\(\int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx\) [546]

Optimal result

Integrand size = 22, antiderivative size = 110 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=-\frac {2 (b B+c C) \arctan \left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}+\frac {B c-b C-a C \cos (x)+a B \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3232, 3203, 632, 210} \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {a B \sin (x)-a C \cos (x)-b C+B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac {2 (b B+c C) \arctan \left (\frac {(a-b) \tan \left (\frac {x}{2}\right )+c}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}} \]

[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 210

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

Rule 632

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 3203

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 3232

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]

Rubi steps \begin{align*} \text {integral}& = \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)) \text {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)) \text {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) \arctan \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] (verified)

Time = 0.51 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=-\frac {2 (b B+c C) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{3/2}}-\frac {b B c+a^2 C-b^2 C+a (b B+c C) \sin (x)}{b \left (-a^2+b^2+c^2\right ) (a+b \cos (x)+c \sin (x))} \]

[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]))

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.87

method	result	size
default	\(-\frac {2 \left (-\frac {\left (B \,a^{2}-a b B -B \,c^{2}-a c C +C b c \right ) \tan \left (\frac {x}{2}\right )}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b}+\frac {b B c +C \,a^{2}-b^{2} C}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b}\right )}{\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} b +2 c \tan \left (\frac {x}{2}\right )+a +b}-\frac {2 \left (B b +C c \right ) \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{\left (a^{2}-b^{2}-c^{2}\right )^{\frac {3}{2}}}\)	\(206\)
risch	\(\frac {-2 i B a b -2 i a c C -2 i B \,{\mathrm e}^{i x} a^{2}+2 i B \,c^{2} {\mathrm e}^{i x}-2 i C b c \,{\mathrm e}^{i x}-2 B b c \,{\mathrm e}^{i x}-2 C \,a^{2} {\mathrm e}^{i x}+2 C \,b^{2} {\mathrm e}^{i x}}{\left (-a^{2}+b^{2}+c^{2}\right ) \left (-i c +b \right ) \left (-i c \,{\mathrm e}^{2 i x}+b \,{\mathrm e}^{2 i x}+i c +2 a \,{\mathrm e}^{i x}+b \right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}-i a^{2} b +i b^{3}+i b \,c^{2}+a b \sqrt {-a^{2}+b^{2}+c^{2}}+a^{2} c -b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}-i a^{2} b +i b^{3}+i b \,c^{2}+a b \sqrt {-a^{2}+b^{2}+c^{2}}+a^{2} c -b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right ) C c}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}+i a^{2} b -i b^{3}-i b \,c^{2}+a b \sqrt {-a^{2}+b^{2}+c^{2}}-a^{2} c +b^{2} c +c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}+i a^{2} b -i b^{3}-i b \,c^{2}+a b \sqrt {-a^{2}+b^{2}+c^{2}}-a^{2} c +b^{2} c +c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right ) C c}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}\)	\(683\)

[In]

int((B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x))^2,x,method=_RETURNVERBOSE)

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 577 vs. \(2 (105) = 210\).

Time = 0.32 (sec) , antiderivative size = 1316, normalized size of antiderivative = 11.96 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\text {Too large to display} \]

[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))]

Sympy [F(-1)]

Timed out. \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x))^2,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(c^2+b^2-a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.86 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right ) + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right )\right )} {\left (B b + C c\right )}}{{\left (a^{2} - b^{2} - c^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, x\right ) - B a b \tan \left (\frac {1}{2} \, x\right ) - C a c \tan \left (\frac {1}{2} \, x\right ) + C b c \tan \left (\frac {1}{2} \, x\right ) - B c^{2} \tan \left (\frac {1}{2} \, x\right ) - C a^{2} + C b^{2} - B b c\right )}}{{\left (a^{3} - a^{2} b - a b^{2} + b^{3} - a c^{2} + b c^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}} \]

[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))

Mupad [B] (verification not implemented)

Time = 27.41 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.84 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {\frac {2\,\left (C\,a^2-C\,b^2+B\,c\,b\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-B\,a^2+C\,a\,c+B\,b\,a+B\,c^2-C\,b\,c\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}}{\left (a-b\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {x}{2}\right )+a+b}-\frac {2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a-2\,b\right )+\frac {2\,\left (-a^2\,c+b^2\,c+c^3\right )}{-a^2+b^2+c^2}}{2\,\sqrt {-a^2+b^2+c^2}}\right )\,\left (B\,b+C\,c\right )}{{\left (-a^2+b^2+c^2\right )}^{3/2}} \]

[In]

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

[Out]

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