3.58 \(\int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=54 \[ \frac {\tan ^{-1}(\sinh (x))}{a}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2747, 3770, 2659, 208} \[ \frac {\tan ^{-1}(\sinh (x))}{a}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]/(a + b*Cosh[x]),x]

[Out]

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

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 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 2747

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

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

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Mathematica [A]  time = 0.05, size = 54, normalized size = 1.00 \[ \frac {2 \left (\frac {b \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+\tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]/(a + b*Cosh[x]),x]

[Out]

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

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fricas [A]  time = 0.63, size = 227, normalized size = 4.20 \[ \left [\frac {\sqrt {a^{2} - b^{2}} b \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) + 2 \, {\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )}{a^{3} - a b^{2}}, \frac {2 \, {\left (\sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )\right )}}{a^{3} - a b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 45, normalized size = 0.83 \[ -\frac {2 \, b \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a} + \frac {2 \, \arctan \left (e^{x}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 51, normalized size = 0.94 \[ -\frac {2 b \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/a*b/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(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*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 = 3.46, size = 286, normalized size = 5.30 \[ \frac {b\,\ln \left (64\,a^4\,b-64\,a^2\,b^3+128\,a^5\,{\mathrm {e}}^x+32\,a\,b^3\,\sqrt {a^2-b^2}-64\,a^3\,b\,\sqrt {a^2-b^2}+32\,a\,b^4\,{\mathrm {e}}^x-128\,a^4\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-160\,a^3\,b^2\,{\mathrm {e}}^x+96\,a^2\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a\,\sqrt {a^2-b^2}}-\frac {b\,\ln \left (64\,a^4\,b-64\,a^2\,b^3+128\,a^5\,{\mathrm {e}}^x-32\,a\,b^3\,\sqrt {a^2-b^2}+64\,a^3\,b\,\sqrt {a^2-b^2}+32\,a\,b^4\,{\mathrm {e}}^x+128\,a^4\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-160\,a^3\,b^2\,{\mathrm {e}}^x-96\,a^2\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a\,\sqrt {a^2-b^2}}-\frac {\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)*(a + b*cosh(x))),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)/(a + b*cosh(x)), x)

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