\(\int \frac {\coth ^2(x)}{a+b \text {csch}(x)} \, dx\) [119]

Optimal result

Integrand size = 13, antiderivative size = 57 \[ \int \frac {\coth ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {x}{a}-\frac {\text {arctanh}(\cosh (x))}{b}+\frac {2 \sqrt {a^2+b^2} \text {arctanh}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a b} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.32 \[ \int \frac {\coth ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {b x+2 \sqrt {-a^2-b^2} \arctan \left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )-a \log \left (\cosh \left (\frac {x}{2}\right )\right )+a \log \left (\sinh \left (\frac {x}{2}\right )\right )}{a b} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.385, Rules used = {3042, 25, 4382, 3042, 4539, 26, 3042, 26, 4257, 4407, 26, 3042, 26, 4318, 3042, 3139, 1083, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\cot (i x)^2}{a+i b \csc (i x)}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\cot (i x)^2}{a+i b \csc (i x)}dx\)

\(\Big \downarrow \) 4382

\(\displaystyle -\int \frac {-\text {csch}^2(x)-1}{a+b \text {csch}(x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4539

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {b-i a \csc (i x)}{a+i b \csc (i x)}dx}{b}+\frac {\int i \csc (i x)dx}{b}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\int \frac {b-i a \csc (i x)}{a+i b \csc (i x)}dx}{b}+\frac {i \int \csc (i x)dx}{b}\)

\(\Big \downarrow \) 4257

\(\displaystyle -\frac {\text {arctanh}(\cosh (x))}{b}+\frac {\int \frac {b-i a \csc (i x)}{a+i b \csc (i x)}dx}{b}\)

\(\Big \downarrow \) 4407

\(\displaystyle -\frac {\text {arctanh}(\cosh (x))}{b}+\frac {\frac {b x}{a}-\frac {i \left (a^2+b^2\right ) \int -\frac {i \text {csch}(x)}{a+b \text {csch}(x)}dx}{a}}{b}\)

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4318

\(\displaystyle \frac {\frac {b x}{a}-\frac {\left (a^2+b^2\right ) \int \frac {1}{\frac {a \sinh (x)}{b}+1}dx}{a b}}{b}-\frac {\text {arctanh}(\cosh (x))}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\text {arctanh}(\cosh (x))}{b}+\frac {\frac {b x}{a}-\frac {\left (a^2+b^2\right ) \int \frac {1}{1-\frac {i a \sin (i x)}{b}}dx}{a b}}{b}\)

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {4 \left (a^2+b^2\right ) \int \frac {1}{4 \left (\frac {a^2}{b^2}+1\right )-\left (\frac {2 a}{b}-2 \tanh \left (\frac {x}{2}\right )\right )^2}d\left (\frac {2 a}{b}-2 \tanh \left (\frac {x}{2}\right )\right )}{a b}+\frac {b x}{a}}{b}-\frac {\text {arctanh}(\cosh (x))}{b}\)

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4382

Int[cot[(c_.) + (d_.)*(x_)]^2*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), 
x_Symbol] :> Int[(-1 + Csc[c + d*x]^2)*(a + b*Csc[c + d*x])^n, x] /; FreeQ[ 
{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0]
 

rule 4407

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
 (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a   Int[Csc[e + f* 
x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0]
 

rule 4539

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_)), x_Symbol] :> Simp[C/b   Int[Csc[e + f*x], x], x] + Simp[1/b   In 
t[(A*b - a*C*Csc[e + f*x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f, A, C}, x]
 

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.47

method	result	size
default	\(\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{b}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}+\frac {\left (2 a^{2}+2 b^{2}\right ) \operatorname {arctanh}\left (\frac {-2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a b \sqrt {a^{2}+b^{2}}}\)	\(84\)
risch	\(\frac {x}{a}+\frac {\sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{x}+\frac {b +\sqrt {a^{2}+b^{2}}}{a}\right )}{b a}-\frac {\sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{x}-\frac {-b +\sqrt {a^{2}+b^{2}}}{a}\right )}{b a}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{b}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{b}\)	\(100\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (53) = 106\).

Time = 0.11 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.47 \[ \int \frac {\coth ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {b x - a \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + a \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + \sqrt {a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}}{a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) + 2 \, {\left (a \cosh \left (x\right ) + b\right )} \sinh \left (x\right ) - a}\right )}{a b} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.58 \[ \int \frac {\coth ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {x}{a} - \frac {\log \left (e^{\left (-x\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{b} - \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{a b} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.56 \[ \int \frac {\coth ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {x}{a} - \frac {\log \left (e^{x} + 1\right )}{b} + \frac {\log \left ({\left | e^{x} - 1 \right |}\right )}{b} - \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{a b} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 316, normalized size of antiderivative = 5.54 \[ \int \frac {\coth ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {x}{a}+\frac {\ln \left (32\,a^2\,b+32\,b^3-32\,b^3\,{\mathrm {e}}^x-32\,a^2\,b\,{\mathrm {e}}^x\right )}{b}-\frac {\ln \left (32\,a^2\,b+32\,b^3+32\,b^3\,{\mathrm {e}}^x+32\,a^2\,b\,{\mathrm {e}}^x\right )}{b}+\frac {\ln \left (128\,b^5\,{\mathrm {e}}^x-64\,a^3\,b^2-64\,a\,b^4-128\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+32\,a^4\,b\,{\mathrm {e}}^x+160\,a^2\,b^3\,{\mathrm {e}}^x+64\,a\,b^3\,\sqrt {a^2+b^2}+32\,a^3\,b\,\sqrt {a^2+b^2}-96\,a^2\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a\,b}-\frac {\ln \left (128\,b^5\,{\mathrm {e}}^x-64\,a^3\,b^2-64\,a\,b^4+128\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}+32\,a^4\,b\,{\mathrm {e}}^x+160\,a^2\,b^3\,{\mathrm {e}}^x-64\,a\,b^3\,\sqrt {a^2+b^2}-32\,a^3\,b\,\sqrt {a^2+b^2}+96\,a^2\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a\,b} \] Input:

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

Output:

x/a + log(32*a^2*b + 32*b^3 - 32*b^3*exp(x) - 32*a^2*b*exp(x))/b - log(32* 
a^2*b + 32*b^3 + 32*b^3*exp(x) + 32*a^2*b*exp(x))/b + (log(128*b^5*exp(x) 
- 64*a^3*b^2 - 64*a*b^4 - 128*b^4*exp(x)*(a^2 + b^2)^(1/2) + 32*a^4*b*exp( 
x) + 160*a^2*b^3*exp(x) + 64*a*b^3*(a^2 + b^2)^(1/2) + 32*a^3*b*(a^2 + b^2 
)^(1/2) - 96*a^2*b^2*exp(x)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a*b) - 
(log(128*b^5*exp(x) - 64*a^3*b^2 - 64*a*b^4 + 128*b^4*exp(x)*(a^2 + b^2)^( 
1/2) + 32*a^4*b*exp(x) + 160*a^2*b^3*exp(x) - 64*a*b^3*(a^2 + b^2)^(1/2) - 
 32*a^3*b*(a^2 + b^2)^(1/2) + 96*a^2*b^2*exp(x)*(a^2 + b^2)^(1/2))*(a^2 + 
b^2)^(1/2))/(a*b)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int \frac {\coth ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {-2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} a i +b i}{\sqrt {a^{2}+b^{2}}}\right ) i +\mathrm {log}\left (e^{x}-1\right ) a -\mathrm {log}\left (e^{x}+1\right ) a +b x}{a b} \] Input:

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

Output:

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