3.253 \(\int \frac{A+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=67 \[ \frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b d \sqrt{a-b} \sqrt{a+b}}+\frac{B x}{b} \]

[Out]

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

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Rubi [A]  time = 0.0777636, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2735, 2659, 205} \[ \frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b d \sqrt{a-b} \sqrt{a+b}}+\frac{B x}{b} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x])/(a + b*Cos[c + d*x]),x]

[Out]

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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

Mathematica [A]  time = 0.11225, size = 68, normalized size = 1.01 \[ \frac{\frac{2 (a B-A b) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+B (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x])/(a + b*Cos[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.098, size = 113, normalized size = 1.7 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{bd}}+2\,{\frac{A}{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 ) }-2\,{\frac{aB}{bd\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((A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x)

[Out]

2/d/b*arctan(tan(1/2*d*x+1/2*c))*B+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)
)*A-2/d/b/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*a*B

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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((A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8237, size = 524, normalized size = 7.82 \begin{align*} \left [\frac{2 \,{\left (B a^{2} - B b^{2}\right )} d x +{\left (B a - A b\right )} \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 - b^{3}\right )} d}, \frac{{\left (B a^{2} - B b^{2}\right )} d x -{\left (B a - A b\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{{\left (a^{2} b - b^{3}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(B*a^2 - B*b^2)*d*x + (B*a - A*b)*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 - b^3)*d), ((B*a^2 - B*b^2)*d*x - (B*a - A*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) +
b)/(sqrt(a^2 - b^2)*sin(d*x + c))))/((a^2*b - b^3)*d)]

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Sympy [A]  time = 127.075, size = 524, normalized size = 7.82 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \left (A + B \cos{\left (c \right )}\right )}{\cos{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{b d} + \frac{B x}{b} - \frac{B \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{b d} & \text{for}\: a = b \\\frac{A}{b d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}} + \frac{B x}{b} + \frac{B}{b d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}} & \text{for}\: a = - b \\\frac{A x + \frac{B \sin{\left (c + d x \right )}}{d}}{a} & \text{for}\: b = 0 \\\frac{x \left (A + B \cos{\left (c \right )}\right )}{a + b \cos{\left (c \right )}} & \text{for}\: d = 0 \\\frac{A b \log{\left (- \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} \right )}}{a b d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b^{2} d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} - \frac{A b \log{\left (\sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} \right )}}{a b d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b^{2} d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} + \frac{B a d x \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}}{a b d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b^{2} d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} - \frac{B a \log{\left (- \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} \right )}}{a b d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b^{2} d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} + \frac{B a \log{\left (\sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} \right )}}{a b d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b^{2} d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} - \frac{B b d x \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}}{a b d \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - b^{2} 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((A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x)

[Out]

Piecewise((zoo*x*(A + B*cos(c))/cos(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (A*tan(c/2 + d*x/2)/(b*d) + B*x/b - B
*tan(c/2 + d*x/2)/(b*d), Eq(a, b)), (A/(b*d*tan(c/2 + d*x/2)) + B*x/b + B/(b*d*tan(c/2 + d*x/2)), Eq(a, -b)),
((A*x + B*sin(c + d*x)/d)/a, Eq(b, 0)), (x*(A + B*cos(c))/(a + b*cos(c)), Eq(d, 0)), (A*b*log(-sqrt(-a/(a - b)
- b/(a - b)) + tan(c/2 + d*x/2))/(a*b*d*sqrt(-a/(a - b) - b/(a - b)) - b**2*d*sqrt(-a/(a - b) - b/(a - b))) -
A*b*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*b*d*sqrt(-a/(a - b) - b/(a - b)) - b**2*d*sqrt(-a
/(a - b) - b/(a - b))) + B*a*d*x*sqrt(-a/(a - b) - b/(a - b))/(a*b*d*sqrt(-a/(a - b) - b/(a - b)) - b**2*d*sqr
t(-a/(a - b) - b/(a - b))) - B*a*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*b*d*sqrt(-a/(a - b)
- b/(a - b)) - b**2*d*sqrt(-a/(a - b) - b/(a - b))) + B*a*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))
/(a*b*d*sqrt(-a/(a - b) - b/(a - b)) - b**2*d*sqrt(-a/(a - b) - b/(a - b))) - B*b*d*x*sqrt(-a/(a - b) - b/(a -
b))/(a*b*d*sqrt(-a/(a - b) - b/(a - b)) - b**2*d*sqrt(-a/(a - b) - b/(a - b))), True))

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Giac [A]  time = 1.58013, size = 136, normalized size = 2.03 \begin{align*} \frac{\frac{{\left (d x + c\right )} B}{b} + \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 )}{\left (B a - A b\right )}}{\sqrt{a^{2} - b^{2}} b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)*B/b + 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)))*(B*a - A*b)/(sqrt(a^2 - b^2)*b))/d