3.454 \(\int \frac{1}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=49 \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d \sqrt{a-b} \sqrt{a+b}} \]

[Out]

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

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Rubi [A]  time = 0.0306421, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2659, 205} \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d \sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(-1),x]

[Out]

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

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

Mathematica [A]  time = 0.035661, size = 48, normalized size = 0.98 \[ -\frac{2 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{d \sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(-1),x]

[Out]

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

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Maple [A]  time = 0.066, size = 44, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{d\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \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(d*x+c)),x)

[Out]

2/d/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((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(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.00423, size = 404, normalized size = 8.24 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \,{\left (a^{2} - b^{2}\right )} d}, \frac{\arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{\sqrt{a^{2} - b^{2}} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 10.2722, size = 172, normalized size = 3.51 \begin{align*} \begin{cases} \frac{\tilde{\infty } x}{\cos{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{1}{b d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}} & \text{for}\: a = - b \\\frac{x}{a + b \cos{\left (c \right )}} & \text{for}\: d = 0 \\\frac{\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{b d} & \text{for}\: a = b \\\frac{\log{\left (- \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} \right )}}{a d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} - \frac{\log{\left (\sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} \right )}}{a d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b d \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(d*x+c)),x)

[Out]

Piecewise((zoo*x/cos(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (1/(b*d*tan(c/2 + d*x/2)), Eq(a, -b)), (x/(a + b*cos
(c)), Eq(d, 0)), (tan(c/2 + d*x/2)/(b*d), Eq(a, b)), (log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a
*d*sqrt(-a/(a - b) - b/(a - b)) - b*d*sqrt(-a/(a - b) - b/(a - b))) - log(sqrt(-a/(a - b) - b/(a - b)) + tan(c
/2 + d*x/2))/(a*d*sqrt(-a/(a - b) - b/(a - b)) - b*d*sqrt(-a/(a - b) - b/(a - b))), True))

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Giac [A]  time = 1.34528, size = 105, normalized size = 2.14 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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