\(\int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx\) [536]

Optimal result

Integrand size = 19, antiderivative size = 113 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {2 (a A-b B) \arctan \left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}+\frac {B c+A c \cos (x)-(A b-a B) \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3234, 3203, 632, 210} \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {2 (a A-b B) \arctan \left (\frac {(a-b) \tan \left (\frac {x}{2}\right )+c}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}+\frac {-\sin (x) (A b-a B)+A c \cos (x)+B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))} \]

[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 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 3234

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]

Rubi steps \begin{align*} \text {integral}& = \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)) \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+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)) \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 (a A-b B) \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+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] (verified)

Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {2 (a A-b B) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{3/2}}+\frac {(a A-b B) c+\left (-a b B+A \left (b^2+c^2\right )\right ) \sin (x)}{b \left (-a^2+b^2+c^2\right ) (a+b \cos (x)+c \sin (x))} \]

[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]))

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.82

method	result	size
default	\(\frac {-\frac {2 \left (A a b -A \,b^{2}-A \,c^{2}-B \,a^{2}+a b B +B \,c^{2}\right ) \tan \left (\frac {x}{2}\right )}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b}+\frac {2 \left (A a -B b \right ) c}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b}}{\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} b +2 c \tan \left (\frac {x}{2}\right )+a +b}+\frac {2 \left (A a -B b \right ) \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{\left (a^{2}-b^{2}-c^{2}\right )^{\frac {3}{2}}}\)	\(206\)
risch	\(\frac {2 i \left (-i A a c \,{\mathrm e}^{i x}+i B b c \,{\mathrm e}^{i x}+A a b \,{\mathrm e}^{i x}-B \,a^{2} {\mathrm e}^{i x}+B \,c^{2} {\mathrm e}^{i x}+A \,b^{2}+A \,c^{2}-a b B \right )}{\left (-a^{2}+b^{2}+c^{2}\right ) \left (-i c +b \right ) \left (-i c \,{\mathrm e}^{2 i x}+b \,{\mathrm e}^{2 i x}+i c +2 a \,{\mathrm e}^{i x}+b \right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}+i a^{2} b -i b^{3}-i b \,c^{2}+a b \sqrt {-a^{2}+b^{2}+c^{2}}-a^{2} c +b^{2} c +c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right ) A a}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}+i a^{2} b -i b^{3}-i b \,c^{2}+a b \sqrt {-a^{2}+b^{2}+c^{2}}-a^{2} c +b^{2} c +c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}-i a^{2} b +i b^{3}+i b \,c^{2}+a b \sqrt {-a^{2}+b^{2}+c^{2}}+a^{2} c -b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right ) A a}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}-i a^{2} b +i b^{3}+i b \,c^{2}+a b \sqrt {-a^{2}+b^{2}+c^{2}}+a^{2} c -b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}\)	\(673\)

[In]

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

[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))/(tan(1/2*x)^2*a-tan(1/2*x)^2*b+2*c*tan(1/2*x)+a+b)+2*(A*a-B*b)/(a^2-b^2-c^2)^(3/2)*arc
tan(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. 559 vs. \(2 (106) = 212\).

Time = 0.32 (sec) , antiderivative size = 1277, normalized size of antiderivative = 11.30 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\text {Too large to display} \]

[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))]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate((A+B*cos(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 = 0.30 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.85 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right ) + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right )\right )} {\left (A a - B b\right )}}{{\left (a^{2} - b^{2} - c^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, x\right ) - A a b \tan \left (\frac {1}{2} \, x\right ) - B a b \tan \left (\frac {1}{2} \, x\right ) + A b^{2} \tan \left (\frac {1}{2} \, x\right ) + A c^{2} \tan \left (\frac {1}{2} \, x\right ) - B c^{2} \tan \left (\frac {1}{2} \, x\right ) + A a c - B b c\right )}}{{\left (a^{3} - a^{2} b - a b^{2} + b^{3} - a c^{2} + b c^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}} \]

[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))

Mupad [B] (verification not implemented)

Time = 26.22 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.81 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a-2\,b\right )+\frac {2\,\left (-a^2\,c+b^2\,c+c^3\right )}{-a^2+b^2+c^2}}{2\,\sqrt {-a^2+b^2+c^2}}\right )\,\left (A\,a-B\,b\right )}{{\left (-a^2+b^2+c^2\right )}^{3/2}}-\frac {\frac {2\,\left (A\,a\,c-B\,b\,c\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A\,b^2+B\,a^2+A\,c^2-B\,c^2-A\,a\,b-B\,a\,b\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}}{\left (a-b\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {x}{2}\right )+a+b} \]

[In]

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

[Out]

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