Optimal. Leaf size=42 \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}} \]
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Rubi [A] time = 0.0748283, 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 \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}} \]
Antiderivative was successfully verified.
[In] Int[(a + b*Cos[x])^(-1),x]
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Rubi in Sympy [A] time = 2.7637, size = 34, normalized size = 0.81 \[ \frac{2 \operatorname{atan}{\left (\frac{\sqrt{a - b} \tan{\left (\frac{x}{2} \right )}}{\sqrt{a + b}} \right )}}{\sqrt{a - b} \sqrt{a + b}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate(1/(a+b*cos(x)),x)
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Mathematica [A] time = 0.0489462, 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]
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Maple [A] time = 0.017, size = 36, normalized size = 0.9 \[ 2\,{\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int(1/(a+b*cos(x)),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(1/(b*cos(x) + a),x, algorithm="maxima")
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Fricas [A] time = 0.250767, size = 1, normalized size = 0.02 \[ \left [\frac{\log \left (-\frac{2 \,{\left (a^{2} b - b^{3} +{\left (a^{3} - a b^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right ) -{\left (2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + 2 \, b^{2}\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right )}{2 \, \sqrt{-a^{2} + b^{2}}}, \frac{\arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right )}{\sqrt{a^{2} - b^{2}}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(1/(b*cos(x) + a),x, algorithm="fricas")
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Sympy [A] time = 16.205, size = 144, normalized size = 3.43 \[ \begin{cases} \tilde{\infty } \left (- \log{\left (\tan{\left (\frac{x}{2} \right )} - 1 \right )} + \log{\left (\tan{\left (\frac{x}{2} \right )} + 1 \right )}\right ) & \text{for}\: a = 0 \wedge b = 0 \\\frac{\tan{\left (\frac{x}{2} \right )}}{b} & \text{for}\: a = b \\\frac{1}{b \tan{\left (\frac{x}{2} \right )}} & \text{for}\: a = - b \\\frac{\log{\left (- \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{x}{2} \right )} \right )}}{a \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} - \frac{\log{\left (\sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{x}{2} \right )} \right )}}{a \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} & \text{otherwise} \end{cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(1/(a+b*cos(x)),x)
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GIAC/XCAS [A] time = 0.203147, size = 82, normalized size = 1.95 \[ -\frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor{\rm sign}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(1/(b*cos(x) + a),x, algorithm="giac")
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