\(\int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx\) [495]

Optimal result

Integrand size = 23, antiderivative size = 237 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\frac {\left (2 a^2 A+A \left (b^2+c^2\right )-3 a (b B+c C)\right ) \arctan \left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{5/2}}+\frac {B c-b C+(A c-a C) \cos (x)-(A b-a B) \sin (x)}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}+\frac {a (B c-b C)+\left (3 a A c-a^2 C-2 c (b B+c C)\right ) \cos (x)-\left (3 a A b-a^2 B-2 b (b B+c C)\right ) \sin (x)}{2 \left (a^2-b^2-c^2\right )^2 (a+b \cos (x)+c \sin (x))} \] Output:

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

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.91 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=-\frac {\left (2 a^2 A+A \left (b^2+c^2\right )-3 a (b B+c C)\right ) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{5/2}}+\frac {-6 a^3 A c-3 a A b^2 c+9 a^2 b B c-3 a A c^3+2 a^4 C-4 a^2 b^2 C+2 b^4 C+5 a^2 c^2 C+4 b^2 c^2 C+2 c^4 C-2 b c \left (2 a^2 A+A \left (b^2+c^2\right )-3 a (b B+c C)\right ) \cos (x)-c \left (-3 a A \left (b^2+c^2\right )+a^2 (b B+c C)+2 \left (b^2+c^2\right ) (b B+c C)\right ) \cos (2 x)-8 a^2 A b^2 \sin (x)+2 A b^4 \sin (x)+4 a^3 b B \sin (x)+2 a b^3 B \sin (x)-12 a^2 A c^2 \sin (x)+2 A b^2 c^2 \sin (x)+8 a b B c^2 \sin (x)+4 a^3 c C \sin (x)+2 a b^2 c C \sin (x)+8 a c^3 C \sin (x)-3 a A b^3 \sin (2 x)+a^2 b^2 B \sin (2 x)+2 b^4 B \sin (2 x)-3 a A b c^2 \sin (2 x)+2 b^2 B c^2 \sin (2 x)+a^2 b c C \sin (2 x)+2 b^3 c C \sin (2 x)+2 b c^3 C \sin (2 x)}{4 b \left (-a^2+b^2+c^2\right )^2 (a+b \cos (x)+c \sin (x))^2} \] Input:

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

Output:

-(((2*a^2*A + A*(b^2 + c^2) - 3*a*(b*B + c*C))*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*a^4*C - 4*a^2*b^2*C + 2*b^4*C + 5*a 
^2*c^2*C + 4*b^2*c^2*C + 2*c^4*C - 2*b*c*(2*a^2*A + A*(b^2 + c^2) - 3*a*(b 
*B + c*C))*Cos[x] - c*(-3*a*A*(b^2 + c^2) + a^2*(b*B + c*C) + 2*(b^2 + c^2 
)*(b*B + c*C))*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] + 4*a^3*c*C*Sin[x] + 2*a*b^2*c*C*Sin[x] + 8*a*c^3*C*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] + a^2*b*c*C*Sin[2*x] + 2*b^3*c*C*Sin[2*x 
] + 2*b*c^3*C*Sin[2*x])/(4*b*(-a^2 + b^2 + c^2)^2*(a + b*Cos[x] + c*Sin[x] 
)^2)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3635, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3635

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3632

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3603

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

(B*c - b*C + (A*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 + A*(b^2 + c^2) - 3*a*(b*B + 
c*C))*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 - b*C) + (3*a*A*c - a^2*C - 2*c*(b*B + c*C))*C 
os[x] - (3*a*A*b - a^2*B - 2*b*(b*B + c*C))*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 3635

Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]) 
^(n_)*((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_) 
]), x_Symbol] :> Simp[(-(c*B - b*C - (a*C - 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*Co 
s[d + e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - b*B - c*C) + (n + 
2)*(a*B - b*A)*Cos[d + e*x] + (n + 2)*(a*C - c*A)*Sin[d + e*x], x], x], x] 
/; FreeQ[{a, b, c, d, e, A, B, C}, 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. \(1079\) vs. \(2(227)=454\).

Time = 0.78 (sec) , antiderivative size = 1080, normalized size of antiderivative = 4.56

method	result	size
default	\(\text {Expression too large to display}\)	\(1080\)
risch	\(\text {Expression too large to display}\)	\(2157\)

Input:

int((A+B*cos(x)+C*sin(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+3*C*a^3*c-6*C*a^2*b*c+3*C*a*b^2*c)/(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*(4*A*a^4*c-12*A*a^3*b*c 
+13*A*a^2*b^2*c+7*A*a^2*c^3-6*A*a*b^3*c-6*A*a*b*c^3+A*b^4*c-A*b^2*c^3-2*A* 
c^5+2*B*a^4*c-9*B*a^3*b*c+14*B*a^2*b^2*c-4*B*a^2*c^3-9*B*a*b^3*c+2*B*b^4*c 
+4*B*b^2*c^3+2*B*c^5-2*C*a^5+2*C*a^4*b+4*C*a^3*b^2-5*C*a^3*c^2-4*C*a^2*b^3 
+14*C*a^2*b*c^2-2*C*a*b^4-13*C*a*b^2*c^2-2*C*a*c^4+2*C*b^5+4*C*b^3*c^2+2*C 
*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+5*C*a^4*c-5*C*a^3*b*c-5*C*a^2*b^2*c 
+4*C*a^2*c^3+5*C*a*b^3*c-4*C*a*b*c^3)/(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*(4*A*a^4*c-3*A*a^2*b^2*c-A*a^2*c^3- 
A*b^4*c-A*b^2*c^3-5*B*a^3*b*c+5*B*a*b^3*c+2*B*a*b*c^3-2*C*a^5+4*C*a^3*b^2- 
C*a^3*c^2-2*C*a*b^4+C*a*b^2*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))/(tan(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-3*C*a*c)/(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)*arctan(1/2*(2*(a-b)*tan(1/2*x)+2*c)/(a^2-b^2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2038 vs. \(2 (225) = 450\).

Time = 0.33 (sec) , antiderivative size = 4240, normalized size of antiderivative = 17.89 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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. 1506 vs. \(2 (225) = 450\).

Time = 0.42 (sec) , antiderivative size = 1506, normalized size of antiderivative = 6.35 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

-(2*A*a^2 - 3*B*a*b + A*b^2 - 3*C*a*c + A*c^2)*(pi*floor(1/2*x/pi + 1/2)*s 
gn(-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*tan(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*tan(1/2*x)^3 + A*b^5*tan(1/2*x)^3 - 2*B*b^5*tan(1/2*x)^3 - 3*C*a^ 
4*c*tan(1/2*x)^3 + 9*C*a^3*b*c*tan(1/2*x)^3 - 9*C*a^2*b^2*c*tan(1/2*x)^3 + 
 3*C*a*b^3*c*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 - 2*C*a^5*tan(1/2*x)^2 
 + 2*C*a^4*b*tan(1/2*x)^2 + 4*C*a^3*b^2*tan(1/2*x)^2 - 4*C*a^2*b^3*tan(1/2 
*x)^2 - 2*C*a*b^4*tan(1/2*x)^2 + 2*C*b^5*tan(1/2*x)^2 + 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*tan(1/2*x)^2 - 9*B*a^3*b*c*ta 
n(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 - 5*C*a^3*c^2*tan(1/2*x)^2 + 14*C*a^2*b*c^2*tan(1 
/2*x)^2 - 13*C*a*b^2*c^2*tan(1/2*x)^2 + 4*C*b^3*c^2*tan(1/2*x)^2 + 7*A*...
 

Mupad [B] (verification not implemented)

Time = 21.37 (sec) , antiderivative size = 1160, normalized size of antiderivative = 4.89 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

- ((2*C*a^5 + A*a^2*c^3 + A*b^2*c^3 - 4*C*a^3*b^2 + C*a^3*c^2 - 4*A*a^4*c 
+ A*b^4*c + 2*C*a*b^4 - 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 - C*a*b^2*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)^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 + 3*C*a^3*c + 2* 
A*a*b*c^2 + 3*C*a*b^2*c - 6*C*a^2*b*c))/((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)^2*(2*B*c^5 - 2*C*a^5 - 2*A*c^5 + 
 2*C*b^5 + 7*A*a^2*c^3 - A*b^2*c^3 - 4*B*a^2*c^3 - 4*C*a^2*b^3 + 4*C*a^3*b 
^2 + 4*B*b^2*c^3 - 5*C*a^3*c^2 + 4*C*b^3*c^2 + 4*A*a^4*c + A*b^4*c + 2*B*a 
^4*c - 2*C*a*b^4 + 2*C*a^4*b + 2*B*b^4*c - 2*C*a*c^4 + 2*C*b*c^4 - 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 - 13*C*a*b^2*c^2 + 14*C*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)*(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 - 4*C*a^2*c^3 - 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 - 5*C*a^4*c + 4*C*a*b*c^3 - 5*C*a*b^3*c + 5*C*a^3*b*c - 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 + 5*C*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...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.59 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\frac {2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) \cos \left (x \right ) a b c +2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) \sin \left (x \right ) a \,c^{2}+2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) a^{2} c +\cos \left (x \right ) a^{2} b^{2}+\cos \left (x \right ) a^{2} c^{2}-\cos \left (x \right ) b^{4}-2 \cos \left (x \right ) b^{2} c^{2}-\cos \left (x \right ) c^{4}+a^{3} b -a \,b^{3}-a b \,c^{2}}{c \left (\cos \left (x \right ) a^{4} b -2 \cos \left (x \right ) a^{2} b^{3}-2 \cos \left (x \right ) a^{2} b \,c^{2}+\cos \left (x \right ) b^{5}+2 \cos \left (x \right ) b^{3} c^{2}+\cos \left (x \right ) b \,c^{4}+\sin \left (x \right ) a^{4} c -2 \sin \left (x \right ) a^{2} b^{2} c -2 \sin \left (x \right ) a^{2} c^{3}+\sin \left (x \right ) b^{4} c +2 \sin \left (x \right ) b^{2} c^{3}+\sin \left (x \right ) c^{5}+a^{5}-2 a^{3} b^{2}-2 a^{3} c^{2}+a \,b^{4}+2 a \,b^{2} c^{2}+a \,c^{4}\right )} \] Input:

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

Output:

(2*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)*a*b*c + 2*sqrt(a**2 - b**2 - c**2)*atan((tan(x/2)*a 
- tan(x/2)*b + c)/sqrt(a**2 - b**2 - c**2))*sin(x)*a*c**2 + 2*sqrt(a**2 - 
b**2 - c**2)*atan((tan(x/2)*a - tan(x/2)*b + c)/sqrt(a**2 - b**2 - c**2))* 
a**2*c + cos(x)*a**2*b**2 + cos(x)*a**2*c**2 - cos(x)*b**4 - 2*cos(x)*b**2 
*c**2 - cos(x)*c**4 + a**3*b - a*b**3 - a*b*c**2)/(c*(cos(x)*a**4*b - 2*co 
s(x)*a**2*b**3 - 2*cos(x)*a**2*b*c**2 + cos(x)*b**5 + 2*cos(x)*b**3*c**2 + 
 cos(x)*b*c**4 + sin(x)*a**4*c - 2*sin(x)*a**2*b**2*c - 2*sin(x)*a**2*c**3 
 + sin(x)*b**4*c + 2*sin(x)*b**2*c**3 + sin(x)*c**5 + a**5 - 2*a**3*b**2 - 
 2*a**3*c**2 + a*b**4 + 2*a*b**2*c**2 + a*c**4))