\(\int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx\) [484]

Optimal result

Integrand size = 19, antiderivative size = 114 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {2 (a A-c C) \arctan \left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}-\frac {b C-(A c-a C) \cos (x)+A b \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))} \] Output:

2*(A*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) 
^(3/2)-(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 
))
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.08 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {2 (a A-c C) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{3/2}}+\frac {a A c-a^2 C+b^2 C+\left (A \left (b^2+c^2\right )-a c C\right ) \sin (x)}{b \left (-a^2+b^2+c^2\right ) (a+b \cos (x)+c \sin (x))} \] Input:

Integrate[(A + C*Sin[x])/(a + b*Cos[x] + c*Sin[x])^2,x]
 

Output:

(2*(a*A - c*C)*ArcTanh[(c + (a - b)*Tan[x/2])/Sqrt[-a^2 + b^2 + c^2]])/(-a 
^2 + b^2 + c^2)^(3/2) + (a*A*c - a^2*C + b^2*C + (A*(b^2 + c^2) - a*c*C)*S 
in[x])/(b*(-a^2 + b^2 + c^2)*(a + b*Cos[x] + c*Sin[x]))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3633, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3633

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3603

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

Int[(A + C*Sin[x])/(a + b*Cos[x] + c*Sin[x])^2,x]
 

Output:

(2*(a*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) - (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]))
 

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 3633

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_) 
]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[-(b*C + (a*C 
- c*A)*Cos[d + e*x] + 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 - c*C)/(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, C} 
, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - c*C, 0]
 

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.82

method	result	size
default	\(\frac {-\frac {2 \left (A a b -A \,b^{2}-A \,c^{2}+a c C -C b c \right ) \tan \left (\frac {x}{2}\right )}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+b \,c^{2}}+\frac {2 \left (A a c -C \,a^{2}+b^{2} C \right )}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+b \,c^{2}}}{\tan \left (\frac {x}{2}\right )^{2} a -b \tan \left (\frac {x}{2}\right )^{2}+2 c \tan \left (\frac {x}{2}\right )+a +b}+\frac {2 \left (A a -C c \right ) \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{\left (a^{2}-b^{2}-c^{2}\right )^{\frac {3}{2}}}\)	\(207\)
risch	\(\frac {2 i \left (-i A \,b^{2}-i A \,c^{2}+i a c C -i A a b \,{\mathrm e}^{i x}+i C b c \,{\mathrm e}^{i x}-A a c \,{\mathrm e}^{i x}+C \,a^{2} {\mathrm e}^{i x}-C \,b^{2} {\mathrm e}^{i x}\right )}{\left (a^{2}-b^{2}-c^{2}\right ) \left (i b +c \right ) \left (-i c \,{\mathrm e}^{2 i x}+b \,{\mathrm e}^{2 i x}+i c +2 a \,{\mathrm e}^{i x}+b \right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}+c^{2}}\, a c +i a^{2} b -i b^{3}-i b \,c^{2}+\sqrt {-a^{2}+b^{2}+c^{2}}\, a b -a^{2} c +b^{2} c +c^{3}}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (b^{2}+c^{2}\right )}\right ) A a}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}+c^{2}}\, a c +i a^{2} b -i b^{3}-i b \,c^{2}+\sqrt {-a^{2}+b^{2}+c^{2}}\, a b -a^{2} c +b^{2} c +c^{3}}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (b^{2}+c^{2}\right )}\right ) C c}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}+c^{2}}\, a c -i a^{2} b +i b^{3}+i b \,c^{2}+\sqrt {-a^{2}+b^{2}+c^{2}}\, a b +a^{2} c -b^{2} c -c^{3}}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (b^{2}+c^{2}\right )}\right ) A a}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}+c^{2}}\, a c -i a^{2} b +i b^{3}+i b \,c^{2}+\sqrt {-a^{2}+b^{2}+c^{2}}\, a b +a^{2} c -b^{2} c -c^{3}}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (b^{2}+c^{2}\right )}\right ) C c}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right )}\)	\(681\)

Input:

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

Output:

2*(-(A*a*b-A*b^2-A*c^2+C*a*c-C*b*c)/(a^3-a^2*b-a*b^2-a*c^2+b^3+b*c^2)*tan( 
1/2*x)+(A*a*c-C*a^2+C*b^2)/(a^3-a^2*b-a*b^2-a*c^2+b^3+b*c^2))/(tan(1/2*x)^ 
2*a-b*tan(1/2*x)^2+2*c*tan(1/2*x)+a+b)+2*(A*a-C*c)/(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. 569 vs. \(2 (107) = 214\).

Time = 0.14 (sec) , antiderivative size = 1301, normalized size of antiderivative = 11.41 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\text {Too large to display} \] Input:

integrate((A+C*sin(x))/(a+b*cos(x)+c*sin(x))^2,x, algorithm="fricas")
 

Output:

[1/2*(2*C*a^4*b - 4*C*a^2*b^3 + 2*C*b^5 + 2*C*b*c^4 - 4*(C*a^2*b - C*b^3)* 
c^2 - (A*a^2*b^2 - C*a*b^2*c + A*a^2*c^2 - C*a*c^3 + (A*a*b^3 - C*b^3*c + 
A*a*b*c^2 - C*b*c^3)*cos(x) + (A*a*b^2*c - C*b^2*c^2 + A*a*c^3 - C*c^4)*si 
n(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)*co 
s(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*(C*a*c^4 - A*c 
^5 + (A*a^2 - 2*A*b^2)*c^3 - (C*a^3 - C*a*b^2)*c^2 + (A*a^2*b^2 - A*b^4)*c 
)*cos(x) - 2*(A*a^2*b^3 - A*b^5 + C*a*b*c^3 - A*b*c^4 + (A*a^2*b - 2*A*b^3 
)*c^2 - (C*a^3*b - C*a*b^3)*c)*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)), (C*a^4*b - 2*C*a^2*b^3 + C*b^5 
 + C*b*c^4 - 2*(C*a^2*b - C*b^3)*c^2 + (A*a^2*b^2 - C*a*b^2*c + A*a^2*c^2 
- C*a*c^3 + (A*a*b^3 - C*b^3*c + A*a*b*c^2 - C*b*c^3)*cos(x) + (A*a*b^2*c 
- C*b^2*c^2 + A*a*c^3 - C*c^4)*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...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+C*sin(x))/(a+b*cos(x)+c*sin(x))^2,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 [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.81 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right ) + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right )\right )} {\left (A a - C c\right )}}{{\left (a^{2} - b^{2} - c^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (A a b \tan \left (\frac {1}{2} \, x\right ) - A b^{2} \tan \left (\frac {1}{2} \, x\right ) + C a c \tan \left (\frac {1}{2} \, x\right ) - C b c \tan \left (\frac {1}{2} \, x\right ) - A c^{2} \tan \left (\frac {1}{2} \, x\right ) + C a^{2} - C b^{2} - A a c\right )}}{{\left (a^{3} - a^{2} b - a b^{2} + b^{3} - a c^{2} + b c^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}} \] Input:

integrate((A+C*sin(x))/(a+b*cos(x)+c*sin(x))^2,x, algorithm="giac")
 

Output:

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*t 
an(1/2*x) + c)/sqrt(a^2 - b^2 - c^2)))*(A*a - C*c)/(a^2 - b^2 - c^2)^(3/2) 
 - 2*(A*a*b*tan(1/2*x) - A*b^2*tan(1/2*x) + C*a*c*tan(1/2*x) - C*b*c*tan(1 
/2*x) - A*c^2*tan(1/2*x) + C*a^2 - C*b^2 - A*a*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 = 16.21 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.79 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a-2\,b\right )+\frac {2\,\left (-a^2\,c+b^2\,c+c^3\right )}{-a^2+b^2+c^2}}{2\,\sqrt {-a^2+b^2+c^2}}\right )\,\left (A\,a-C\,c\right )}{{\left (-a^2+b^2+c^2\right )}^{3/2}}-\frac {\frac {2\,\left (-C\,a^2+A\,c\,a+C\,b^2\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A\,b^2+C\,b\,c-A\,a\,b+A\,c^2-C\,a\,c\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}}{\left (a-b\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {x}{2}\right )+a+b} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 550, normalized size of antiderivative = 4.82 \[ \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx=\frac {2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) \cos \left (x \right ) a^{2} b c -2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) \cos \left (x \right ) b \,c^{3}+2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) \sin \left (x \right ) a^{2} c^{2}-2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) \sin \left (x \right ) c^{4}+2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) a^{3} c -2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) a \,c^{3}+\cos \left (x \right ) a^{3} b^{2}-\cos \left (x \right ) a \,b^{4}-\cos \left (x \right ) a \,b^{2} c^{2}+a^{4} b -a^{2} b^{3}-2 a^{2} b \,c^{2}+b^{3} c^{2}+b \,c^{4}}{c \left (\cos \left (x \right ) a^{4} b -2 \cos \left (x \right ) a^{2} b^{3}-2 \cos \left (x \right ) a^{2} b \,c^{2}+\cos \left (x \right ) b^{5}+2 \cos \left (x \right ) b^{3} c^{2}+\cos \left (x \right ) b \,c^{4}+\sin \left (x \right ) a^{4} c -2 \sin \left (x \right ) a^{2} b^{2} c -2 \sin \left (x \right ) a^{2} c^{3}+\sin \left (x \right ) b^{4} c +2 \sin \left (x \right ) b^{2} c^{3}+\sin \left (x \right ) c^{5}+a^{5}-2 a^{3} b^{2}-2 a^{3} c^{2}+a \,b^{4}+2 a \,b^{2} c^{2}+a \,c^{4}\right )} \] Input:

int((A+C*sin(x))/(a+b*cos(x)+c*sin(x))^2,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**2*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))*cos(x)*b*c**3 + 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**2*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))*sin(x)*c**4 + 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**3*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))*a*c**3 + cos(x)*a**3*b**2 - cos(x)*a*b**4 - cos(x)*a*b**2*c** 
2 + a**4*b - a**2*b**3 - 2*a**2*b*c**2 + b**3*c**2 + b*c**4)/(c*(cos(x)*a* 
*4*b - 2*cos(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))