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

Optimal. Leaf size=42 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}} \]

[Out]

2*arctan((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 = 0.06, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3831, 2659, 205} \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]/(a + b*Sech[x]),x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[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 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 41, normalized size = 0.98 \[ -\frac {2 \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]/(a + b*Sech[x]),x]

[Out]

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

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fricas [A]  time = 0.41, size = 165, normalized size = 3.93 \[ \left [-\frac {\sqrt {-a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) - a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) + a}\right )}{a^{2} - b^{2}}, -\frac {2 \, \arctan \left (-\frac {a \cosh \relax (x) + a \sinh \relax (x) + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 32, normalized size = 0.76 \[ \frac {2 \, \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 36, normalized size = 0.86 \[ \frac {2 \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*sech(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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 0.12, size = 43, normalized size = 1.02 \[ \frac {2\,\mathrm {atan}\left (\frac {b}{\sqrt {a^2-b^2}}+\frac {a\,{\mathrm {e}}^x}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)*(a + b/cosh(x))),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)/(a + b*sech(x)), x)

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