3.205 \(\int \frac {A+B \text {csch}(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=99 \[ \frac {b B \log (a+b \cosh (x))}{a^2-b^2}+\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (1-\cosh (x))}{2 (a+b)}-\frac {B \log (\cosh (x)+1)}{2 (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {4225, 4401, 2659, 208, 2668, 706, 31, 633} \[ \frac {b B \log (a+b \cosh (x))}{a^2-b^2}+\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (1-\cosh (x))}{2 (a+b)}-\frac {B \log (\cosh (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Csch[x])/(a + b*Cosh[x]),x]

[Out]

(2*A*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) + (B*Log[1 - Cosh[x]])/(2*(a + b)
) - (B*Log[1 + Cosh[x]])/(2*(a - b)) + (b*B*Log[a + b*Cosh[x]])/(a^2 - b^2)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int \frac {A+B \text {csch}(x)}{a+b \cosh (x)} \, dx &=-\left (i \int \frac {\text {csch}(x) (i B+i A \sinh (x))}{a+b \cosh (x)} \, dx\right )\\ &=\int \left (\frac {A}{a+b \cosh (x)}+\frac {B \text {csch}(x)}{a+b \cosh (x)}\right ) \, dx\\ &=A \int \frac {1}{a+b \cosh (x)} \, dx+B \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx\\ &=(2 A) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )-(b B) \operatorname {Subst}\left (\int \frac {1}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \cosh (x)\right )\\ &=\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {(b B) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \cosh (x)\right )}{a^2-b^2}+\frac {(b B) \operatorname {Subst}\left (\int \frac {-a+x}{b^2-x^2} \, dx,x,b \cosh (x)\right )}{a^2-b^2}\\ &=\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {b B \log (a+b \cosh (x))}{a^2-b^2}+\frac {B \operatorname {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \cosh (x)\right )}{2 (a-b)}-\frac {B \operatorname {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \cosh (x)\right )}{2 (a+b)}\\ &=\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (1-\cosh (x))}{2 (a+b)}-\frac {B \log (1+\cosh (x))}{2 (a-b)}+\frac {b B \log (a+b \cosh (x))}{a^2-b^2}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 81, normalized size = 0.82 \[ \frac {B \left (b \log (a+b \cosh (x))+a \log \left (\tanh \left (\frac {x}{2}\right )\right )-b \log (\sinh (x))\right )}{a^2-b^2}-\frac {2 A \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Csch[x])/(a + b*Cosh[x]),x]

[Out]

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

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fricas [A]  time = 3.25, size = 298, normalized size = 3.01 \[ \left [\frac {B b \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + \sqrt {a^{2} - b^{2}} A \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) - {\left (B a + B b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + {\left (B a - B b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right )}{a^{2} - b^{2}}, \frac {B b \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - 2 \, \sqrt {-a^{2} + b^{2}} A \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) - {\left (B a + B b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + {\left (B a - B b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right )}{a^{2} - b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(B*b*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) + sqrt(a^2 - b^2)*A*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*
b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*co
sh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)) - (B*a + B*b)*log(cosh(x) + sinh(x) + 1)
+ (B*a - B*b)*log(cosh(x) + sinh(x) - 1))/(a^2 - b^2), (B*b*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - 2*sq
rt(-a^2 + b^2)*A*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) - (B*a + B*b)*log(cosh(x) +
sinh(x) + 1) + (B*a - B*b)*log(cosh(x) + sinh(x) - 1))/(a^2 - b^2)]

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giac [A]  time = 0.15, size = 90, normalized size = 0.91 \[ \frac {B b \log \left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )}{a^{2} - b^{2}} + \frac {2 \, A \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}}} - \frac {B \log \left (e^{x} + 1\right )}{a - b} + \frac {B \log \left ({\left | e^{x} - 1 \right |}\right )}{a + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b*log(b*e^(2*x) + 2*a*e^x + b)/(a^2 - b^2) + 2*A*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/sqrt(-a^2 + b^2) - B*l
og(e^x + 1)/(a - b) + B*log(abs(e^x - 1))/(a + b)

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maple [A]  time = 0.12, size = 138, normalized size = 1.39 \[ \frac {B b \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}{\left (a +b \right ) \left (a -b \right )}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) A a}{\left (a +b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) A b}{\left (a +b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a +b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*csch(x))/(a+b*cosh(x)),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 3.33, size = 983, normalized size = 9.93 \[ \frac {\ln \left (\frac {\left (\frac {32\,\left (A^2\,a^2\,b+2\,{\mathrm {e}}^x\,A^2\,a\,b^2+A^2\,b^3-8\,{\mathrm {e}}^x\,A\,B\,a^2\,b-4\,A\,B\,a\,b^2+2\,{\mathrm {e}}^x\,A\,B\,b^3+4\,{\mathrm {e}}^x\,B^2\,a^3+2\,B^2\,a^2\,b+3\,{\mathrm {e}}^x\,B^2\,a\,b^2\right )}{b^5}+\frac {\left (\frac {32\,\left (2\,B\,b^4+B\,a^2\,b^2-4\,A\,a^4\,{\mathrm {e}}^x-A\,b^4\,{\mathrm {e}}^x+2\,A\,a\,b^3-2\,A\,a^3\,b+6\,B\,a\,b^3\,{\mathrm {e}}^x-3\,B\,a^3\,b\,{\mathrm {e}}^x+5\,A\,a^2\,b^2\,{\mathrm {e}}^x\right )}{b^5}-\frac {32\,\left (A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b^3+B\,a^2\,b\right )\,\left (4\,{\mathrm {e}}^x\,a^5+3\,a^4\,b-5\,{\mathrm {e}}^x\,a^3\,b^2-3\,a^2\,b^3+{\mathrm {e}}^x\,a\,b^4\right )}{b^5\,\left (a^4-2\,a^2\,b^2+b^4\right )}\right )\,\left (A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b^3+B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4}\right )\,\left (A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b^3+B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {32\,\left ({\mathrm {e}}^x\,A^2\,B\,a\,b+A^2\,B\,b^2-4\,{\mathrm {e}}^x\,A\,B^2\,a^2-2\,A\,B^2\,a\,b+{\mathrm {e}}^x\,A\,B^2\,b^2+4\,{\mathrm {e}}^x\,B^3\,a\,b+2\,B^3\,b^2\right )}{b^5}\right )\,\left (A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b^3+B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {B\,\ln \left ({\mathrm {e}}^x+1\right )}{a-b}-\frac {\ln \left (-\frac {32\,\left ({\mathrm {e}}^x\,A^2\,B\,a\,b+A^2\,B\,b^2-4\,{\mathrm {e}}^x\,A\,B^2\,a^2-2\,A\,B^2\,a\,b+{\mathrm {e}}^x\,A\,B^2\,b^2+4\,{\mathrm {e}}^x\,B^3\,a\,b+2\,B^3\,b^2\right )}{b^5}-\frac {\left (\frac {32\,\left (A^2\,a^2\,b+2\,{\mathrm {e}}^x\,A^2\,a\,b^2+A^2\,b^3-8\,{\mathrm {e}}^x\,A\,B\,a^2\,b-4\,A\,B\,a\,b^2+2\,{\mathrm {e}}^x\,A\,B\,b^3+4\,{\mathrm {e}}^x\,B^2\,a^3+2\,B^2\,a^2\,b+3\,{\mathrm {e}}^x\,B^2\,a\,b^2\right )}{b^5}-\frac {\left (\frac {32\,\left (2\,B\,b^4+B\,a^2\,b^2-4\,A\,a^4\,{\mathrm {e}}^x-A\,b^4\,{\mathrm {e}}^x+2\,A\,a\,b^3-2\,A\,a^3\,b+6\,B\,a\,b^3\,{\mathrm {e}}^x-3\,B\,a^3\,b\,{\mathrm {e}}^x+5\,A\,a^2\,b^2\,{\mathrm {e}}^x\right )}{b^5}+\frac {32\,\left (B\,b^3+A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^2\,b\right )\,\left (4\,{\mathrm {e}}^x\,a^5+3\,a^4\,b-5\,{\mathrm {e}}^x\,a^3\,b^2-3\,a^2\,b^3+{\mathrm {e}}^x\,a\,b^4\right )}{b^5\,\left (a^4-2\,a^2\,b^2+b^4\right )}\right )\,\left (B\,b^3+A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4}\right )\,\left (B\,b^3+A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4}\right )\,\left (B\,b^3+A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {B\,\ln \left ({\mathrm {e}}^x-1\right )}{a+b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/sinh(x))/(a + b*cosh(x)),x)

[Out]

(log((((32*(A^2*b^3 + A^2*a^2*b + 2*B^2*a^2*b + 4*B^2*a^3*exp(x) + 3*B^2*a*b^2*exp(x) - 4*A*B*a*b^2 + 2*A*B*b^
3*exp(x) + 2*A^2*a*b^2*exp(x) - 8*A*B*a^2*b*exp(x)))/b^5 + (((32*(2*B*b^4 + B*a^2*b^2 - 4*A*a^4*exp(x) - A*b^4
*exp(x) + 2*A*a*b^3 - 2*A*a^3*b + 6*B*a*b^3*exp(x) - 3*B*a^3*b*exp(x) + 5*A*a^2*b^2*exp(x)))/b^5 - (32*(A*((a
+ b)^3*(a - b)^3)^(1/2) - B*b^3 + B*a^2*b)*(3*a^4*b - 3*a^2*b^3 + 4*a^5*exp(x) + a*b^4*exp(x) - 5*a^3*b^2*exp(
x)))/(b^5*(a^4 + b^4 - 2*a^2*b^2)))*(A*((a + b)^3*(a - b)^3)^(1/2) - B*b^3 + B*a^2*b))/(a^4 + b^4 - 2*a^2*b^2)
)*(A*((a + b)^3*(a - b)^3)^(1/2) - B*b^3 + B*a^2*b))/(a^4 + b^4 - 2*a^2*b^2) - (32*(2*B^3*b^2 + A^2*B*b^2 - 2*
A*B^2*a*b + 4*B^3*a*b*exp(x) - 4*A*B^2*a^2*exp(x) + A*B^2*b^2*exp(x) + A^2*B*a*b*exp(x)))/b^5)*(A*((a + b)^3*(
a - b)^3)^(1/2) - B*b^3 + B*a^2*b))/(a^4 + b^4 - 2*a^2*b^2) - (B*log(exp(x) + 1))/(a - b) - (log(- (32*(2*B^3*
b^2 + A^2*B*b^2 - 2*A*B^2*a*b + 4*B^3*a*b*exp(x) - 4*A*B^2*a^2*exp(x) + A*B^2*b^2*exp(x) + A^2*B*a*b*exp(x)))/
b^5 - (((32*(A^2*b^3 + A^2*a^2*b + 2*B^2*a^2*b + 4*B^2*a^3*exp(x) + 3*B^2*a*b^2*exp(x) - 4*A*B*a*b^2 + 2*A*B*b
^3*exp(x) + 2*A^2*a*b^2*exp(x) - 8*A*B*a^2*b*exp(x)))/b^5 - (((32*(2*B*b^4 + B*a^2*b^2 - 4*A*a^4*exp(x) - A*b^
4*exp(x) + 2*A*a*b^3 - 2*A*a^3*b + 6*B*a*b^3*exp(x) - 3*B*a^3*b*exp(x) + 5*A*a^2*b^2*exp(x)))/b^5 + (32*(B*b^3
+ A*((a + b)^3*(a - b)^3)^(1/2) - B*a^2*b)*(3*a^4*b - 3*a^2*b^3 + 4*a^5*exp(x) + a*b^4*exp(x) - 5*a^3*b^2*exp
(x)))/(b^5*(a^4 + b^4 - 2*a^2*b^2)))*(B*b^3 + A*((a + b)^3*(a - b)^3)^(1/2) - B*a^2*b))/(a^4 + b^4 - 2*a^2*b^2
))*(B*b^3 + A*((a + b)^3*(a - b)^3)^(1/2) - B*a^2*b))/(a^4 + b^4 - 2*a^2*b^2))*(B*b^3 + A*((a + b)^3*(a - b)^3
)^(1/2) - B*a^2*b))/(a^4 + b^4 - 2*a^2*b^2) + (B*log(exp(x) - 1))/(a + b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \operatorname {csch}{\relax (x )}}{a + b \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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