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

Optimal. Leaf size=37 \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{1-a} \tan \left (\frac{x}{2}\right )}{\sqrt{a+1}}\right )}{\sqrt{1-a^2}} \]

[Out]

(2*ArcTan[(Sqrt[1 - a]*Tan[x/2])/Sqrt[1 + a]])/Sqrt[1 - a^2]

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Rubi [A]  time = 0.0183064, antiderivative size = 37, 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} \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{1-a} \tan \left (\frac{x}{2}\right )}{\sqrt{a+1}}\right )}{\sqrt{1-a^2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + a*Cos[x])^(-1),x]

[Out]

(2*ArcTan[(Sqrt[1 - a]*Tan[x/2])/Sqrt[1 + a]])/Sqrt[1 - a^2]

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

Mathematica [A]  time = 0.0234304, size = 31, normalized size = 0.84 \[ \frac{2 \tanh ^{-1}\left (\frac{(a-1) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-1}}\right )}{\sqrt{a^2-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + a*Cos[x])^(-1),x]

[Out]

(2*ArcTanh[((-1 + a)*Tan[x/2])/Sqrt[-1 + a^2]])/Sqrt[-1 + a^2]

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Maple [A]  time = 0.012, size = 30, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{\sqrt{ \left ( 1+a \right ) \left ( a-1 \right ) }}{\it Artanh} \left ({\frac{ \left ( a-1 \right ) \tan \left ( x/2 \right ) }{\sqrt{ \left ( 1+a \right ) \left ( a-1 \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/((1+a)*(a-1))^(1/2)*arctanh((a-1)*tan(1/2*x)/((1+a)*(a-1))^(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/(1+a*cos(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.16777, size = 306, normalized size = 8.27 \begin{align*} \left [\frac{\log \left (-\frac{{\left (a^{2} - 2\right )} \cos \left (x\right )^{2} - 2 \, \sqrt{a^{2} - 1}{\left (a + \cos \left (x\right )\right )} \sin \left (x\right ) - 2 \, a^{2} - 2 \, a \cos \left (x\right ) + 1}{a^{2} \cos \left (x\right )^{2} + 2 \, a \cos \left (x\right ) + 1}\right )}{2 \, \sqrt{a^{2} - 1}}, -\frac{\sqrt{-a^{2} + 1} \arctan \left (\frac{\sqrt{-a^{2} + 1}{\left (a + \cos \left (x\right )\right )}}{{\left (a^{2} - 1\right )} \sin \left (x\right )}\right )}{a^{2} - 1}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 5.919, size = 110, normalized size = 2.97 \begin{align*} \begin{cases} - \frac{1}{\tan{\left (\frac{x}{2} \right )}} & \text{for}\: a = -1 \\\tan{\left (\frac{x}{2} \right )} & \text{for}\: a = 1 \\- \frac{\log{\left (- \sqrt{\frac{a}{a - 1} + \frac{1}{a - 1}} + \tan{\left (\frac{x}{2} \right )} \right )}}{a \sqrt{\frac{a}{a - 1} + \frac{1}{a - 1}} - \sqrt{\frac{a}{a - 1} + \frac{1}{a - 1}}} + \frac{\log{\left (\sqrt{\frac{a}{a - 1} + \frac{1}{a - 1}} + \tan{\left (\frac{x}{2} \right )} \right )}}{a \sqrt{\frac{a}{a - 1} + \frac{1}{a - 1}} - \sqrt{\frac{a}{a - 1} + \frac{1}{a - 1}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-1/tan(x/2), Eq(a, -1)), (tan(x/2), Eq(a, 1)), (-log(-sqrt(a/(a - 1) + 1/(a - 1)) + tan(x/2))/(a*sq
rt(a/(a - 1) + 1/(a - 1)) - sqrt(a/(a - 1) + 1/(a - 1))) + log(sqrt(a/(a - 1) + 1/(a - 1)) + tan(x/2))/(a*sqrt
(a/(a - 1) + 1/(a - 1)) - sqrt(a/(a - 1) + 1/(a - 1))), True))

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Giac [A]  time = 1.09308, size = 72, 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\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) - \tan \left (\frac{1}{2} \, x\right )}{\sqrt{-a^{2} + 1}}\right )\right )}}{\sqrt{-a^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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