3.118 \(\int \frac {\text {sech}(x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=50 \[ \frac {b \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}+\frac {\tan ^{-1}(\sinh (x))}{a} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {3518, 3110, 3770, 3074, 206} \[ \frac {b \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}+\frac {\tan ^{-1}(\sinh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]/(a + b*Coth[x]),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3770

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 60, normalized size = 1.20 \[ \frac {2 \left (\tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )-\frac {b \tan ^{-1}\left (\frac {a+b \tanh \left (\frac {x}{2}\right )}{\sqrt {b-a} \sqrt {a+b}}\right )}{\sqrt {b-a} \sqrt {a+b}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]/(a + b*Coth[x]),x]

[Out]

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

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fricas [A]  time = 0.46, size = 200, normalized size = 4.00 \[ \left [\frac {\sqrt {a^{2} - b^{2}} b \log \left (\frac {{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} + a - b}{{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} - a + b}\right ) + 2 \, {\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )}{a^{3} - a b^{2}}, -\frac {2 \, {\left (\sqrt {-a^{2} + b^{2}} b \arctan \left (\frac {\sqrt {-a^{2} + b^{2}}}{{\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x)}\right ) - {\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )\right )}}{a^{3} - a b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(a+b*coth(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.12, size = 48, normalized size = 0.96 \[ -\frac {2 \, b \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a} + \frac {2 \, \arctan \left (e^{x}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(a+b*coth(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 54, normalized size = 1.08 \[ -\frac {2 b \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a \sqrt {-a^{2}+b^{2}}}+\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)/(a+b*coth(x)),x)

[Out]

-2/a*b/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))+2/a*arctan(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(sech(x)/(a+b*coth(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 = 3.22, size = 164, normalized size = 3.28 \[ \frac {b\,\ln \left (32\,a\,b^2\,{\mathrm {e}}^x+32\,a^2\,b\,{\mathrm {e}}^x+32\,a\,b\,\sqrt {a^2-b^2}\right )}{a\,\sqrt {a^2-b^2}}-\frac {b\,\ln \left (32\,a\,b^2\,{\mathrm {e}}^x+32\,a^2\,b\,{\mathrm {e}}^x-32\,a\,b\,\sqrt {a^2-b^2}\right )}{a\,\sqrt {a^2-b^2}}+\frac {\ln \left (32\,a\,b\,{\mathrm {e}}^x-32\,a^2\,{\mathrm {e}}^x+a\,b\,32{}\mathrm {i}-a^2\,32{}\mathrm {i}\right )\,1{}\mathrm {i}}{a}-\frac {\ln \left (32\,a^2\,{\mathrm {e}}^x-32\,a\,b\,{\mathrm {e}}^x+a\,b\,32{}\mathrm {i}-a^2\,32{}\mathrm {i}\right )\,1{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)*(a + b*coth(x))),x)

[Out]

(log(a*b*32i - a^2*32i - 32*a^2*exp(x) + 32*a*b*exp(x))*1i)/a - (log(a*b*32i - a^2*32i + 32*a^2*exp(x) - 32*a*
b*exp(x))*1i)/a - (b*log(32*a*b^2*exp(x) + 32*a^2*b*exp(x) - 32*a*b*(a^2 - b^2)^(1/2)))/(a*(a^2 - b^2)^(1/2))
+ (b*log(32*a*b^2*exp(x) + 32*a^2*b*exp(x) + 32*a*b*(a^2 - b^2)^(1/2)))/(a*(a^2 - b^2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(a+b*coth(x)),x)

[Out]

Integral(sech(x)/(a + b*coth(x)), x)

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