\(\int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx\) [490]

Optimal result

Integrand size = 22, antiderivative size = 197 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=-\frac {3 a (b B+c C) \arctan \left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{5/2}}+\frac {B c-b C-a C \cos (x)+a B \sin (x)}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}+\frac {a (B c-b C)-\left (2 b B c+\left (a^2+2 c^2\right ) C\right ) \cos (x)+\left (a^2 B+2 b (b B+c C)\right ) \sin (x)}{2 \left (a^2-b^2-c^2\right )^2 (a+b \cos (x)+c \sin (x))} \] Output:

-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*cos(x)+a*B*sin(x))/(a^2-b^2-c^2)/(a+b*cos(x)+c* 
sin(x))^2+1/2*(a*(B*c-C*b)-(2*b*B*c+(a^2+2*c^2)*C)*cos(x)+(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 = 0.81 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.58 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\frac {3 a (b B+c C) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{5/2}}+\frac {9 a^2 b B c+2 a^4 C-4 a^2 b^2 C+2 b^4 C+5 a^2 c^2 C+4 b^2 c^2 C+2 c^4 C+6 a b c (b B+c C) \cos (x)-c \left (a^2+2 \left (b^2+c^2\right )\right ) (b B+c C) \cos (2 x)+4 a^3 b B \sin (x)+2 a b^3 B \sin (x)+8 a b B c^2 \sin (x)+4 a^3 c C \sin (x)+2 a b^2 c C \sin (x)+8 a c^3 C \sin (x)+a^2 b^2 B \sin (2 x)+2 b^4 B \sin (2 x)+2 b^2 B c^2 \sin (2 x)+a^2 b c C \sin (2 x)+2 b^3 c C \sin (2 x)+2 b c^3 C \sin (2 x)}{4 b \left (-a^2+b^2+c^2\right )^2 (a+b \cos (x)+c \sin (x))^2} \] Input:

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

Output:

(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) + (9*a^2*b*B*c + 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 + 6*a*b*c*(b*B + c*C)*Cos[x] - c*(a^2 
 + 2*(b^2 + c^2))*(b*B + c*C)*Cos[2*x] + 4*a^3*b*B*Sin[x] + 2*a*b^3*B*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] + a^2*b^2*B*Sin[2*x] + 2*b^4*B*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 = 0.66 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3635, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3635

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3632

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3603

\(\displaystyle \frac {a B \sin (x)-a C \cos (x)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}-\frac {\frac {6 a (b B+c C) \int \frac {1}{(a-b) \tan ^2\left (\frac {x}{2}\right )+2 c \tan \left (\frac {x}{2}\right )+a+b}d\tan \left (\frac {x}{2}\right )}{a^2-b^2-c^2}-\frac {\sin (x) \left (a^2 B+2 b (b B+c C)\right )-\cos (x) \left (a^2 C+2 c (b B+c C)\right )+a (B c-b C)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {a B \sin (x)-a C \cos (x)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}-\frac {-\frac {12 a (b B+c C) \int \frac {1}{-\left (2 c+2 (a-b) \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2-c^2\right )}d\left (2 c+2 (a-b) \tan \left (\frac {x}{2}\right )\right )}{a^2-b^2-c^2}-\frac {\sin (x) \left (a^2 B+2 b (b B+c C)\right )-\cos (x) \left (a^2 C+2 c (b B+c C)\right )+a (B c-b C)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {a B \sin (x)-a C \cos (x)-b C+B c}{2 \left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))^2}-\frac {\frac {6 a (b B+c C) \arctan \left (\frac {2 (a-b) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}-\frac {\sin (x) \left (a^2 B+2 b (b B+c C)\right )-\cos (x) \left (a^2 C+2 c (b B+c C)\right )+a (B c-b C)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}}{2 \left (a^2-b^2-c^2\right )}\)

Input:

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

Output:

(B*c - b*C - a*C*Cos[x] + a*B*Sin[x])/(2*(a^2 - b^2 - c^2)*(a + b*Cos[x] + 
 c*Sin[x])^2) - ((6*a*(b*B + c*C)*ArcTan[(2*c + 2*(a - b)*Tan[x/2])/(2*Sqr 
t[a^2 - b^2 - c^2])])/(a^2 - b^2 - c^2)^(3/2) - (a*(B*c - b*C) - (a^2*C + 
2*c*(b*B + c*C))*Cos[x] + (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 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. \(794\) vs. \(2(187)=374\).

Time = 0.71 (sec) , antiderivative size = 795, normalized size of antiderivative = 4.04

method	result	size
default	\(-\frac {2 \left (-\frac {\left (2 B \,a^{4}-3 B \,a^{3} b +2 B \,a^{2} b^{2}-4 B \,a^{2} c^{2}-3 B a \,b^{3}+2 B \,b^{4}+4 B \,b^{2} c^{2}+2 B \,c^{4}-3 C \,a^{3} c +6 C \,a^{2} b c -3 C a \,b^{2} c \right ) \tan \left (\frac {x}{2}\right )^{3}}{2 \left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a -b \right )}-\frac {\left (2 B \,a^{4} c -9 B \,a^{3} b c +14 B \,a^{2} b^{2} c -4 B \,a^{2} c^{3}-9 B a \,b^{3} c +2 B \,b^{4} c +4 B \,b^{2} c^{3}+2 B \,c^{5}-2 C \,a^{5}+2 C \,a^{4} b +4 C \,a^{3} b^{2}-5 C \,a^{3} c^{2}-4 C \,a^{2} b^{3}+14 C \,a^{2} b \,c^{2}-2 C a \,b^{4}-13 C a \,b^{2} c^{2}-2 C a \,c^{4}+2 C \,b^{5}+4 C \,b^{3} c^{2}+2 C b \,c^{4}\right ) \tan \left (\frac {x}{2}\right )^{2}}{2 \left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 B \,a^{5}-3 B \,a^{4} b +B \,a^{3} b^{2}-4 B \,a^{3} c^{2}+B \,a^{2} b^{3}-8 B \,a^{2} b \,c^{2}-3 B a \,b^{4}+8 B a \,b^{2} c^{2}+2 B a \,c^{4}+2 B \,b^{5}+4 B \,b^{3} c^{2}+2 B b \,c^{4}-5 C \,a^{4} c +5 C \,a^{3} b c +5 C \,a^{2} b^{2} c -4 C \,a^{2} c^{3}-5 C a \,b^{3} c +4 C a b \,c^{3}\right ) \tan \left (\frac {x}{2}\right )}{2 \left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {a \left (5 B \,a^{2} b c -5 B \,b^{3} c -2 B b \,c^{3}+2 C \,a^{4}-4 C \,a^{2} b^{2}+C \,a^{2} c^{2}+2 C \,b^{4}-C \,b^{2} c^{2}\right )}{2 \left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {x}{2}\right )^{2} a -b \tan \left (\frac {x}{2}\right )^{2}+2 c \tan \left (\frac {x}{2}\right )+a +b \right )^{2}}-\frac {3 a \left (B b +C c \right ) \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \sqrt {a^{2}-b^{2}-c^{2}}}\)	\(795\)
risch	\(\frac {i \left (-2 C \,c^{4}+2 \,{\mathrm e}^{2 i x} a^{4} C +2 C \,c^{4} {\mathrm e}^{2 i x}+2 C \,b^{4} {\mathrm e}^{2 i x}+2 i B \,b^{4}-B \,a^{2} b c +2 i C \,b^{3} c +2 i C b \,c^{3}-4 C \,a^{2} b^{2} {\mathrm e}^{2 i x}+5 C \,a^{2} c^{2} {\mathrm e}^{2 i x}+4 C \,b^{2} c^{2} {\mathrm e}^{2 i x}+i B \,a^{2} b^{2}+2 i {\mathrm e}^{2 i x} a^{4} B +2 i B \,b^{4} {\mathrm e}^{2 i x}+2 i B \,c^{4} {\mathrm e}^{2 i x}+2 i B \,b^{2} c^{2}+6 C a b \,c^{2} {\mathrm e}^{3 i x}-3 i C a \,c^{3} {\mathrm e}^{3 i x}+4 i C \,a^{3} c \,{\mathrm e}^{i x}+5 i C a \,c^{3} {\mathrm e}^{i x}+3 i B a \,b^{3} {\mathrm e}^{3 i x}+5 i B \,a^{2} b^{2} {\mathrm e}^{2 i x}-4 i B \,a^{2} c^{2} {\mathrm e}^{2 i x}+4 i B \,b^{2} c^{2} {\mathrm e}^{2 i x}+6 B a \,b^{2} c \,{\mathrm e}^{3 i x}+9 B \,a^{2} b c \,{\mathrm e}^{2 i x}+i C \,a^{2} b c +4 i B \,a^{3} b \,{\mathrm e}^{i x}+5 i B a \,b^{3} {\mathrm e}^{i x}-2 B \,b^{3} c -2 B b \,c^{3}+5 i B a b \,c^{2} {\mathrm e}^{i x}-3 i B a b \,c^{2} {\mathrm e}^{3 i x}+3 i C a \,b^{2} c \,{\mathrm e}^{3 i x}+9 i C \,a^{2} b c \,{\mathrm e}^{2 i x}+5 i C a \,b^{2} c \,{\mathrm e}^{i x}-C \,a^{2} c^{2}-2 C \,b^{2} c^{2}\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 )^{2} \left (a^{2}-b^{2}-c^{2}\right )^{2} \left (i b +c \right )}-\frac {3 \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}+c^{2}}\, a c -i a^{2} b +i b^{3}+i b \,c^{2}+\sqrt {-a^{2}+b^{2}+c^{2}}\, a b +a^{2} c -b^{2} c -c^{3}}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (b^{2}+c^{2}\right )}\right ) a b B}{2 \sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}+c^{2}}\, a c -i a^{2} b +i b^{3}+i b \,c^{2}+\sqrt {-a^{2}+b^{2}+c^{2}}\, a b +a^{2} c -b^{2} c -c^{3}}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (b^{2}+c^{2}\right )}\right ) a c C}{2 \sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}+c^{2}}\, a c +i a^{2} b -i b^{3}-i b \,c^{2}+\sqrt {-a^{2}+b^{2}+c^{2}}\, a b -a^{2} c +b^{2} c +c^{3}}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (b^{2}+c^{2}\right )}\right ) a b B}{2 \sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}+c^{2}}\, a c +i a^{2} b -i b^{3}-i b \,c^{2}+\sqrt {-a^{2}+b^{2}+c^{2}}\, a b -a^{2} c +b^{2} c +c^{3}}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (b^{2}+c^{2}\right )}\right ) a c C}{2 \sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )^{2}}\)	\(1055\)

Input:

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

Output:

-2*(-1/2*(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^4-2*a^2*b^2-2*a^2*c^ 
2+b^4+2*b^2*c^2+c^4)/(a-b)*tan(1/2*x)^3-1/2*(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*(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*a*(5*B*a^2*b*c-5*B*b^3*c-2*B*b*c^3+2*C*a^ 
4-4*C*a^2*b^2+C*a^2*c^2+2*C*b^4-C*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)^2-3*a*(B*b+C*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-c^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1551 vs. \(2 (187) = 374\).

Time = 0.28 (sec) , antiderivative size = 3264, normalized size of antiderivative = 16.57 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((B*cos(x)+C*sin(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. 1034 vs. \(2 (187) = 374\).

Time = 0.37 (sec) , antiderivative size = 1034, normalized size of antiderivative = 5.25 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

3*(B*a*b + C*a*c)*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*t 
an(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 - 5*B*a^4*b*tan(1/2*x)^3 + 5*B*a^3*b^2*tan(1/2*x)^3 - 5*B*a^2*b^3* 
tan(1/2*x)^3 + 5*B*a*b^4*tan(1/2*x)^3 - 2*B*b^5*tan(1/2*x)^3 - 3*C*a^4*c*t 
an(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 - 4*B*a^3*c^2*tan(1/2*x)^3 + 4*B*a^2*b*c^2*tan(1/2*x) 
^3 + 4*B*a*b^2*c^2*tan(1/2*x)^3 - 4*B*b^3*c^2*tan(1/2*x)^3 + 2*B*a*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 + 2*B*a^4*c*tan(1/2*x)^2 - 9*B*a^3*b* 
c*tan(1/2*x)^2 + 14*B*a^2*b^2*c*tan(1/2*x)^2 - 9*B*a*b^3*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 - 4*B*a^2*c 
^3*tan(1/2*x)^2 + 4*B*b^2*c^3*tan(1/2*x)^2 - 2*C*a*c^4*tan(1/2*x)^2 + 2*C* 
b*c^4*tan(1/2*x)^2 + 2*B*c^5*tan(1/2*x)^2 + 2*B*a^5*tan(1/2*x) - 3*B*a^4*b 
*tan(1/2*x) + B*a^3*b^2*tan(1/2*x) + B*a^2*b^3*tan(1/2*x) - 3*B*a*b^4*tan( 
1/2*x) + 2*B*b^5*tan(1/2*x) - 5*C*a^4*c*tan(1/2*x) + 5*C*a^3*b*c*tan(1/2*x 
) + 5*C*a^2*b^2*c*tan(1/2*x) - 5*C*a*b^3*c*tan(1/2*x) - 4*B*a^3*c^2*tan(1/ 
2*x) - 8*B*a^2*b*c^2*tan(1/2*x) + 8*B*a*b^2*c^2*tan(1/2*x) + 4*B*b^3*c^...
 

Mupad [B] (verification not implemented)

Time = 20.43 (sec) , antiderivative size = 923, normalized size of antiderivative = 4.69 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((tan(x/2)^3*(2*B*a^4 + 2*B*b^4 + 2*B*c^4 + 2*B*a^2*b^2 - 4*B*a^2*c^2 + 4* 
B*b^2*c^2 - 3*B*a*b^3 - 3*B*a^3*b - 3*C*a^3*c - 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)) - (2*C*a 
^5 - 4*C*a^3*b^2 + C*a^3*c^2 + 2*C*a*b^4 - 2*B*a*b*c^3 - 5*B*a*b^3*c + 5*B 
*a^3*b*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)^2*(2*B*c^5 - 2*C*a^5 + 2*C*b^5 - 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 + 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 - 9*B*a 
*b^3*c - 9*B*a^3*b*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)*(2*B*a^5 + 2*B*b^5 + B*a^2*b^3 + B*a^3*b^2 - 4*B*a^3*c^2 + 4*B*b^3*c^2 
 - 4*C*a^2*c^3 - 3*B*a*b^4 - 3*B*a^4*b + 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 + 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/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)) + (3*a*atanh((3*a*(B*b + C*c)*(tan(x/2)*(2*a - 2*b) + (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)/(a^4 + b^4 + c^4 - 
 2*a^2*b^2 - 2*a^2*c^2 + 2*b^2*c^2))*(a^4 + b^4 + c^4 - 2*a^2*b^2 - 2*a^2* 
c^2 + 2*b^2*c^2))/(2*(3*B*a*b + 3*C*a*c)*(b^2 - a^2 + c^2)^(5/2)))*(B*b...
 

Reduce [B] (verification not implemented)

Time = 2.41 (sec) , antiderivative size = 14520, normalized size of antiderivative = 73.71 \[ \int \frac {B \cos (x)+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 24*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**2*b**5*c**2 - 24*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**2*b**3*c**4 + 24*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**6*c**2 + 
24*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**4*c**4 - 24*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** 
3*b**5*c - 24*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**3*b**3*c**3 + 24*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**6*c + 24*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**4*c**3 + 12*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**2*b**6*c - 48*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**2 
*b**4*c**3 - 60*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**2*b**2*c**5 - 12*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**7*c + 48*sqrt(a**2 - b**2 - c**2)*atan((...