3.536 \(\int \frac{A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx\)

Optimal. Leaf size=113 \[ \frac{2 (a A-b B) \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}+\frac{-\sin (x) (A b-a B)+A c \cos (x)+B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))} \]

[Out]

(2*(a*A - b*B)*ArcTan[(c + (a - b)*Tan[x/2])/Sqrt[a^2 - b^2 - c^2]])/(a^2 - b^2 - c^2)^(3/2) + (B*c + A*c*Cos[
x] - (A*b - a*B)*Sin[x])/((a^2 - b^2 - c^2)*(a + b*Cos[x] + c*Sin[x]))

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Rubi [A]  time = 0.106341, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3155, 3124, 618, 204} \[ \frac{2 (a A-b B) \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}+\frac{-\sin (x) (A b-a B)+A c \cos (x)+B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[x])/(a + b*Cos[x] + c*Sin[x])^2,x]

[Out]

(2*(a*A - b*B)*ArcTan[(c + (a - b)*Tan[x/2])/Sqrt[a^2 - b^2 - c^2]])/(a^2 - b^2 - c^2)^(3/2) + (B*c + A*c*Cos[
x] - (A*b - a*B)*Sin[x])/((a^2 - b^2 - c^2)*(a + b*Cos[x] + c*Sin[x]))

Rule 3155

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.))/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)])^2, x_Symbol] :> Simp[(c*B + c*A*Cos[d + e*x] + (a*B - b*A)*Sin[d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Cos
[d + e*x] + c*Sin[d + e*x])), x] + Dist[(a*A - b*B)/(a^2 - b^2 - c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d + e
*x]), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B, 0]

Rule 3124

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[(2*f)/e, Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx &=\frac{B c+A c \cos (x)-(A b-a B) \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}+\frac{(a A-b B) \int \frac{1}{a+b \cos (x)+c \sin (x)} \, dx}{a^2-b^2-c^2}\\ &=\frac{B c+A c \cos (x)-(A b-a B) \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}+\frac{(2 (a A-b B)) \operatorname{Subst}\left (\int \frac{1}{a+b+2 c x+(a-b) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{a^2-b^2-c^2}\\ &=\frac{B c+A c \cos (x)-(A b-a B) \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac{(4 (a A-b B)) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2-c^2\right )-x^2} \, dx,x,2 c+2 (a-b) \tan \left (\frac{x}{2}\right )\right )}{a^2-b^2-c^2}\\ &=\frac{2 (a A-b B) \tan ^{-1}\left (\frac{c+(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}+\frac{B c+A c \cos (x)-(A b-a B) \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.287751, size = 118, normalized size = 1.04 \[ \frac{\sin (x) \left (A \left (b^2+c^2\right )-a b B\right )+c (a A-b B)}{b \left (-a^2+b^2+c^2\right ) (a+b \cos (x)+c \sin (x))}+\frac{2 (a A-b B) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[x])/(a + b*Cos[x] + c*Sin[x])^2,x]

[Out]

(2*(a*A - b*B)*ArcTanh[(c + (a - b)*Tan[x/2])/Sqrt[-a^2 + b^2 + c^2]])/(-a^2 + b^2 + c^2)^(3/2) + ((a*A - b*B)
*c + (-(a*b*B) + A*(b^2 + c^2))*Sin[x])/(b*(-a^2 + b^2 + c^2)*(a + b*Cos[x] + c*Sin[x]))

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Maple [B]  time = 0.094, size = 254, normalized size = 2.3 \begin{align*} 2\,{\frac{1}{a \left ( \tan \left ( x/2 \right ) \right ) ^{2}-b \left ( \tan \left ( x/2 \right ) \right ) ^{2}+2\,c\tan \left ( x/2 \right ) +a+b} \left ( -{\frac{ \left ( aAb-A{b}^{2}-A{c}^{2}-{a}^{2}B+abB+B{c}^{2} \right ) \tan \left ( x/2 \right ) }{{a}^{3}-{a}^{2}b-a{b}^{2}-a{c}^{2}+{b}^{3}+b{c}^{2}}}+{\frac{ \left ( aA-bB \right ) c}{{a}^{3}-{a}^{2}b-a{b}^{2}-a{c}^{2}+{b}^{3}+b{c}^{2}}} \right ) }+2\,{\frac{aA}{ \left ({a}^{2}-{b}^{2}-{c}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }-2\,{\frac{bB}{ \left ({a}^{2}-{b}^{2}-{c}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(x))/(a+b*cos(x)+c*sin(x))^2,x)

[Out]

2*(-(A*a*b-A*b^2-A*c^2-B*a^2+B*a*b+B*c^2)/(a^3-a^2*b-a*b^2-a*c^2+b^3+b*c^2)*tan(1/2*x)+(A*a-B*b)*c/(a^3-a^2*b-
a*b^2-a*c^2+b^3+b*c^2))/(a*tan(1/2*x)^2-b*tan(1/2*x)^2+2*c*tan(1/2*x)+a+b)+2/(a^2-b^2-c^2)^(3/2)*arctan(1/2*(2
*(a-b)*tan(1/2*x)+2*c)/(a^2-b^2-c^2)^(1/2))*a*A-2/(a^2-b^2-c^2)^(3/2)*arctan(1/2*(2*(a-b)*tan(1/2*x)+2*c)/(a^2
-b^2-c^2)^(1/2))*b*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(x))/(a+b*cos(x)+c*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.88085, size = 2691, normalized size = 23.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x))/(a+b*cos(x)+c*sin(x))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*B*c^5 - 4*(B*a^2 - B*b^2)*c^3 + (A*a^2*b^2 - B*a*b^3 + (A*a^2 - B*a*b)*c^2 + (A*a*b^3 - B*b^4 + (A*a*
b - B*b^2)*c^2)*cos(x) + ((A*a - B*b)*c^3 + (A*a*b^2 - B*b^3)*c)*sin(x))*sqrt(-a^2 + b^2 + c^2)*log(-(a^2*b^2
- 2*b^4 - c^4 - (a^2 + 3*b^2)*c^2 - (2*a^2*b^2 - b^4 - 2*a^2*c^2 + c^4)*cos(x)^2 - 2*(a*b^3 + a*b*c^2)*cos(x)
- 2*(a*b^2*c + a*c^3 - (b*c^3 - (2*a^2*b - b^3)*c)*cos(x))*sin(x) + 2*(2*a*b*c*cos(x)^2 - a*b*c + (b^2*c + c^3
)*cos(x) - (b^3 + b*c^2 + (a*b^2 - a*c^2)*cos(x))*sin(x))*sqrt(-a^2 + b^2 + c^2))/(2*a*b*cos(x) + (b^2 - c^2)*
cos(x)^2 + a^2 + c^2 + 2*(b*c*cos(x) + a*c)*sin(x))) + 2*(B*a^4 - 2*B*a^2*b^2 + B*b^4)*c + 2*(A*c^5 - (A*a^2 +
B*a*b - 2*A*b^2)*c^3 + (B*a^3*b - A*a^2*b^2 - B*a*b^3 + A*b^4)*c)*cos(x) - 2*(B*a^3*b^2 - A*a^2*b^3 - B*a*b^4
+ A*b^5 + A*b*c^4 - (A*a^2*b + B*a*b^2 - 2*A*b^3)*c^2)*sin(x))/(a^5*b^2 - 2*a^3*b^4 + a*b^6 + a*c^6 - (2*a^3
- 3*a*b^2)*c^4 + (a^5 - 4*a^3*b^2 + 3*a*b^4)*c^2 + (a^4*b^3 - 2*a^2*b^5 + b^7 + b*c^6 - (2*a^2*b - 3*b^3)*c^4
+ (a^4*b - 4*a^2*b^3 + 3*b^5)*c^2)*cos(x) + (c^7 - (2*a^2 - 3*b^2)*c^5 + (a^4 - 4*a^2*b^2 + 3*b^4)*c^3 + (a^4*
b^2 - 2*a^2*b^4 + b^6)*c)*sin(x)), -(B*c^5 - 2*(B*a^2 - B*b^2)*c^3 - (A*a^2*b^2 - B*a*b^3 + (A*a^2 - B*a*b)*c^
2 + (A*a*b^3 - B*b^4 + (A*a*b - B*b^2)*c^2)*cos(x) + ((A*a - B*b)*c^3 + (A*a*b^2 - B*b^3)*c)*sin(x))*sqrt(a^2
- b^2 - c^2)*arctan(-(a*b*cos(x) + a*c*sin(x) + b^2 + c^2)*sqrt(a^2 - b^2 - c^2)/((c^3 - (a^2 - b^2)*c)*cos(x)
+ (a^2*b - b^3 - b*c^2)*sin(x))) + (B*a^4 - 2*B*a^2*b^2 + B*b^4)*c + (A*c^5 - (A*a^2 + B*a*b - 2*A*b^2)*c^3 +
(B*a^3*b - A*a^2*b^2 - B*a*b^3 + A*b^4)*c)*cos(x) - (B*a^3*b^2 - A*a^2*b^3 - B*a*b^4 + A*b^5 + A*b*c^4 - (A*a
^2*b + B*a*b^2 - 2*A*b^3)*c^2)*sin(x))/(a^5*b^2 - 2*a^3*b^4 + a*b^6 + a*c^6 - (2*a^3 - 3*a*b^2)*c^4 + (a^5 - 4
*a^3*b^2 + 3*a*b^4)*c^2 + (a^4*b^3 - 2*a^2*b^5 + b^7 + b*c^6 - (2*a^2*b - 3*b^3)*c^4 + (a^4*b - 4*a^2*b^3 + 3*
b^5)*c^2)*cos(x) + (c^7 - (2*a^2 - 3*b^2)*c^5 + (a^4 - 4*a^2*b^2 + 3*b^4)*c^3 + (a^4*b^2 - 2*a^2*b^4 + b^6)*c)
*sin(x))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x))/(a+b*cos(x)+c*sin(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.17346, size = 282, normalized size = 2.5 \begin{align*} -\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 ) + c}{\sqrt{a^{2} - b^{2} - c^{2}}}\right )\right )}{\left (A a - B b\right )}}{{\left (a^{2} - b^{2} - c^{2}\right )}^{\frac{3}{2}}} + \frac{2 \,{\left (B a^{2} \tan \left (\frac{1}{2} \, x\right ) - A a b \tan \left (\frac{1}{2} \, x\right ) - B a b \tan \left (\frac{1}{2} \, x\right ) + A b^{2} \tan \left (\frac{1}{2} \, x\right ) + A c^{2} \tan \left (\frac{1}{2} \, x\right ) - B c^{2} \tan \left (\frac{1}{2} \, x\right ) + A a c - B b c\right )}}{{\left (a^{3} - a^{2} b - a b^{2} + b^{3} - a c^{2} + b c^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} - b \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, c \tan \left (\frac{1}{2} \, x\right ) + a + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x))/(a+b*cos(x)+c*sin(x))^2,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) + c)/sqrt(a^2 - b^2 - c^2)
))*(A*a - B*b)/(a^2 - b^2 - c^2)^(3/2) + 2*(B*a^2*tan(1/2*x) - A*a*b*tan(1/2*x) - B*a*b*tan(1/2*x) + A*b^2*tan
(1/2*x) + A*c^2*tan(1/2*x) - B*c^2*tan(1/2*x) + A*a*c - B*b*c)/((a^3 - a^2*b - a*b^2 + b^3 - a*c^2 + b*c^2)*(a
*tan(1/2*x)^2 - b*tan(1/2*x)^2 + 2*c*tan(1/2*x) + a + b))