\(\int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx\) [476]

Optimal result

Integrand size = 22, antiderivative size = 85 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx=-\frac {(b B+c C) \text {arctanh}\left (\frac {c \cos (x)-b \sin (x)}{\sqrt {b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}}-\frac {B c-b C+A c \cos (x)-A b \sin (x)}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx=\frac {2 (b B+c C) \text {arctanh}\left (\frac {-c+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}}+\frac {b (-B c+b C)+A \left (b^2+c^2\right ) \sin (x)}{b \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3632

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3553

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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]
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {-\frac {2 \left (A \,b^{2}+A \,c^{2}-B \,c^{2}+C b c \right ) \tan \left (\frac {x}{2}\right )}{b \left (b^{2}+c^{2}\right )}+\frac {2 \left (B c -b C \right )}{b^{2}+c^{2}}}{b \tan \left (\frac {x}{2}\right )^{2}-2 c \tan \left (\frac {x}{2}\right )-b}+\frac {2 \left (B b +C c \right ) \operatorname {arctanh}\left (\frac {2 b \tan \left (\frac {x}{2}\right )-2 c}{2 \sqrt {b^{2}+c^{2}}}\right )}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(124\)
risch	\(-\frac {2 i \left (A c -i A b +B c \,{\mathrm e}^{i x}-C b \,{\mathrm e}^{i x}\right )}{\left (-i b +c \right ) \left (i b +c \right ) \left (c \,{\mathrm e}^{2 i x}+i b \,{\mathrm e}^{2 i x}-c +i b \right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i b^{3}+i b \,c^{2}-b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\right ) B b}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i b^{3}+i b \,c^{2}-b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\right ) C c}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {\ln \left ({\mathrm e}^{i x}-\frac {i b^{3}+i b \,c^{2}-b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\right ) B b}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {\ln \left ({\mathrm e}^{i x}-\frac {i b^{3}+i b \,c^{2}-b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\right ) C c}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(293\)

Input:

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

Output:

2*(-(A*b^2+A*c^2-B*c^2+C*b*c)/b/(b^2+c^2)*tan(1/2*x)+(B*c-C*b)/(b^2+c^2))/ 
(b*tan(1/2*x)^2-2*c*tan(1/2*x)-b)+2*(B*b+C*c)/(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. 226 vs. \(2 (81) = 162\).

Time = 0.09 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.66 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx=\frac {2 \, C b^{3} - 2 \, B b^{2} c + 2 \, C b c^{2} - 2 \, B c^{3} + \sqrt {b^{2} + c^{2}} {\left ({\left (B b^{2} + C b c\right )} \cos \left (x\right ) + {\left (B b c + C c^{2}\right )} \sin \left (x\right )\right )} \log \left (-\frac {2 \, b c \cos \left (x\right ) \sin \left (x\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} - 2 \, b^{2} - c^{2} + 2 \, \sqrt {b^{2} + c^{2}} {\left (c \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, b c \cos \left (x\right ) \sin \left (x\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} + c^{2}}\right ) - 2 \, {\left (A b^{2} c + A c^{3}\right )} \cos \left (x\right ) + 2 \, {\left (A b^{3} + A b c^{2}\right )} \sin \left (x\right )}{2 \, {\left ({\left (b^{5} + 2 \, b^{3} c^{2} + b c^{4}\right )} \cos \left (x\right ) + {\left (b^{4} c + 2 \, b^{2} c^{3} + c^{5}\right )} \sin \left (x\right )\right )}} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (81) = 162\).

Time = 0.13 (sec) , antiderivative size = 286, normalized size of antiderivative = 3.36 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx=-B {\left (\frac {b \log \left (\frac {c - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {b^{2} + c^{2}}}{c - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {b^{2} + c^{2}}}\right )}{{\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b c + \frac {c^{2} \sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}}{b^{4} + b^{2} c^{2} + \frac {2 \, {\left (b^{3} c + b c^{3}\right )} \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {{\left (b^{4} + b^{2} c^{2}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}}\right )} - C {\left (\frac {c \log \left (\frac {c - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {b^{2} + c^{2}}}{c - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {b^{2} + c^{2}}}\right )}{{\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b + \frac {c \sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}}{b^{3} + b c^{2} + \frac {2 \, {\left (b^{2} c + c^{3}\right )} \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {{\left (b^{3} + b c^{2}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}}\right )} - \frac {A}{c^{2} \tan \left (x\right ) + b c} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.76 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx=-\frac {{\left (B b + C c\right )} \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, c - 2 \, \sqrt {b^{2} + c^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, c + 2 \, \sqrt {b^{2} + c^{2}} \right |}}\right )}{{\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (A b^{2} \tan \left (\frac {1}{2} \, x\right ) + C b c \tan \left (\frac {1}{2} \, x\right ) + A c^{2} \tan \left (\frac {1}{2} \, x\right ) - B c^{2} \tan \left (\frac {1}{2} \, x\right ) + C b^{2} - B b c\right )}}{{\left (b^{3} + b c^{2}\right )} {\left (b \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, c \tan \left (\frac {1}{2} \, x\right ) - b\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 15.97 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.66 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx=-\frac {\frac {2\,\left (B\,c-C\,b\right )}{b^2+c^2}-\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A\,b^2+A\,c^2-B\,c^2+C\,b\,c\right )}{b\,\left (b^2+c^2\right )}}{-b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {x}{2}\right )+b}+\frac {\mathrm {atan}\left (\frac {b^2\,c\,1{}\mathrm {i}+c^3\,1{}\mathrm {i}-b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (b^2+c^2\right )\,1{}\mathrm {i}}{{\left (b^2+c^2\right )}^{3/2}}\right )\,\left (B\,b+C\,c\right )\,2{}\mathrm {i}}{{\left (b^2+c^2\right )}^{3/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.52 \[ \int \frac {A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx=\frac {-2 \sqrt {b^{2}+c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) b i -c i}{\sqrt {b^{2}+c^{2}}}\right ) \cos \left (x \right ) b c i -2 \sqrt {b^{2}+c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) b i -c i}{\sqrt {b^{2}+c^{2}}}\right ) \sin \left (x \right ) c^{2} i -\cos \left (x \right ) a \,b^{2}-\cos \left (x \right ) a \,c^{2}}{c \left (\cos \left (x \right ) b^{3}+\cos \left (x \right ) b \,c^{2}+\sin \left (x \right ) b^{2} c +\sin \left (x \right ) c^{3}\right )} \] Input:

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

Output:

( - 2*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/sqrt(b**2 + c**2))*cos(x 
)*b*c*i - 2*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/sqrt(b**2 + c**2)) 
*sin(x)*c**2*i - cos(x)*a*b**2 - cos(x)*a*c**2)/(c*(cos(x)*b**3 + cos(x)*b 
*c**2 + sin(x)*b**2*c + sin(x)*c**3))