\(\int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx\) [480]

Optimal result

Integrand size = 19, antiderivative size = 200 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\frac {\left (2 a^2 A-3 a b B+A \left (b^2+c^2\right )\right ) \arctan \left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{5/2}}+\frac {B c+A c \cos (x)-(A b-a B) \sin (x)}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}+\frac {a B c+(3 a A-2 b B) c \cos (x)-\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (x)}{2 \left (a^2-b^2-c^2\right )^2 (a+b \cos (x)+c \sin (x))} \] Output:

(2*a^2*A-3*a*b*B+A*(b^2+c^2))*arctan((c+(a-b)*tan(1/2*x))/(a^2-b^2-c^2)^(1 
/2))/(a^2-b^2-c^2)^(5/2)+1/2*(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))^2+1/2*(a*B*c+(3*A*a-2*B*b)*c*cos(x)-(3*A*a*b-B*a^ 
2-2*B*b^2)*sin(x))/(a^2-b^2-c^2)^2/(a+b*cos(x)+c*sin(x))
 

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.63 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=-\frac {\left (2 a^2 A-3 a b B+A \left (b^2+c^2\right )\right ) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{5/2}}+\frac {-6 a^3 A c-3 a A b^2 c+9 a^2 b B c-3 a A c^3-2 b c \left (2 a^2 A-3 a b B+A \left (b^2+c^2\right )\right ) \cos (x)+c \left (-a^2 b B+3 a A \left (b^2+c^2\right )-2 b B \left (b^2+c^2\right )\right ) \cos (2 x)-8 a^2 A b^2 \sin (x)+2 A b^4 \sin (x)+4 a^3 b B \sin (x)+2 a b^3 B \sin (x)-12 a^2 A c^2 \sin (x)+2 A b^2 c^2 \sin (x)+8 a b B c^2 \sin (x)-3 a A b^3 \sin (2 x)+a^2 b^2 B \sin (2 x)+2 b^4 B \sin (2 x)-3 a A b c^2 \sin (2 x)+2 b^2 B c^2 \sin (2 x)}{4 b \left (-a^2+b^2+c^2\right )^2 (a+b \cos (x)+c \sin (x))^2} \] Input:

Integrate[(A + B*Cos[x])/(a + b*Cos[x] + c*Sin[x])^3,x]
 

Output:

-(((2*a^2*A - 3*a*b*B + A*(b^2 + c^2))*ArcTanh[(c + (a - b)*Tan[x/2])/Sqrt 
[-a^2 + b^2 + c^2]])/(-a^2 + b^2 + c^2)^(5/2)) + (-6*a^3*A*c - 3*a*A*b^2*c 
 + 9*a^2*b*B*c - 3*a*A*c^3 - 2*b*c*(2*a^2*A - 3*a*b*B + A*(b^2 + c^2))*Cos 
[x] + c*(-(a^2*b*B) + 3*a*A*(b^2 + c^2) - 2*b*B*(b^2 + c^2))*Cos[2*x] - 8* 
a^2*A*b^2*Sin[x] + 2*A*b^4*Sin[x] + 4*a^3*b*B*Sin[x] + 2*a*b^3*B*Sin[x] - 
12*a^2*A*c^2*Sin[x] + 2*A*b^2*c^2*Sin[x] + 8*a*b*B*c^2*Sin[x] - 3*a*A*b^3* 
Sin[2*x] + a^2*b^2*B*Sin[2*x] + 2*b^4*B*Sin[2*x] - 3*a*A*b*c^2*Sin[2*x] + 
2*b^2*B*c^2*Sin[2*x])/(4*b*(-a^2 + b^2 + c^2)^2*(a + b*Cos[x] + c*Sin[x])^ 
2)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 3637, 25, 3042, 3632, 3042, 3603, 1083, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3}dx\)

\(\Big \downarrow \) 3637

\(\displaystyle \frac {-\sin (x) (A b-a B)+A c \cos (x)+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}-\frac {\int -\frac {2 (a A-b B)-(A b-a B) \cos (x)-A c \sin (x)}{(a+b \cos (x)+c \sin (x))^2}dx}{2 \left (a^2-b^2-c^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {2 (a A-b B)-(A b-a B) \cos (x)-A c \sin (x)}{(a+b \cos (x)+c \sin (x))^2}dx}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+A c \cos (x)+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {2 (a A-b B)-(A b-a B) \cos (x)-A c \sin (x)}{(a+b \cos (x)+c \sin (x))^2}dx}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+A c \cos (x)+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\)

\(\Big \downarrow \) 3632

\(\displaystyle \frac {\frac {\left (2 a^2 A-3 a b B+A \left (b^2+c^2\right )\right ) \int \frac {1}{a+b \cos (x)+c \sin (x)}dx}{a^2-b^2-c^2}+\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b^2 B\right )+c \cos (x) (3 a A-2 b B)+a B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+A c \cos (x)+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (2 a^2 A-3 a b B+A \left (b^2+c^2\right )\right ) \int \frac {1}{a+b \cos (x)+c \sin (x)}dx}{a^2-b^2-c^2}+\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b^2 B\right )+c \cos (x) (3 a A-2 b B)+a B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+A c \cos (x)+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\)

\(\Big \downarrow \) 3603

\(\displaystyle \frac {\frac {2 \left (2 a^2 A-3 a b B+A \left (b^2+c^2\right )\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {x}{2}\right )+2 c \tan \left (\frac {x}{2}\right )+a+b}d\tan \left (\frac {x}{2}\right )}{a^2-b^2-c^2}+\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b^2 B\right )+c \cos (x) (3 a A-2 b B)+a B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+A c \cos (x)+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b^2 B\right )+c \cos (x) (3 a A-2 b B)+a B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac {4 \left (2 a^2 A-3 a b B+A \left (b^2+c^2\right )\right ) \int \frac {1}{-\left (2 c+2 (a-b) \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2-c^2\right )}d\left (2 c+2 (a-b) \tan \left (\frac {x}{2}\right )\right )}{a^2-b^2-c^2}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+A c \cos (x)+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {2 \left (2 a^2 A-3 a b B+A \left (b^2+c^2\right )\right ) \arctan \left (\frac {2 (a-b) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}+\frac {-\sin (x) \left (a^2 (-B)+3 a A b-2 b^2 B\right )+c \cos (x) (3 a A-2 b B)+a B c}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}+\frac {-\sin (x) (A b-a B)+A c \cos (x)+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}\)

Input:

Int[(A + B*Cos[x])/(a + b*Cos[x] + c*Sin[x])^3,x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 217

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 1083

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3603

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(-1), x_Symbol] :> Module[{f = FreeFactors[Tan[(d + e*x)/2], x]}, Simp[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 3632

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] + 
 Simp[(a*A - b*B - c*C)/(a^2 - b^2 - c^2)   Int[1/(a + b*Cos[d + e*x] + c*S 
in[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 3637

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(852\) vs. \(2(190)=380\).

Time = 0.74 (sec) , antiderivative size = 853, normalized size of antiderivative = 4.26

method	result	size
default	\(\frac {-\frac {\left (4 A \,a^{3} b -7 A \,a^{2} b^{2}-5 A \,a^{2} c^{2}+2 A a \,b^{3}+2 A a b \,c^{2}+A \,b^{4}+3 A \,b^{2} c^{2}+2 A \,c^{4}-2 B \,a^{4}+3 B \,a^{3} b -2 B \,a^{2} b^{2}+4 B \,a^{2} c^{2}+3 B a \,b^{3}-2 B \,b^{4}-4 B \,b^{2} c^{2}-2 B \,c^{4}\right ) \tan \left (\frac {x}{2}\right )^{3}}{\left (a -b \right ) \left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right )}+\frac {c \left (4 A \,a^{4}-12 A \,a^{3} b +13 A \,a^{2} b^{2}+7 A \,a^{2} c^{2}-6 A a \,b^{3}-6 A a b \,c^{2}+A \,b^{4}-A \,b^{2} c^{2}-2 A \,c^{4}+2 B \,a^{4}-9 B \,a^{3} b +14 B \,a^{2} b^{2}-4 B \,a^{2} c^{2}-9 B a \,b^{3}+2 B \,b^{4}+4 B \,b^{2} c^{2}+2 B \,c^{4}\right ) \tan \left (\frac {x}{2}\right )^{2}}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (4 A \,a^{4} b -5 A \,a^{3} b^{2}-11 A \,a^{3} c^{2}-3 A \,a^{2} b^{3}+3 A \,a^{2} b \,c^{2}+5 A a \,b^{4}+7 A a \,b^{2} c^{2}+2 A a \,c^{4}-A \,b^{5}+A \,b^{3} c^{2}+2 A b \,c^{4}-2 B \,a^{5}+3 B \,a^{4} b -B \,a^{3} b^{2}+4 B \,a^{3} c^{2}-B \,a^{2} b^{3}+8 B \,a^{2} b \,c^{2}+3 B a \,b^{4}-8 B a \,b^{2} c^{2}-2 B a \,c^{4}-2 B \,b^{5}-4 B \,b^{3} c^{2}-2 B b \,c^{4}\right ) \tan \left (\frac {x}{2}\right )}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {c \left (4 A \,a^{4}-3 A \,a^{2} b^{2}-A \,a^{2} c^{2}-A \,b^{4}-A \,b^{2} c^{2}-5 B \,a^{3} b +5 B a \,b^{3}+2 B a b \,c^{2}\right )}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {x}{2}\right )^{2} a -b \tan \left (\frac {x}{2}\right )^{2}+2 c \tan \left (\frac {x}{2}\right )+a +b \right )^{2}}+\frac {\left (2 a^{2} A +A \,b^{2}+A \,c^{2}-3 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^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \sqrt {a^{2}-b^{2}-c^{2}}}\)	\(853\)
risch	\(\text {Expression too large to display}\)	\(1643\)

Input:

int((A+B*cos(x))/(a+b*cos(x)+c*sin(x))^3,x,method=_RETURNVERBOSE)
 

Output:

2*(-1/2*(4*A*a^3*b-7*A*a^2*b^2-5*A*a^2*c^2+2*A*a*b^3+2*A*a*b*c^2+A*b^4+3*A 
*b^2*c^2+2*A*c^4-2*B*a^4+3*B*a^3*b-2*B*a^2*b^2+4*B*a^2*c^2+3*B*a*b^3-2*B*b 
^4-4*B*b^2*c^2-2*B*c^4)/(a-b)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^4)* 
tan(1/2*x)^3+1/2*c*(4*A*a^4-12*A*a^3*b+13*A*a^2*b^2+7*A*a^2*c^2-6*A*a*b^3- 
6*A*a*b*c^2+A*b^4-A*b^2*c^2-2*A*c^4+2*B*a^4-9*B*a^3*b+14*B*a^2*b^2-4*B*a^2 
*c^2-9*B*a*b^3+2*B*b^4+4*B*b^2*c^2+2*B*c^4)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2 
*b^2*c^2+c^4)/(a^2-2*a*b+b^2)*tan(1/2*x)^2-1/2*(4*A*a^4*b-5*A*a^3*b^2-11*A 
*a^3*c^2-3*A*a^2*b^3+3*A*a^2*b*c^2+5*A*a*b^4+7*A*a*b^2*c^2+2*A*a*c^4-A*b^5 
+A*b^3*c^2+2*A*b*c^4-2*B*a^5+3*B*a^4*b-B*a^3*b^2+4*B*a^3*c^2-B*a^2*b^3+8*B 
*a^2*b*c^2+3*B*a*b^4-8*B*a*b^2*c^2-2*B*a*c^4-2*B*b^5-4*B*b^3*c^2-2*B*b*c^4 
)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^4)/(a^2-2*a*b+b^2)*tan(1/2*x)+1 
/2*c*(4*A*a^4-3*A*a^2*b^2-A*a^2*c^2-A*b^4-A*b^2*c^2-5*B*a^3*b+5*B*a*b^3+2* 
B*a*b*c^2)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^4)/(a^2-2*a*b+b^2))/(t 
an(1/2*x)^2*a-b*tan(1/2*x)^2+2*c*tan(1/2*x)+a+b)^2+(2*A*a^2+A*b^2+A*c^2-3* 
B*a*b)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^4)/(a^2-b^2-c^2)^(1/2)*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. 1620 vs. \(2 (187) = 374\).

Time = 0.27 (sec) , antiderivative size = 3402, normalized size of antiderivative = 17.01 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(x))/(a+b*cos(x)+c*sin(x))^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(x))/(a+b*cos(x)+c*sin(x))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((A+B*cos(x))/(a+b*cos(x)+c*sin(x))^3,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(c^2+b^2-a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1162 vs. \(2 (187) = 374\).

Time = 0.38 (sec) , antiderivative size = 1162, normalized size of antiderivative = 5.81 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(x))/(a+b*cos(x)+c*sin(x))^3,x, algorithm="giac")
 

Output:

-(2*A*a^2 - 3*B*a*b + A*b^2 + A*c^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^4 - 2*a^2*b^2 + b^4 - 2*a^2*c^2 + 2*b^2*c^2 + c^4)*sqrt(a^2 - b^2 - c^2 
)) + (2*B*a^5*tan(1/2*x)^3 - 4*A*a^4*b*tan(1/2*x)^3 - 5*B*a^4*b*tan(1/2*x) 
^3 + 11*A*a^3*b^2*tan(1/2*x)^3 + 5*B*a^3*b^2*tan(1/2*x)^3 - 9*A*a^2*b^3*ta 
n(1/2*x)^3 - 5*B*a^2*b^3*tan(1/2*x)^3 + A*a*b^4*tan(1/2*x)^3 + 5*B*a*b^4*t 
an(1/2*x)^3 + A*b^5*tan(1/2*x)^3 - 2*B*b^5*tan(1/2*x)^3 + 5*A*a^3*c^2*tan( 
1/2*x)^3 - 4*B*a^3*c^2*tan(1/2*x)^3 - 7*A*a^2*b*c^2*tan(1/2*x)^3 + 4*B*a^2 
*b*c^2*tan(1/2*x)^3 - A*a*b^2*c^2*tan(1/2*x)^3 + 4*B*a*b^2*c^2*tan(1/2*x)^ 
3 + 3*A*b^3*c^2*tan(1/2*x)^3 - 4*B*b^3*c^2*tan(1/2*x)^3 - 2*A*a*c^4*tan(1/ 
2*x)^3 + 2*B*a*c^4*tan(1/2*x)^3 + 2*A*b*c^4*tan(1/2*x)^3 - 2*B*b*c^4*tan(1 
/2*x)^3 + 4*A*a^4*c*tan(1/2*x)^2 + 2*B*a^4*c*tan(1/2*x)^2 - 12*A*a^3*b*c*t 
an(1/2*x)^2 - 9*B*a^3*b*c*tan(1/2*x)^2 + 13*A*a^2*b^2*c*tan(1/2*x)^2 + 14* 
B*a^2*b^2*c*tan(1/2*x)^2 - 6*A*a*b^3*c*tan(1/2*x)^2 - 9*B*a*b^3*c*tan(1/2* 
x)^2 + A*b^4*c*tan(1/2*x)^2 + 2*B*b^4*c*tan(1/2*x)^2 + 7*A*a^2*c^3*tan(1/2 
*x)^2 - 4*B*a^2*c^3*tan(1/2*x)^2 - 6*A*a*b*c^3*tan(1/2*x)^2 - A*b^2*c^3*ta 
n(1/2*x)^2 + 4*B*b^2*c^3*tan(1/2*x)^2 - 2*A*c^5*tan(1/2*x)^2 + 2*B*c^5*tan 
(1/2*x)^2 + 2*B*a^5*tan(1/2*x) - 4*A*a^4*b*tan(1/2*x) - 3*B*a^4*b*tan(1/2* 
x) + 5*A*a^3*b^2*tan(1/2*x) + B*a^3*b^2*tan(1/2*x) + 3*A*a^2*b^3*tan(1/2*x 
) + B*a^2*b^3*tan(1/2*x) - 5*A*a*b^4*tan(1/2*x) - 3*B*a*b^4*tan(1/2*x) ...
 

Mupad [B] (verification not implemented)

Time = 21.39 (sec) , antiderivative size = 946, normalized size of antiderivative = 4.73 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx =\text {Too large to display} \] Input:

int((A + B*cos(x))/(a + b*cos(x) + c*sin(x))^3,x)
 

Output:

- ((A*a^2*c^3 + A*b^2*c^3 - 4*A*a^4*c + A*b^4*c - 2*B*a*b*c^3 - 5*B*a*b^3* 
c + 5*B*a^3*b*c + 3*A*a^2*b^2*c)/((a - b)^2*(a^4 + b^4 + c^4 - 2*a^2*b^2 - 
 2*a^2*c^2 + 2*b^2*c^2)) - (tan(x/2)*(A*b^5 + 2*B*a^5 + 2*B*b^5 + 3*A*a^2* 
b^3 + 5*A*a^3*b^2 + 11*A*a^3*c^2 + B*a^2*b^3 + B*a^3*b^2 - A*b^3*c^2 - 4*B 
*a^3*c^2 + 4*B*b^3*c^2 - 5*A*a*b^4 - 4*A*a^4*b - 2*A*a*c^4 - 3*B*a*b^4 - 3 
*B*a^4*b - 2*A*b*c^4 + 2*B*a*c^4 + 2*B*b*c^4 - 7*A*a*b^2*c^2 - 3*A*a^2*b*c 
^2 + 8*B*a*b^2*c^2 - 8*B*a^2*b*c^2))/((a - b)^2*(a^4 + b^4 + c^4 - 2*a^2*b 
^2 - 2*a^2*c^2 + 2*b^2*c^2)) - (tan(x/2)^2*(2*B*c^5 - 2*A*c^5 + 7*A*a^2*c^ 
3 - A*b^2*c^3 - 4*B*a^2*c^3 + 4*B*b^2*c^3 + 4*A*a^4*c + A*b^4*c + 2*B*a^4* 
c + 2*B*b^4*c - 6*A*a*b*c^3 - 6*A*a*b^3*c - 12*A*a^3*b*c - 9*B*a*b^3*c - 9 
*B*a^3*b*c + 13*A*a^2*b^2*c + 14*B*a^2*b^2*c))/((a - b)^2*(a^4 + b^4 + c^4 
 - 2*a^2*b^2 - 2*a^2*c^2 + 2*b^2*c^2)) + (tan(x/2)^3*(A*b^4 - 2*B*a^4 + 2* 
A*c^4 - 2*B*b^4 - 2*B*c^4 - 7*A*a^2*b^2 - 5*A*a^2*c^2 - 2*B*a^2*b^2 + 3*A* 
b^2*c^2 + 4*B*a^2*c^2 - 4*B*b^2*c^2 + 2*A*a*b^3 + 4*A*a^3*b + 3*B*a*b^3 + 
3*B*a^3*b + 2*A*a*b*c^2))/((a - b)*(a^4 + b^4 + c^4 - 2*a^2*b^2 - 2*a^2*c^ 
2 + 2*b^2*c^2)))/(tan(x/2)^4*(a^2 - 2*a*b + b^2) + 2*a*b + tan(x/2)*(4*a*c 
 + 4*b*c) + tan(x/2)^3*(4*a*c - 4*b*c) + a^2 + b^2 + tan(x/2)^2*(2*a^2 - 2 
*b^2 + 4*c^2)) - (atanh((2*a^4*c + 2*b^4*c + 2*c^5 - 4*a^2*c^3 + 4*b^2*c^3 
 - 4*a^2*b^2*c)/(2*(b^2 - a^2 + c^2)^(5/2)) + (tan(x/2)*(2*a - 2*b)*(a^4 + 
 b^4 + c^4 - 2*a^2*b^2 - 2*a^2*c^2 + 2*b^2*c^2))/(2*(b^2 - a^2 + c^2)^(...
 

Reduce [B] (verification not implemented)

Time = 4.77 (sec) , antiderivative size = 9162, normalized size of antiderivative = 45.81 \[ \int \frac {A+B \cos (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx =\text {Too large to display} \] Input:

int((A+B*cos(x))/(a+b*cos(x)+c*sin(x))^3,x)
 

Output:

(16*sqrt(a**2 - b**2 - c**2)*atan((tan(x/2)*a - tan(x/2)*b + c)/sqrt(a**2 
- b**2 - c**2))*cos(x)**3*sin(x)*a**3*b**3*c**2 - 16*sqrt(a**2 - b**2 - c* 
*2)*atan((tan(x/2)*a - tan(x/2)*b + c)/sqrt(a**2 - b**2 - c**2))*cos(x)**3 
*sin(x)*a*b**5*c**2 + 8*sqrt(a**2 - b**2 - c**2)*atan((tan(x/2)*a - tan(x/ 
2)*b + c)/sqrt(a**2 - b**2 - c**2))*cos(x)**3*sin(x)*a*b**3*c**4 + 16*sqrt 
(a**2 - b**2 - c**2)*atan((tan(x/2)*a - tan(x/2)*b + c)/sqrt(a**2 - b**2 - 
 c**2))*cos(x)**3*a**4*b**3*c - 16*sqrt(a**2 - b**2 - c**2)*atan((tan(x/2) 
*a - tan(x/2)*b + c)/sqrt(a**2 - b**2 - c**2))*cos(x)**3*a**2*b**5*c + 8*s 
qrt(a**2 - b**2 - c**2)*atan((tan(x/2)*a - tan(x/2)*b + c)/sqrt(a**2 - b** 
2 - c**2))*cos(x)**3*a**2*b**3*c**3 - 8*sqrt(a**2 - b**2 - c**2)*atan((tan 
(x/2)*a - tan(x/2)*b + c)/sqrt(a**2 - b**2 - c**2))*cos(x)**2*sin(x)**2*a* 
*3*b**4*c + 40*sqrt(a**2 - b**2 - c**2)*atan((tan(x/2)*a - tan(x/2)*b + c) 
/sqrt(a**2 - b**2 - c**2))*cos(x)**2*sin(x)**2*a**3*b**2*c**3 + 8*sqrt(a** 
2 - b**2 - c**2)*atan((tan(x/2)*a - tan(x/2)*b + c)/sqrt(a**2 - b**2 - c** 
2))*cos(x)**2*sin(x)**2*a*b**6*c - 44*sqrt(a**2 - b**2 - c**2)*atan((tan(x 
/2)*a - tan(x/2)*b + c)/sqrt(a**2 - b**2 - c**2))*cos(x)**2*sin(x)**2*a*b* 
*4*c**3 + 20*sqrt(a**2 - b**2 - c**2)*atan((tan(x/2)*a - tan(x/2)*b + c)/s 
qrt(a**2 - b**2 - c**2))*cos(x)**2*sin(x)**2*a*b**2*c**5 + 80*sqrt(a**2 - 
b**2 - c**2)*atan((tan(x/2)*a - tan(x/2)*b + c)/sqrt(a**2 - b**2 - c**2))* 
cos(x)**2*sin(x)*a**4*b**2*c**2 - 80*sqrt(a**2 - b**2 - c**2)*atan((tan...