\(\int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx\) [485]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.80 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=-\frac {\left (2 a^2 A+A \left (b^2+c^2\right )-3 a c C\right ) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{5/2}}+\frac {-6 a^3 A c-3 a A b^2 c-3 a A c^3+2 a^4 C-4 a^2 b^2 C+2 b^4 C+5 a^2 c^2 C+4 b^2 c^2 C+2 c^4 C-2 b c \left (2 a^2 A+A \left (b^2+c^2\right )-3 a c C\right ) \cos (x)-c \left (-3 a A \left (b^2+c^2\right )+a^2 c C+2 c \left (b^2+c^2\right ) C\right ) \cos (2 x)-8 a^2 A b^2 \sin (x)+2 A b^4 \sin (x)-12 a^2 A c^2 \sin (x)+2 A b^2 c^2 \sin (x)+4 a^3 c C \sin (x)+2 a b^2 c C \sin (x)+8 a c^3 C \sin (x)-3 a A b^3 \sin (2 x)-3 a A b c^2 \sin (2 x)+a^2 b c C \sin (2 x)+2 b^3 c C \sin (2 x)+2 b c^3 C \sin (2 x)}{4 b \left (-a^2+b^2+c^2\right )^2 (a+b \cos (x)+c \sin (x))^2} \] Input:

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

Output:

-(((2*a^2*A + A*(b^2 + c^2) - 3*a*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 
 - 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*c*C)*Cos[x] - c*(-3*a*A* 
(b^2 + c^2) + a^2*c*C + 2*c*(b^2 + c^2)*C)*Cos[2*x] - 8*a^2*A*b^2*Sin[x] + 
 2*A*b^4*Sin[x] - 12*a^2*A*c^2*Sin[x] + 2*A*b^2*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] - 3*a*A*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 = 0.69 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 3636, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3636

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3632

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3603

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

-1/2*(b*C - (A*c - a*C)*Cos[x] + A*b*Sin[x])/((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*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 - (3*a*A*c - a^2*C - 2*c^2*C)*Cos[x] + b*(3*a*A - 2*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 3636

Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]) 
^(n_)*((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[(b*C + (a* 
C - c*A)*Cos[d + e*x] + 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 - c*C) - (n + 2)*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, 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. \(831\) vs. \(2(190)=380\).

Time = 0.72 (sec) , antiderivative size = 832, normalized size of antiderivative = 4.16

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}+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}}{\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 (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 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}}{\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}+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 )}{\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 {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}-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}}{\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 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}}}\)	\(832\)
risch	\(\text {Expression too large to display}\)	\(1656\)

Input:

int((A+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+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*(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* 
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+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^3-2*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*t 
an(1/2*x)^2+2*c*tan(1/2*x)+a+b)^2+(2*A*a^2+A*b^2+A*c^2-3*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-c^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1676 vs. \(2 (187) = 374\).

Time = 0.30 (sec) , antiderivative size = 3513, normalized size of antiderivative = 17.56 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+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. 1054 vs. \(2 (187) = 374\).

Time = 0.39 (sec) , antiderivative size = 1054, normalized size of antiderivative = 5.27 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

-(2*A*a^2 + A*b^2 - 3*C*a*c + 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 
)) - (4*A*a^4*b*tan(1/2*x)^3 - 11*A*a^3*b^2*tan(1/2*x)^3 + 9*A*a^2*b^3*tan 
(1/2*x)^3 - A*a*b^4*tan(1/2*x)^3 - A*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 + 7*A*a^2*b*c^2*tan(1/2*x)^3 + A* 
a*b^2*c^2*tan(1/2*x)^3 - 3*A*b^3*c^2*tan(1/2*x)^3 + 2*A*a*c^4*tan(1/2*x)^3 
 - 2*A*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 + 12*A*a^3*b*c*tan(1/ 
2*x)^2 - 13*A*a^2*b^2*c*tan(1/2*x)^2 + 6*A*a*b^3*c*tan(1/2*x)^2 - A*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*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*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*A*c^5*tan(1/2*x)^2 + 4*A*a^4*b*tan(1 
/2*x) - 5*A*a^3*b^2*tan(1/2*x) - 3*A*a^2*b^3*tan(1/2*x) + 5*A*a*b^4*tan(1/ 
2*x) - A*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) - 11*A*a^3*c^2*tan(1/2*x 
) + 3*A*a^2*b*c^2*tan(1/2*x) + 7*A*a*b^2*c^2*tan(1/2*x) + A*b^3*c^2*tan...
 

Mupad [B] (verification not implemented)

Time = 19.72 (sec) , antiderivative size = 912, normalized size of antiderivative = 4.56 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx =\text {Too large to display} \] Input:

int((A + 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 + 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)*(A*b^3*c^2 - 3*A* 
a^2*b^3 - 5*A*a^3*b^2 - 11*A*a^3*c^2 - A*b^5 + 4*C*a^2*c^3 + 5*A*a*b^4 + 4 
*A*a^4*b + 2*A*a*c^4 + 2*A*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 - 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)^2*(2*A* 
c^5 + 2*C*a^5 - 2*C*b^5 - 7*A*a^2*c^3 + A*b^2*c^3 + 4*C*a^2*b^3 - 4*C*a^3* 
b^2 + 5*C*a^3*c^2 - 4*C*b^3*c^2 - 4*A*a^4*c - A*b^4*c + 2*C*a*b^4 - 2*C*a^ 
4*b + 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 - 1 
3*A*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)^3*(A*b^4 + 2*A*c^4 - 
 7*A*a^2*b^2 - 5*A*a^2*c^2 + 3*A*b^2*c^2 + 2*A*a*b^3 + 4*A*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)^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)^(5/2)))*(2*A*a^2 + A*b^2 + A*c^2 - 3*C*a*c))/(b^2 - a^...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 4229, normalized size of antiderivative = 21.14 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^3} \, dx =\text {Too large to display} \] Input:

int((A+C*sin(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)*sin(x)*a**4*b*c**2 - 16*sqrt(a**2 - b**2 - c**2)*at 
an((tan(x/2)*a - tan(x/2)*b + c)/sqrt(a**2 - b**2 - c**2))*cos(x)*sin(x)*a 
**3*b**2*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)*sin(x)*a**2*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)*sin(x)*a**2*b*c**4 - 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)*sin(x)*a*b**4*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)*sin(x)*a*b**2*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)*a**5*b*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)*a**4*b**2*c + 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)*a**3*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)*a**3*b*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)*a**2*b**4*c + 16*sqr 
t(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**2*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))*sin(x)**2*a**4*b**2*c + ...