3.1.95 \(\int \frac {\cosh ^2(x)}{a+b \text {csch}(x)} \, dx\) [95]

3.1.95.1 Optimal result

Integrand size = 13, antiderivative size = 77 \[ \int \frac {\cosh ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {\left (a^2+2 b^2\right ) x}{2 a^3}+\frac {2 b \sqrt {a^2+b^2} \text {arctanh}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^3}-\frac {\cosh (x) (2 b-a \sinh (x))}{2 a^2} \]

output

1/2*(a^2+2*b^2)*x/a^3-1/2*cosh(x)*(2*b-a*sinh(x))/a^2+2*b*arctanh((a-b*tan 
h(1/2*x))/(a^2+b^2)^(1/2))*(a^2+b^2)^(1/2)/a^3
 

3.1.95.2 Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04 \[ \int \frac {\cosh ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {2 a^2 x+4 b^2 x+8 b \sqrt {-a^2-b^2} \arctan \left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )-4 a b \cosh (x)+a^2 \sinh (2 x)}{4 a^3} \]

input

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

output

(2*a^2*x + 4*b^2*x + 8*b*Sqrt[-a^2 - b^2]*ArcTan[(a - b*Tanh[x/2])/Sqrt[-a 
^2 - b^2]] - 4*a*b*Cosh[x] + a^2*Sinh[2*x])/(4*a^3)
 

3.1.95.3 Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.54 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.077, Rules used = {3042, 4360, 26, 26, 3042, 26, 3344, 26, 3042, 3214, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

\(\displaystyle \int \frac {i \sinh (x) \cosh ^2(x)}{i a \sinh (x)+i b}dx\)

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3344

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle -i \left (-\frac {i \left (-\frac {8 b \left (a^2+b^2\right ) \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 a-2 b \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 a-2 b \tanh \left (\frac {x}{2}\right )\right )}{a}-\frac {x \left (a^2+2 b^2\right )}{a}\right )}{2 a^2}-\frac {i \cosh (x) (2 b-a \sinh (x))}{2 a^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle -i \left (-\frac {i \left (-\frac {4 b \sqrt {a^2+b^2} \text {arctanh}\left (\frac {2 a-2 b \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{a}-\frac {x \left (a^2+2 b^2\right )}{a}\right )}{2 a^2}-\frac {i \cosh (x) (2 b-a \sinh (x))}{2 a^2}\right )\)

input

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

output

(-I)*(((-1/2*I)*(-(((a^2 + 2*b^2)*x)/a) - (4*b*Sqrt[a^2 + b^2]*ArcTanh[(2* 
a - 2*b*Tanh[x/2])/(2*Sqrt[a^2 + b^2])])/a))/a^2 - ((I/2)*Cosh[x]*(2*b - a 
*Sinh[x]))/a^2)
 

3.1.95.3.1 Defintions of rubi rules used

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 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/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 3344

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g* 
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* 
p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( 
(p - 1)/(b^2*(m + p)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si 
n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 
2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 
, 0] && IntegerQ[2*m]
 

rule 4360

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 

3.1.95.4 Maple [A] (verified)

Time = 3.48 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.58

method	result	size
risch	\(\frac {x}{2 a}+\frac {x \,b^{2}}{a^{3}}+\frac {{\mathrm e}^{2 x}}{8 a}-\frac {b \,{\mathrm e}^{x}}{2 a^{2}}-\frac {b \,{\mathrm e}^{-x}}{2 a^{2}}-\frac {{\mathrm e}^{-2 x}}{8 a}+\frac {\sqrt {a^{2}+b^{2}}\, b \ln \left ({\mathrm e}^{x}+\frac {b +\sqrt {a^{2}+b^{2}}}{a}\right )}{a^{3}}-\frac {\sqrt {a^{2}+b^{2}}\, b \ln \left ({\mathrm e}^{x}-\frac {-b +\sqrt {a^{2}+b^{2}}}{a}\right )}{a^{3}}\)	\(122\)
default	\(\frac {2 b \sqrt {a^{2}+b^{2}}\, \operatorname {arctanh}\left (\frac {-2 b \tanh \left (\frac {x}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-a +2 b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (a^{2}+2 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a^{3}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-a^{2}-2 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 a^{3}}\)	\(150\)

input

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

output

1/2*x/a+x/a^3*b^2+1/8/a*exp(x)^2-1/2*b/a^2*exp(x)-1/2*b/a^2/exp(x)-1/8/a/e 
xp(x)^2+(a^2+b^2)^(1/2)*b/a^3*ln(exp(x)+(b+(a^2+b^2)^(1/2))/a)-(a^2+b^2)^( 
1/2)*b/a^3*ln(exp(x)-(-b+(a^2+b^2)^(1/2))/a)
 

3.1.95.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (68) = 136\).

Time = 0.26 (sec) , antiderivative size = 304, normalized size of antiderivative = 3.95 \[ \int \frac {\cosh ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {a^{2} \cosh \left (x\right )^{4} + a^{2} \sinh \left (x\right )^{4} - 4 \, a b \cosh \left (x\right )^{3} + 4 \, {\left (a^{2} + 2 \, b^{2}\right )} x \cosh \left (x\right )^{2} + 4 \, {\left (a^{2} \cosh \left (x\right ) - a b\right )} \sinh \left (x\right )^{3} - 4 \, a b \cosh \left (x\right ) + 2 \, {\left (3 \, a^{2} \cosh \left (x\right )^{2} - 6 \, a b \cosh \left (x\right ) + 2 \, {\left (a^{2} + 2 \, b^{2}\right )} x\right )} \sinh \left (x\right )^{2} + 8 \, {\left (b \cosh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) \sinh \left (x\right ) + b \sinh \left (x\right )^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}}{a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) + 2 \, {\left (a \cosh \left (x\right ) + b\right )} \sinh \left (x\right ) - a}\right ) - a^{2} + 4 \, {\left (a^{2} \cosh \left (x\right )^{3} - 3 \, a b \cosh \left (x\right )^{2} + 2 \, {\left (a^{2} + 2 \, b^{2}\right )} x \cosh \left (x\right ) - a b\right )} \sinh \left (x\right )}{8 \, {\left (a^{3} \cosh \left (x\right )^{2} + 2 \, a^{3} \cosh \left (x\right ) \sinh \left (x\right ) + a^{3} \sinh \left (x\right )^{2}\right )}} \]

input

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

output

1/8*(a^2*cosh(x)^4 + a^2*sinh(x)^4 - 4*a*b*cosh(x)^3 + 4*(a^2 + 2*b^2)*x*c 
osh(x)^2 + 4*(a^2*cosh(x) - a*b)*sinh(x)^3 - 4*a*b*cosh(x) + 2*(3*a^2*cosh 
(x)^2 - 6*a*b*cosh(x) + 2*(a^2 + 2*b^2)*x)*sinh(x)^2 + 8*(b*cosh(x)^2 + 2* 
b*cosh(x)*sinh(x) + b*sinh(x)^2)*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^2 + 4*(a^2*cosh(x)^3 - 
 3*a*b*cosh(x)^2 + 2*(a^2 + 2*b^2)*x*cosh(x) - a*b)*sinh(x))/(a^3*cosh(x)^ 
2 + 2*a^3*cosh(x)*sinh(x) + a^3*sinh(x)^2)
 

3.1.95.6 Sympy [F]

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

input

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

output

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

3.1.95.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.58 \[ \int \frac {\cosh ^2(x)}{a+b \text {csch}(x)} \, dx=-\frac {{\left (4 \, b e^{\left (-x\right )} - a\right )} e^{\left (2 \, x\right )}}{8 \, a^{2}} - \frac {4 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )}}{8 \, a^{2}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} x}{2 \, a^{3}} - \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} \]

input

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

output

-1/8*(4*b*e^(-x) - a)*e^(2*x)/a^2 - 1/8*(4*b*e^(-x) + a*e^(-2*x))/a^2 + 1/ 
2*(a^2 + 2*b^2)*x/a^3 - (a^2*b + b^3)*log((a*e^(-x) - b - sqrt(a^2 + b^2)) 
/(a*e^(-x) - b + sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^3)
 

3.1.95.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.57 \[ \int \frac {\cosh ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {a e^{\left (2 \, x\right )} - 4 \, b e^{x}}{8 \, a^{2}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} x}{2 \, a^{3}} - \frac {{\left (4 \, a b e^{x} + a^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, a^{3}} - \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} \]

input

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

output

1/8*(a*e^(2*x) - 4*b*e^x)/a^2 + 1/2*(a^2 + 2*b^2)*x/a^3 - 1/8*(4*a*b*e^x + 
 a^2)*e^(-2*x)/a^3 - (a^2*b + b^3)*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)))/(sqrt(a^2 + b^2)*a^3)
 

3.1.95.9 Mupad [B] (verification not implemented)

Time = 2.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.06 \[ \int \frac {\cosh ^2(x)}{a+b \text {csch}(x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,a}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a}-\frac {b\,{\mathrm {e}}^x}{2\,a^2}-\frac {b\,{\mathrm {e}}^{-x}}{2\,a^2}+\frac {x\,\left (a^2+2\,b^2\right )}{2\,a^3}-\frac {b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^x\,\left (a^2+b^2\right )}{a^4}-\frac {2\,b\,\left (a-b\,{\mathrm {e}}^x\right )\,\sqrt {a^2+b^2}}{a^4}\right )\,\sqrt {a^2+b^2}}{a^3}+\frac {b\,\ln \left (\frac {2\,b\,\left (a-b\,{\mathrm {e}}^x\right )\,\sqrt {a^2+b^2}}{a^4}+\frac {2\,b\,{\mathrm {e}}^x\,\left (a^2+b^2\right )}{a^4}\right )\,\sqrt {a^2+b^2}}{a^3} \]

input

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

output

exp(2*x)/(8*a) - exp(-2*x)/(8*a) - (b*exp(x))/(2*a^2) - (b*exp(-x))/(2*a^2 
) + (x*(a^2 + 2*b^2))/(2*a^3) - (b*log((2*b*exp(x)*(a^2 + b^2))/a^4 - (2*b 
*(a - b*exp(x))*(a^2 + b^2)^(1/2))/a^4)*(a^2 + b^2)^(1/2))/a^3 + (b*log((2 
*b*(a - b*exp(x))*(a^2 + b^2)^(1/2))/a^4 + (2*b*exp(x)*(a^2 + b^2))/a^4)*( 
a^2 + b^2)^(1/2))/a^3