3.3 \(\int \frac{1}{a+b \cos (x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0423913, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2659, 205} \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[x])^(-1),x]

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0312415, size = 41, normalized size = 0.98 \[ -\frac{2 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[x])^(-1),x]

[Out]

(-2*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2]

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Maple [A]  time = 0.01, size = 36, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(x)),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07912, size = 328, normalized size = 7.81 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right )}{2 \,{\left (a^{2} - b^{2}\right )}}, \frac{\arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right )}{\sqrt{a^{2} - b^{2}}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)),x, algorithm="fricas")

[Out]

[-1/2*sqrt(-a^2 + b^2)*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) -
a^2 + 2*b^2)/(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2))/(a^2 - b^2), arctan(-(a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x)
))/sqrt(a^2 - b^2)]

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Sympy [A]  time = 8.781, size = 144, normalized size = 3.43 \begin{align*} \begin{cases} \tilde{\infty } \left (- \log{\left (\tan{\left (\frac{x}{2} \right )} - 1 \right )} + \log{\left (\tan{\left (\frac{x}{2} \right )} + 1 \right )}\right ) & \text{for}\: a = 0 \wedge b = 0 \\\frac{1}{b \tan{\left (\frac{x}{2} \right )}} & \text{for}\: a = - b \\\frac{\tan{\left (\frac{x}{2} \right )}}{b} & \text{for}\: a = b \\\frac{\log{\left (- \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{x}{2} \right )} \right )}}{a \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} - \frac{\log{\left (\sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{x}{2} \right )} \right )}}{a \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)),x)

[Out]

Piecewise((zoo*(-log(tan(x/2) - 1) + log(tan(x/2) + 1)), Eq(a, 0) & Eq(b, 0)), (1/(b*tan(x/2)), Eq(a, -b)), (t
an(x/2)/b, Eq(a, b)), (log(-sqrt(-a/(a - b) - b/(a - b)) + tan(x/2))/(a*sqrt(-a/(a - b) - b/(a - b)) - b*sqrt(
-a/(a - b) - b/(a - b))) - log(sqrt(-a/(a - b) - b/(a - b)) + tan(x/2))/(a*sqrt(-a/(a - b) - b/(a - b)) - b*sq
rt(-a/(a - b) - b/(a - b))), True))

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Giac [A]  time = 1.08842, size = 82, normalized size = 1.95 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)),x, algorithm="giac")

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))/sqrt(a^
2 - b^2)