3.184 \(\int \frac {\coth ^2(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=77 \[ -\frac {a \coth (x)}{a^2-b^2}+\frac {b \text {csch}(x)}{a^2-b^2}+\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

2*a^2*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/(a-b)^(3/2)/(a+b)^(3/2)-a*coth(x)/(a^2-b^2)+b*csch(x)/(a^2-
b^2)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2727, 3767, 8, 2606, 2659, 208} \[ -\frac {a \coth (x)}{a^2-b^2}+\frac {b \text {csch}(x)}{a^2-b^2}+\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^2/(a + b*Cosh[x]),x]

[Out]

(2*a^2*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)) - (a*Coth[x])/(a^2 - b^2) +
(b*Csch[x])/(a^2 - b^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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 2727

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a/(a^2 - b^
2), Int[(g*Tan[e + f*x])^p/Sin[e + f*x]^2, x], x] + (-Dist[(b*g)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 1)/Cos
[e + f*x], x], x] - Dist[(a^2*g^2)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 2)/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*p] && GtQ[p, 1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {\coth ^2(x)}{a+b \cosh (x)} \, dx &=\frac {a \int \text {csch}^2(x) \, dx}{a^2-b^2}+\frac {a^2 \int \frac {1}{a+b \cosh (x)} \, dx}{a^2-b^2}-\frac {b \int \coth (x) \text {csch}(x) \, dx}{a^2-b^2}\\ &=-\frac {(i a) \operatorname {Subst}(\int 1 \, dx,x,-i \coth (x))}{a^2-b^2}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^2-b^2}+\frac {(i b) \operatorname {Subst}(\int 1 \, dx,x,-i \text {csch}(x))}{a^2-b^2}\\ &=\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}-\frac {a \coth (x)}{a^2-b^2}+\frac {b \text {csch}(x)}{a^2-b^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 77, normalized size = 1.00 \[ \frac {2 a^2 \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac {\tanh \left (\frac {x}{2}\right )}{2 (a-b)}-\frac {\coth \left (\frac {x}{2}\right )}{2 (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) - Coth[x/2]/(2*(a + b)) - Tanh[x/2]/(2
*(a - b))

________________________________________________________________________________________

fricas [B]  time = 0.41, size = 470, normalized size = 6.10 \[ \left [\frac {2 \, a^{3} - 2 \, a b^{2} + {\left (a^{2} \cosh \relax (x)^{2} + 2 \, a^{2} \cosh \relax (x) \sinh \relax (x) + a^{2} \sinh \relax (x)^{2} - a^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) - 2 \, {\left (a^{2} b - b^{3}\right )} \cosh \relax (x) - 2 \, {\left (a^{2} b - b^{3}\right )} \sinh \relax (x)}{a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{2}}, \frac {2 \, {\left (a^{3} - a b^{2} + {\left (a^{2} \cosh \relax (x)^{2} + 2 \, a^{2} \cosh \relax (x) \sinh \relax (x) + a^{2} \sinh \relax (x)^{2} - a^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) - {\left (a^{2} b - b^{3}\right )} \cosh \relax (x) - {\left (a^{2} b - b^{3}\right )} \sinh \relax (x)\right )}}{a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a^3 - 2*a*b^2 + (a^2*cosh(x)^2 + 2*a^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 - a^2)*sqrt(a^2 - b^2)*log((b^2*cos
h(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*co
sh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)) - 2*(a^2*b
- b^3)*cosh(x) - 2*(a^2*b - b^3)*sinh(x))/(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 - 2*(a^4
- 2*a^2*b^2 + b^4)*cosh(x)*sinh(x) - (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^2), 2*(a^3 - a*b^2 + (a^2*cosh(x)^2 + 2*a
^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 - a^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a
)/(a^2 - b^2)) - (a^2*b - b^3)*cosh(x) - (a^2*b - b^3)*sinh(x))/(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^
4)*cosh(x)^2 - 2*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x) - (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^2)]

________________________________________________________________________________________

giac [A]  time = 0.14, size = 76, normalized size = 0.99 \[ \frac {2 \, a^{2} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {2 \, {\left (b e^{x} - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (e^{\left (2 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.09, size = 78, normalized size = 1.01 \[ -\frac {\tanh \left (\frac {x}{2}\right )}{2 \left (a -b \right )}-\frac {1}{2 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )}+\frac {2 a^{2} \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(coth(x)^2/(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?

________________________________________________________________________________________

mupad [B]  time = 1.34, size = 337, normalized size = 4.38 \[ -\frac {\frac {2\,a}{a^2-b^2}-\frac {2\,b\,{\mathrm {e}}^x}{a^2-b^2}}{{\mathrm {e}}^{2\,x}-1}-\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2\,a^2}{b^2\,{\left (a^2-b^2\right )}^2\,\sqrt {a^4}}+\frac {2\,\left (a^3\,\sqrt {a^4}-a\,b^2\,\sqrt {a^4}\right )}{a\,b^2\,\left (a^2-b^2\right )\,\sqrt {-{\left (a^2-b^2\right )}^3}\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}\right )-\frac {2\,\left (b^3\,\sqrt {a^4}-a^2\,b\,\sqrt {a^4}\right )}{a\,b^2\,\left (a^2-b^2\right )\,\sqrt {-{\left (a^2-b^2\right )}^3}\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}\right )\,\left (\frac {b^3\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{2}-\frac {a^2\,b\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{2}\right )\right )\,\sqrt {a^4}}{\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^2/(a + b*cosh(x)),x)

[Out]

- ((2*a)/(a^2 - b^2) - (2*b*exp(x))/(a^2 - b^2))/(exp(2*x) - 1) - (2*atan((exp(x)*((2*a^2)/(b^2*(a^2 - b^2)^2*
(a^4)^(1/2)) + (2*(a^3*(a^4)^(1/2) - a*b^2*(a^4)^(1/2)))/(a*b^2*(a^2 - b^2)*(-(a^2 - b^2)^3)^(1/2)*(b^6 - a^6
- 3*a^2*b^4 + 3*a^4*b^2)^(1/2))) - (2*(b^3*(a^4)^(1/2) - a^2*b*(a^4)^(1/2)))/(a*b^2*(a^2 - b^2)*(-(a^2 - b^2)^
3)^(1/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)))*((b^3*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/2 - (a^2
*b*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/2))*(a^4)^(1/2))/(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{2}{\relax (x )}}{a + b \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________