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

Optimal. Leaf size=110 \[ \frac{a B \sin (x)-a C \cos (x)-b C+B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac{2 (b B+c C) \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0963721, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3153, 3124, 618, 204} \[ \frac{a B \sin (x)-a C \cos (x)-b C+B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac{2 (b B+c C) \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3153

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[(c*B - b*C - (a*C - 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 - c*C)/(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, C}, x] && NeQ[
a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B - c*C, 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{B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx &=\frac{B c-b C-a C \cos (x)+a B \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac{(b B+c C) \int \frac{1}{a+b \cos (x)+c \sin (x)} \, dx}{a^2-b^2-c^2}\\ &=\frac{B c-b C-a C \cos (x)+a B \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac{(2 (b B+c C)) \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-b C-a C \cos (x)+a B \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}+\frac{(4 (b B+c C)) \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 (b B+c C) \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-b C-a C \cos (x)+a B \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.35717, size = 116, normalized size = 1.05 \[ -\frac{a^2 C+a \sin (x) (b B+c C)-b^2 C+b B c}{b \left (-a^2+b^2+c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac{2 (b B+c C) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.1, size = 255, 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 ({a}^{2}B-abB-B{c}^{2}-acC+Cbc \right ) \tan \left ( x/2 \right ) }{{a}^{3}-{a}^{2}b-a{b}^{2}-a{c}^{2}+{b}^{3}+b{c}^{2}}}+{\frac{bBc+{a}^{2}C-{b}^{2}C}{{a}^{3}-{a}^{2}b-a{b}^{2}-a{c}^{2}+{b}^{3}+b{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 ) }-2\,{\frac{Cc}{ \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((B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x))^2,x)

[Out]

-2*(-(B*a^2-B*a*b-B*c^2-C*a*c+C*b*c)/(a^3-a^2*b-a*b^2-a*c^2+b^3+b*c^2)*tan(1/2*x)+(B*b*c+C*a^2-C*b^2)/(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))*b*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))*C*c

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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((B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.79401, size = 2799, normalized size = 25.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*C*a^4*b - 4*C*a^2*b^3 + 2*C*b^5 + 2*C*b*c^4 - 2*B*c^5 + 4*(B*a^2 - B*b^2)*c^3 - 4*(C*a^2*b - C*b^3)*c^
2 + (B*a*b^3 + C*a*b^2*c + B*a*b*c^2 + C*a*c^3 + (B*b^4 + C*b^3*c + B*b^2*c^2 + C*b*c^3)*cos(x) + (B*b^3*c + C
*b^2*c^2 + B*b*c^3 + C*c^4)*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*(B*a*b*c^3 + C*a*c^4 - (C*a^3 - C*a*b^2)*c^2 - (B*a^3*
b - B*a*b^3)*c)*cos(x) + 2*(B*a^3*b^2 - B*a*b^4 - B*a*b^2*c^2 - C*a*b*c^3 + (C*a^3*b - C*a*b^3)*c)*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)), (C*a^4*b - 2*C*a^2*b^3 + C*b^5 +
C*b*c^4 - B*c^5 + 2*(B*a^2 - B*b^2)*c^3 - 2*(C*a^2*b - C*b^3)*c^2 - (B*a*b^3 + C*a*b^2*c + B*a*b*c^2 + C*a*c^
3 + (B*b^4 + C*b^3*c + B*b^2*c^2 + C*b*c^3)*cos(x) + (B*b^3*c + C*b^2*c^2 + B*b*c^3 + C*c^4)*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 + (B*a*b*c^3 + C*a*c^4 - (C*a^3 - C*a*b^2)
*c^2 - (B*a^3*b - B*a*b^3)*c)*cos(x) + (B*a^3*b^2 - B*a*b^4 - B*a*b^2*c^2 - C*a*b*c^3 + (C*a^3*b - C*a*b^3)*c)
*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((B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.17682, size = 277, normalized size = 2.52 \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 (B b + C c\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 ) - B a b \tan \left (\frac{1}{2} \, x\right ) - C a c \tan \left (\frac{1}{2} \, x\right ) + C b c \tan \left (\frac{1}{2} \, x\right ) - B c^{2} \tan \left (\frac{1}{2} \, x\right ) - C a^{2} + C b^{2} - 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((B*cos(x)+C*sin(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))
)*(B*b + C*c)/(a^2 - b^2 - c^2)^(3/2) + 2*(B*a^2*tan(1/2*x) - B*a*b*tan(1/2*x) - C*a*c*tan(1/2*x) + C*b*c*tan(
1/2*x) - B*c^2*tan(1/2*x) - C*a^2 + C*b^2 - 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))