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

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Rubi [A]  time = 0.04, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.84, size = 137, normalized size = 3.26 \[ \left [-\frac {\sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \relax (x) + {\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \relax (x) + b\right )} \sin \relax (x) - a^{2} + 2 \, b^{2}}{b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}}\right )}{2 \, {\left (a^{2} - b^{2}\right )}}, \frac {\arctan \left (-\frac {a \cos \relax (x) + b}{\sqrt {a^{2} - b^{2}} \sin \relax (x)}\right )}{\sqrt {a^{2} - b^{2}}}\right ] \]

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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giac [A]  time = 0.83, size = 61, normalized size = 1.45 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}}} \]

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)

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maple [A]  time = 0.02, size = 36, normalized size = 0.86 \[ \frac {2 \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(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 = 0.59, size = 38, normalized size = 0.90 \[ \frac {2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a-2\,b\right )}{2\,\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(x)),x)

[Out]

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

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sympy [A]  time = 4.30, size = 144, normalized size = 3.43 \[ \begin {cases} \tilde {\infty } \left (- \log {\left (\tan {\left (\frac {x}{2} \right )} - 1 \right )} + \log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {1}{b \tan {\left (\frac {x}{2} \right )}} & \text {for}\: a = - b \\\frac {\tan {\left (\frac {x}{2} \right )}}{b} & \text {for}\: a = b \\\frac {\log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )}}{a \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {\log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {x}{2} \right )} \right )}}{a \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} & \text {otherwise} \end {cases} \]

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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