\(\int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx\) [477]

Optimal result

Integrand size = 22, antiderivative size = 129 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx=-\frac {A \text {arctanh}\left (\frac {c \cos (x)-b \sin (x)}{\sqrt {b^2+c^2}}\right )}{2 \left (b^2+c^2\right )^{3/2}}-\frac {B c-b C+A c \cos (x)-A b \sin (x)}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}-\frac {c (b B+c C) \cos (x)-b (b B+c C) \sin (x)}{\left (b^2+c^2\right )^2 (b \cos (x)+c \sin (x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx=\frac {A \text {arctanh}\left (\frac {-c+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}}+\frac {b^2 C+c^2 C-A b c \cos (x)-c (b B+c C) \cos (2 x)+A b^2 \sin (x)+b^2 B \sin (2 x)+b c C \sin (2 x)}{2 b \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {3042, 3635, 3042, 3632, 3042, 3553, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3635

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3632

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3553

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {A \text {arctanh}\left (\frac {c \cos (x)-b \sin (x)}{\sqrt {b^2+c^2}}\right )}{\sqrt {b^2+c^2}}-\frac {2 (c \cos (x) (b B+c C)-b \sin (x) (b B+c C))}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}}{2 \left (b^2+c^2\right )}-\frac {-A b \sin (x)+A c \cos (x)-b C+B c}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}\)

Input:

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

Output:

-1/2*(B*c - b*C + A*c*Cos[x] - A*b*Sin[x])/((b^2 + c^2)*(b*Cos[x] + c*Sin[ 
x])^2) + (-((A*ArcTanh[(c*Cos[x] - b*Sin[x])/Sqrt[b^2 + c^2]])/Sqrt[b^2 + 
c^2]) - (2*(c*(b*B + c*C)*Cos[x] - b*(b*B + c*C)*Sin[x]))/((b^2 + c^2)*(b* 
Cos[x] + c*Sin[x])))/(2*(b^2 + c^2))
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3553

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x 
_Symbol] :> Simp[-d^(-1)   Subst[Int[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + 
d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^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 [A] (verified)

Time = 0.38 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.69

method	result	size
default	\(-\frac {2 \left (-\frac {\left (A \,b^{2}+2 A \,c^{2}-2 B \,b^{2}-2 B \,c^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}}{2 b \left (b^{2}+c^{2}\right )}-\frac {\left (A \,b^{2} c -2 c^{3} A +2 B \,b^{2} c +2 B \,c^{3}+2 C \,b^{3}+2 C b \,c^{2}\right ) \tan \left (\frac {x}{2}\right )^{2}}{2 \left (b^{2}+c^{2}\right ) b^{2}}-\frac {\left (A \,b^{2}-2 A \,c^{2}+2 B \,b^{2}+2 B \,c^{2}\right ) \tan \left (\frac {x}{2}\right )}{2 \left (b^{2}+c^{2}\right ) b}+\frac {A c}{2 b^{2}+2 c^{2}}\right )}{\left (b \tan \left (\frac {x}{2}\right )^{2}-2 c \tan \left (\frac {x}{2}\right )-b \right )^{2}}+\frac {A \,\operatorname {arctanh}\left (\frac {2 b \tan \left (\frac {x}{2}\right )-2 c}{2 \sqrt {b^{2}+c^{2}}}\right )}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(218\)
risch	\(-\frac {i \left (2 C \,b^{2} {\mathrm e}^{2 i x}-2 b B c -2 C \,c^{2}-2 A b c \,{\mathrm e}^{3 i x}-i A \,b^{2} {\mathrm e}^{3 i x}+i A \,c^{2} {\mathrm e}^{3 i x}+2 i B \,c^{2} {\mathrm e}^{2 i x}+2 C \,c^{2} {\mathrm e}^{2 i x}+i A \,b^{2} {\mathrm e}^{i x}+i A \,c^{2} {\mathrm e}^{i x}+2 i C b c +2 i B \,b^{2}+2 i B \,b^{2} {\mathrm e}^{2 i x}\right )}{\left (c \,{\mathrm e}^{2 i x}+i b \,{\mathrm e}^{2 i x}-c +i b \right )^{2} \left (-i b +c \right ) \left (i b +c \right )^{2}}+\frac {A \ln \left ({\mathrm e}^{i x}+\frac {i b^{3}+i b \,c^{2}-b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\right )}{2 \left (b^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {A \ln \left ({\mathrm e}^{i x}-\frac {i b^{3}+i b \,c^{2}-b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\right )}{2 \left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(285\)

Input:

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

Output:

-2*(-1/2*(A*b^2+2*A*c^2-2*B*b^2-2*B*c^2)/b/(b^2+c^2)*tan(1/2*x)^3-1/2*(A*b 
^2*c-2*A*c^3+2*B*b^2*c+2*B*c^3+2*C*b^3+2*C*b*c^2)/(b^2+c^2)/b^2*tan(1/2*x) 
^2-1/2*(A*b^2-2*A*c^2+2*B*b^2+2*B*c^2)/(b^2+c^2)/b*tan(1/2*x)+1/2*A*c/(b^2 
+c^2))/(b*tan(1/2*x)^2-2*c*tan(1/2*x)-b)^2+A/(b^2+c^2)^(3/2)*arctanh(1/2*( 
2*b*tan(1/2*x)-2*c)/(b^2+c^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (121) = 242\).

Time = 0.10 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.41 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx=\frac {2 \, C b^{3} + 2 \, B b^{2} c + 6 \, C b c^{2} - 2 \, B c^{3} - 8 \, {\left (B b^{2} c + C b c^{2}\right )} \cos \left (x\right )^{2} + {\left (2 \, A b c \cos \left (x\right ) \sin \left (x\right ) + A c^{2} + {\left (A b^{2} - A c^{2}\right )} \cos \left (x\right )^{2}\right )} \sqrt {b^{2} + c^{2}} \log \left (-\frac {2 \, b c \cos \left (x\right ) \sin \left (x\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} - 2 \, b^{2} - c^{2} + 2 \, \sqrt {b^{2} + c^{2}} {\left (c \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, b c \cos \left (x\right ) \sin \left (x\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} + c^{2}}\right ) - 2 \, {\left (A b^{2} c + A c^{3}\right )} \cos \left (x\right ) + 2 \, {\left (A b^{3} + A b c^{2} + 2 \, {\left (B b^{3} + C b^{2} c - B b c^{2} - C c^{3}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{4 \, {\left (b^{4} c^{2} + 2 \, b^{2} c^{4} + c^{6} + {\left (b^{6} + b^{4} c^{2} - b^{2} c^{4} - c^{6}\right )} \cos \left (x\right )^{2} + 2 \, {\left (b^{5} c + 2 \, b^{3} c^{3} + b c^{5}\right )} \cos \left (x\right ) \sin \left (x\right )\right )}} \] Input:

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

Output:

1/4*(2*C*b^3 + 2*B*b^2*c + 6*C*b*c^2 - 2*B*c^3 - 8*(B*b^2*c + C*b*c^2)*cos 
(x)^2 + (2*A*b*c*cos(x)*sin(x) + A*c^2 + (A*b^2 - A*c^2)*cos(x)^2)*sqrt(b^ 
2 + c^2)*log(-(2*b*c*cos(x)*sin(x) + (b^2 - c^2)*cos(x)^2 - 2*b^2 - c^2 + 
2*sqrt(b^2 + c^2)*(c*cos(x) - b*sin(x)))/(2*b*c*cos(x)*sin(x) + (b^2 - c^2 
)*cos(x)^2 + c^2)) - 2*(A*b^2*c + A*c^3)*cos(x) + 2*(A*b^3 + A*b*c^2 + 2*( 
B*b^3 + C*b^2*c - B*b*c^2 - C*c^3)*cos(x))*sin(x))/(b^4*c^2 + 2*b^2*c^4 + 
c^6 + (b^6 + b^4*c^2 - b^2*c^4 - c^6)*cos(x)^2 + 2*(b^5*c + 2*b^3*c^3 + b* 
c^5)*cos(x)*sin(x))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 451 vs. \(2 (121) = 242\).

Time = 0.14 (sec) , antiderivative size = 451, normalized size of antiderivative = 3.50 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*A*(2*(b^2*c - (b^3 - 2*b*c^2)*sin(x)/(cos(x) + 1) - (b^2*c - 2*c^3)*s 
in(x)^2/(cos(x) + 1)^2 - (b^3 + 2*b*c^2)*sin(x)^3/(cos(x) + 1)^3)/(b^6 + b 
^4*c^2 + 4*(b^5*c + b^3*c^3)*sin(x)/(cos(x) + 1) - 2*(b^6 - b^4*c^2 - 2*b^ 
2*c^4)*sin(x)^2/(cos(x) + 1)^2 - 4*(b^5*c + b^3*c^3)*sin(x)^3/(cos(x) + 1) 
^3 + (b^6 + b^4*c^2)*sin(x)^4/(cos(x) + 1)^4) + log((c - b*sin(x)/(cos(x) 
+ 1) + sqrt(b^2 + c^2))/(c - b*sin(x)/(cos(x) + 1) - sqrt(b^2 + c^2)))/(b^ 
2 + c^2)^(3/2)) + 2*B*(b*sin(x)/(cos(x) + 1) + c*sin(x)^2/(cos(x) + 1)^2 - 
 b*sin(x)^3/(cos(x) + 1)^3)/(b^4 + 4*b^3*c*sin(x)/(cos(x) + 1) - 4*b^3*c*s 
in(x)^3/(cos(x) + 1)^3 + b^4*sin(x)^4/(cos(x) + 1)^4 - 2*(b^4 - 2*b^2*c^2) 
*sin(x)^2/(cos(x) + 1)^2) + 2*C*sin(x)^2/((b^3 + 4*b^2*c*sin(x)/(cos(x) + 
1) - 4*b^2*c*sin(x)^3/(cos(x) + 1)^3 + b^3*sin(x)^4/(cos(x) + 1)^4 - 2*(b^ 
3 - 2*b*c^2)*sin(x)^2/(cos(x) + 1)^2)*(cos(x) + 1)^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (121) = 242\).

Time = 0.39 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.09 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx=\frac {A \log \left (\frac {{\left | -2 \, b \tan \left (\frac {1}{2} \, x\right ) + 2 \, c - 2 \, \sqrt {b^{2} + c^{2}} \right |}}{{\left | -2 \, b \tan \left (\frac {1}{2} \, x\right ) + 2 \, c + 2 \, \sqrt {b^{2} + c^{2}} \right |}}\right )}{2 \, {\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}} + \frac {A b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, B b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, A b c^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, B b c^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, C b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + A b^{2} c \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, B b^{2} c \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, C b c^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, A c^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, B c^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + A b^{3} \tan \left (\frac {1}{2} \, x\right ) + 2 \, B b^{3} \tan \left (\frac {1}{2} \, x\right ) - 2 \, A b c^{2} \tan \left (\frac {1}{2} \, x\right ) + 2 \, B b c^{2} \tan \left (\frac {1}{2} \, x\right ) - A b^{2} c}{{\left (b^{4} + b^{2} c^{2}\right )} {\left (b \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, c \tan \left (\frac {1}{2} \, x\right ) - b\right )}^{2}} \] Input:

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

Output:

1/2*A*log(abs(-2*b*tan(1/2*x) + 2*c - 2*sqrt(b^2 + c^2))/abs(-2*b*tan(1/2* 
x) + 2*c + 2*sqrt(b^2 + c^2)))/(b^2 + c^2)^(3/2) + (A*b^3*tan(1/2*x)^3 - 2 
*B*b^3*tan(1/2*x)^3 + 2*A*b*c^2*tan(1/2*x)^3 - 2*B*b*c^2*tan(1/2*x)^3 + 2* 
C*b^3*tan(1/2*x)^2 + A*b^2*c*tan(1/2*x)^2 + 2*B*b^2*c*tan(1/2*x)^2 + 2*C*b 
*c^2*tan(1/2*x)^2 - 2*A*c^3*tan(1/2*x)^2 + 2*B*c^3*tan(1/2*x)^2 + A*b^3*ta 
n(1/2*x) + 2*B*b^3*tan(1/2*x) - 2*A*b*c^2*tan(1/2*x) + 2*B*b*c^2*tan(1/2*x 
) - A*b^2*c)/((b^4 + b^2*c^2)*(b*tan(1/2*x)^2 - 2*c*tan(1/2*x) - b)^2)
 

Mupad [B] (verification not implemented)

Time = 16.72 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.05 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx=\frac {\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A\,b^2-2\,A\,c^2+2\,B\,b^2+2\,B\,c^2\right )}{b\,\left (b^2+c^2\right )}-\frac {A\,c}{b^2+c^2}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,B\,c^3-2\,A\,c^3+2\,C\,b^3+A\,b^2\,c+2\,B\,b^2\,c+2\,C\,b\,c^2\right )}{b^2\,\left (b^2+c^2\right )}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (A\,b^2+2\,A\,c^2-2\,B\,b^2-2\,B\,c^2\right )}{b\,\left (b^2+c^2\right )}}{b^2-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,b^2-4\,c^2\right )+b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,b\,c\,\mathrm {tan}\left (\frac {x}{2}\right )-4\,b\,c\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}+\frac {A\,\mathrm {atan}\left (\frac {b^2\,c\,1{}\mathrm {i}+c^3\,1{}\mathrm {i}-b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (b^2+c^2\right )\,1{}\mathrm {i}}{{\left (b^2+c^2\right )}^{3/2}}\right )\,1{}\mathrm {i}}{{\left (b^2+c^2\right )}^{3/2}} \] Input:

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

Output:

((tan(x/2)*(A*b^2 - 2*A*c^2 + 2*B*b^2 + 2*B*c^2))/(b*(b^2 + c^2)) - (A*c)/ 
(b^2 + c^2) + (tan(x/2)^2*(2*B*c^3 - 2*A*c^3 + 2*C*b^3 + A*b^2*c + 2*B*b^2 
*c + 2*C*b*c^2))/(b^2*(b^2 + c^2)) + (tan(x/2)^3*(A*b^2 + 2*A*c^2 - 2*B*b^ 
2 - 2*B*c^2))/(b*(b^2 + c^2)))/(b^2 - tan(x/2)^2*(2*b^2 - 4*c^2) + b^2*tan 
(x/2)^4 + 4*b*c*tan(x/2) - 4*b*c*tan(x/2)^3) + (A*atan((b^2*c*1i + c^3*1i 
- b*tan(x/2)*(b^2 + c^2)*1i)/(b^2 + c^2)^(3/2))*1i)/(b^2 + c^2)^(3/2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 1368, normalized size of antiderivative = 10.60 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 8*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/sqrt(b**2 + c**2))*cos(x 
)**3*sin(x)*a*b**4*c**2*i + 4*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/ 
sqrt(b**2 + c**2))*cos(x)**2*sin(x)**2*a*b**5*c*i - 20*sqrt(b**2 + c**2)*a 
tan((tan(x/2)*b*i - c*i)/sqrt(b**2 + c**2))*cos(x)**2*sin(x)**2*a*b**3*c** 
3*i - 4*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/sqrt(b**2 + c**2))*cos 
(x)**2*a*b**5*c*i + 8*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/sqrt(b** 
2 + c**2))*cos(x)*sin(x)**3*a*b**4*c**2*i - 16*sqrt(b**2 + c**2)*atan((tan 
(x/2)*b*i - c*i)/sqrt(b**2 + c**2))*cos(x)*sin(x)**3*a*b**2*c**4*i - 8*sqr 
t(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/sqrt(b**2 + c**2))*cos(x)*sin(x)* 
a*b**4*c**2*i + 4*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/sqrt(b**2 + 
c**2))*sin(x)**4*a*b**3*c**3*i - 4*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - 
c*i)/sqrt(b**2 + c**2))*sin(x)**4*a*b*c**5*i - 4*sqrt(b**2 + c**2)*atan((t 
an(x/2)*b*i - c*i)/sqrt(b**2 + c**2))*sin(x)**2*a*b**3*c**3*i + 2*cos(x)** 
3*sin(x)*a*b**6*c + 2*cos(x)**3*sin(x)*a*b**4*c**3 - 2*cos(x)**3*a*b**5*c* 
*2 - 2*cos(x)**3*a*b**3*c**4 - cos(x)**2*sin(x)**2*a*b**7 + 4*cos(x)**2*si 
n(x)**2*a*b**5*c**2 + 5*cos(x)**2*sin(x)**2*a*b**3*c**4 + 2*cos(x)**2*sin( 
x)**2*b**8 + 6*cos(x)**2*sin(x)**2*b**6*c**2 + 6*cos(x)**2*sin(x)**2*b**4* 
c**4 + 2*cos(x)**2*sin(x)**2*b**2*c**6 + 2*cos(x)**2*sin(x)*a*b**6*c - 2*c 
os(x)**2*sin(x)*a*b**4*c**3 - 4*cos(x)**2*sin(x)*a*b**2*c**5 + cos(x)**2*a 
*b**7 + cos(x)**2*a*b**5*c**2 - 2*cos(x)*sin(x)**3*a*b**6*c + 2*cos(x)*...