )\right )\right )\right ) \sinh (x)\right )}{2 c^2} \] Input:

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

Output:

(Csch[x/2]*Sech[x/2]*(-(b*c*Cosh[x]) + ((2*(a*c^2 + b*d^2)*ArcTan[(d - c*T 
anh[x/2])/Sqrt[-c^2 - d^2]])/Sqrt[-c^2 - d^2] + b*d*(Log[Cosh[x/2]] - Log[ 
Sinh[x/2]]))*Sinh[x]))/(2*c^2)
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.69 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4721, 25, 25, 3042, 25, 3535, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4721

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

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 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 3535

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[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*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, 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]
 

rule 4721

Int[(csc[(a_.) + (b_.)*(x_)]^2*(C_.) + (A_))*(u_), x_Symbol] :> Int[Activat 
eTrig[u]*((C + A*Sin[a + b*x]^2)/Sin[a + b*x]^2), x] /; FreeQ[{a, b, A, C}, 
 x] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 1.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25

method	result	size
default	\(-\frac {b \tanh \left (\frac {x}{2}\right )}{2 c}-\frac {\left (-4 a \,c^{2}-4 b \,d^{2}\right ) \operatorname {arctanh}\left (\frac {2 c \tanh \left (\frac {x}{2}\right )-2 d}{2 \sqrt {c^{2}+d^{2}}}\right )}{2 c^{2} \sqrt {c^{2}+d^{2}}}-\frac {b}{2 c \tanh \left (\frac {x}{2}\right )}-\frac {b d \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{c^{2}}\)	\(86\)
parts	\(\frac {2 a \,\operatorname {arctanh}\left (\frac {2 c \tanh \left (\frac {x}{2}\right )-2 d}{2 \sqrt {c^{2}+d^{2}}}\right )}{\sqrt {c^{2}+d^{2}}}+b \left (-\frac {\tanh \left (\frac {x}{2}\right )}{2 c}+\frac {2 d^{2} \operatorname {arctanh}\left (\frac {2 c \tanh \left (\frac {x}{2}\right )-2 d}{2 \sqrt {c^{2}+d^{2}}}\right )}{c^{2} \sqrt {c^{2}+d^{2}}}-\frac {1}{2 \tanh \left (\frac {x}{2}\right ) c}-\frac {d \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{c^{2}}\right )\)	\(111\)
risch	\(-\frac {2 b}{c \left ({\mathrm e}^{2 x}-1\right )}+\frac {b d \ln \left (1+{\mathrm e}^{x}\right )}{c^{2}}-\frac {b d \ln \left (-1+{\mathrm e}^{x}\right )}{c^{2}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c -c^{2}-d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) a}{\sqrt {c^{2}+d^{2}}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c -c^{2}-d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) b \,d^{2}}{\sqrt {c^{2}+d^{2}}\, c^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) a}{\sqrt {c^{2}+d^{2}}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) b \,d^{2}}{\sqrt {c^{2}+d^{2}}\, c^{2}}\)	\(245\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 401 vs. \(2 (65) = 130\).

Time = 0.33 (sec) , antiderivative size = 401, normalized size of antiderivative = 5.81 \[ \int \frac {a+b \text {csch}^2(x)}{c+d \sinh (x)} \, dx=\frac {2 \, b c^{3} + 2 \, b c d^{2} + {\left (a c^{2} + b d^{2} - {\left (a c^{2} + b d^{2}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (a c^{2} + b d^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (a c^{2} + b d^{2}\right )} \sinh \left (x\right )^{2}\right )} \sqrt {c^{2} + d^{2}} \log \left (\frac {d^{2} \cosh \left (x\right )^{2} + d^{2} \sinh \left (x\right )^{2} + 2 \, c d \cosh \left (x\right ) + 2 \, c^{2} + d^{2} + 2 \, {\left (d^{2} \cosh \left (x\right ) + c d\right )} \sinh \left (x\right ) - 2 \, \sqrt {c^{2} + d^{2}} {\left (d \cosh \left (x\right ) + d \sinh \left (x\right ) + c\right )}}{d \cosh \left (x\right )^{2} + d \sinh \left (x\right )^{2} + 2 \, c \cosh \left (x\right ) + 2 \, {\left (d \cosh \left (x\right ) + c\right )} \sinh \left (x\right ) - d}\right ) + {\left (b c^{2} d + b d^{3} - {\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (b c^{2} d + b d^{3}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - {\left (b c^{2} d + b d^{3} - {\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (b c^{2} d + b d^{3}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{c^{4} + c^{2} d^{2} - {\left (c^{4} + c^{2} d^{2}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (c^{4} + c^{2} d^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (c^{4} + c^{2} d^{2}\right )} \sinh \left (x\right )^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (65) = 130\).

Time = 0.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.29 \[ \int \frac {a+b \text {csch}^2(x)}{c+d \sinh (x)} \, dx=b {\left (\frac {d^{2} \log \left (\frac {d e^{\left (-x\right )} - c - \sqrt {c^{2} + d^{2}}}{d e^{\left (-x\right )} - c + \sqrt {c^{2} + d^{2}}}\right )}{\sqrt {c^{2} + d^{2}} c^{2}} + \frac {d \log \left (e^{\left (-x\right )} + 1\right )}{c^{2}} - \frac {d \log \left (e^{\left (-x\right )} - 1\right )}{c^{2}} + \frac {2}{c e^{\left (-2 \, x\right )} - c}\right )} + \frac {a \log \left (\frac {d e^{\left (-x\right )} - c - \sqrt {c^{2} + d^{2}}}{d e^{\left (-x\right )} - c + \sqrt {c^{2} + d^{2}}}\right )}{\sqrt {c^{2} + d^{2}}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.58 \[ \int \frac {a+b \text {csch}^2(x)}{c+d \sinh (x)} \, dx=\frac {b d \log \left (e^{x} + 1\right )}{c^{2}} - \frac {b d \log \left ({\left | e^{x} - 1 \right |}\right )}{c^{2}} + \frac {{\left (a c^{2} + b d^{2}\right )} \log \left (\frac {{\left | 2 \, d e^{x} + 2 \, c - 2 \, \sqrt {c^{2} + d^{2}} \right |}}{{\left | 2 \, d e^{x} + 2 \, c + 2 \, \sqrt {c^{2} + d^{2}} \right |}}\right )}{\sqrt {c^{2} + d^{2}} c^{2}} - \frac {2 \, b}{c {\left (e^{\left (2 \, x\right )} - 1\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.18 (sec) , antiderivative size = 613, normalized size of antiderivative = 8.88 \[ \int \frac {a+b \text {csch}^2(x)}{c+d \sinh (x)} \, dx=\frac {b\,d\,\ln \left ({\mathrm {e}}^x+1\right )}{c^2}-\frac {b\,d\,\ln \left ({\mathrm {e}}^x-1\right )}{c^2}-\frac {2\,b}{c\,\left ({\mathrm {e}}^{2\,x}-1\right )}-\frac {\ln \left (\frac {\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,\left (a^2\,c^4+2\,a\,b\,c^2\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3\,d+2\,b^2\,c^2\,d^2-3\,{\mathrm {e}}^x\,b^2\,c\,d^3+2\,b^2\,d^4\right )}{c^2\,d^4}-\frac {\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,c\,\left (4\,a\,c^3\,{\mathrm {e}}^x-2\,b\,d^3-2\,a\,c^2\,d+a\,c\,d^2\,{\mathrm {e}}^x+3\,b\,c\,d^2\,{\mathrm {e}}^x\right )}{d^5}+\frac {32\,\left (a\,c^2+b\,d^2\right )\,\left (-4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2+2\,d^3\right )}{d^5\,\sqrt {c^2+d^2}}\right )}{c^2\,\sqrt {c^2+d^2}}\right )}{c^2\,\sqrt {c^2+d^2}}-\frac {32\,b\,\left (a\,c^2+b\,d^2\right )\,\left (2\,b\,d+a\,c\,{\mathrm {e}}^x-4\,b\,c\,{\mathrm {e}}^x\right )}{c^3\,d^3}\right )\,\left (a\,c^2+b\,d^2\right )\,\sqrt {c^2+d^2}}{c^4+c^2\,d^2}+\frac {\ln \left (-\frac {\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,\left (a^2\,c^4+2\,a\,b\,c^2\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3\,d+2\,b^2\,c^2\,d^2-3\,{\mathrm {e}}^x\,b^2\,c\,d^3+2\,b^2\,d^4\right )}{c^2\,d^4}+\frac {\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,c\,\left (4\,a\,c^3\,{\mathrm {e}}^x-2\,b\,d^3-2\,a\,c^2\,d+a\,c\,d^2\,{\mathrm {e}}^x+3\,b\,c\,d^2\,{\mathrm {e}}^x\right )}{d^5}-\frac {32\,\left (a\,c^2+b\,d^2\right )\,\left (-4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2+2\,d^3\right )}{d^5\,\sqrt {c^2+d^2}}\right )}{c^2\,\sqrt {c^2+d^2}}\right )}{c^2\,\sqrt {c^2+d^2}}-\frac {32\,b\,\left (a\,c^2+b\,d^2\right )\,\left (2\,b\,d+a\,c\,{\mathrm {e}}^x-4\,b\,c\,{\mathrm {e}}^x\right )}{c^3\,d^3}\right )\,\left (a\,c^2+b\,d^2\right )\,\sqrt {c^2+d^2}}{c^4+c^2\,d^2} \] Input:

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

Output:

(b*d*log(exp(x) + 1))/c^2 - (b*d*log(exp(x) - 1))/c^2 - (2*b)/(c*(exp(2*x) 
 - 1)) - (log(((a*c^2 + b*d^2)*((32*(a^2*c^4 + 2*b^2*d^4 + 2*b^2*c^2*d^2 - 
 3*b^2*c*d^3*exp(x) - 4*b^2*c^3*d*exp(x) + 2*a*b*c^2*d^2))/(c^2*d^4) - ((a 
*c^2 + b*d^2)*((32*c*(4*a*c^3*exp(x) - 2*b*d^3 - 2*a*c^2*d + a*c*d^2*exp(x 
) + 3*b*c*d^2*exp(x)))/d^5 + (32*(a*c^2 + b*d^2)*(3*c^2*d + 2*d^3 - 4*c^3* 
exp(x) - 3*c*d^2*exp(x)))/(d^5*(c^2 + d^2)^(1/2))))/(c^2*(c^2 + d^2)^(1/2) 
)))/(c^2*(c^2 + d^2)^(1/2)) - (32*b*(a*c^2 + b*d^2)*(2*b*d + a*c*exp(x) - 
4*b*c*exp(x)))/(c^3*d^3))*(a*c^2 + b*d^2)*(c^2 + d^2)^(1/2))/(c^4 + c^2*d^ 
2) + (log(- ((a*c^2 + b*d^2)*((32*(a^2*c^4 + 2*b^2*d^4 + 2*b^2*c^2*d^2 - 3 
*b^2*c*d^3*exp(x) - 4*b^2*c^3*d*exp(x) + 2*a*b*c^2*d^2))/(c^2*d^4) + ((a*c 
^2 + b*d^2)*((32*c*(4*a*c^3*exp(x) - 2*b*d^3 - 2*a*c^2*d + a*c*d^2*exp(x) 
+ 3*b*c*d^2*exp(x)))/d^5 - (32*(a*c^2 + b*d^2)*(3*c^2*d + 2*d^3 - 4*c^3*ex 
p(x) - 3*c*d^2*exp(x)))/(d^5*(c^2 + d^2)^(1/2))))/(c^2*(c^2 + d^2)^(1/2))) 
)/(c^2*(c^2 + d^2)^(1/2)) - (32*b*(a*c^2 + b*d^2)*(2*b*d + a*c*exp(x) - 4* 
b*c*exp(x)))/(c^3*d^3))*(a*c^2 + b*d^2)*(c^2 + d^2)^(1/2))/(c^4 + c^2*d^2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 333, normalized size of antiderivative = 4.83 \[ \int \frac {a+b \text {csch}^2(x)}{c+d \sinh (x)} \, dx=\frac {2 e^{2 x} \sqrt {c^{2}+d^{2}}\, \mathit {atan} \left (\frac {e^{x} d i +c i}{\sqrt {c^{2}+d^{2}}}\right ) a \,c^{2} i +2 e^{2 x} \sqrt {c^{2}+d^{2}}\, \mathit {atan} \left (\frac {e^{x} d i +c i}{\sqrt {c^{2}+d^{2}}}\right ) b \,d^{2} i -2 \sqrt {c^{2}+d^{2}}\, \mathit {atan} \left (\frac {e^{x} d i +c i}{\sqrt {c^{2}+d^{2}}}\right ) a \,c^{2} i -2 \sqrt {c^{2}+d^{2}}\, \mathit {atan} \left (\frac {e^{x} d i +c i}{\sqrt {c^{2}+d^{2}}}\right ) b \,d^{2} i -e^{2 x} \mathrm {log}\left (e^{x}-1\right ) b \,c^{2} d -e^{2 x} \mathrm {log}\left (e^{x}-1\right ) b \,d^{3}+e^{2 x} \mathrm {log}\left (e^{x}+1\right ) b \,c^{2} d +e^{2 x} \mathrm {log}\left (e^{x}+1\right ) b \,d^{3}-2 e^{2 x} b \,c^{3}-2 e^{2 x} b c \,d^{2}+\mathrm {log}\left (e^{x}-1\right ) b \,c^{2} d +\mathrm {log}\left (e^{x}-1\right ) b \,d^{3}-\mathrm {log}\left (e^{x}+1\right ) b \,c^{2} d -\mathrm {log}\left (e^{x}+1\right ) b \,d^{3}}{c^{2} \left (e^{2 x} c^{2}+e^{2 x} d^{2}-c^{2}-d^{2}\right )} \] Input:

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

Output:

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