\(\int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx\) [18]

Optimal result

Integrand size = 31, antiderivative size = 87 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx=\frac {B x}{b}+\frac {2 (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} e}-\frac {C \log (a+b \cos (d+e x))}{b e} \] Output:

B*x/b+2*(A*b-B*a)*arctan((a-b)^(1/2)*tan(1/2*e*x+1/2*d)/(a+b)^(1/2))/(a-b) 
^(1/2)/b/(a+b)^(1/2)/e-C*ln(a+b*cos(e*x+d))/b/e
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx=\frac {B (d+e x)+\frac {2 (-A b+a B) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-C \log (a+b \cos (d+e x))}{b e} \] Input:

Integrate[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + b*Cos[d + e*x]),x]
 

Output:

(B*(d + e*x) + (2*(-(A*b) + a*B)*ArcTanh[((a - b)*Tan[(d + e*x)/2])/Sqrt[- 
a^2 + b^2]])/Sqrt[-a^2 + b^2] - C*Log[a + b*Cos[d + e*x]])/(b*e)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4877, 3042, 3147, 16, 3214, 3042, 3138, 218}

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.

\(\displaystyle \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)}dx\)

\(\Big \downarrow \) 4877

\(\displaystyle \int \frac {A+B \cos (d+e x)}{a+b \cos (d+e x)}dx+C \int \frac {\sin (d+e x)}{a+b \cos (d+e x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (d+e x+\frac {\pi }{2}\right )}{a+b \sin \left (d+e x+\frac {\pi }{2}\right )}dx+C \int \frac {\cos \left (d+e x-\frac {\pi }{2}\right )}{a-b \sin \left (d+e x-\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3147

\(\displaystyle \int \frac {A+B \sin \left (d+e x+\frac {\pi }{2}\right )}{a+b \sin \left (d+e x+\frac {\pi }{2}\right )}dx-\frac {C \int \frac {1}{a+b \cos (d+e x)}d(b \cos (d+e x))}{b e}\)

\(\Big \downarrow \) 16

\(\displaystyle \int \frac {A+B \sin \left (d+e x+\frac {\pi }{2}\right )}{a+b \sin \left (d+e x+\frac {\pi }{2}\right )}dx-\frac {C \log (a+b \cos (d+e x))}{b e}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {(A b-a B) \int \frac {1}{a+b \cos (d+e x)}dx}{b}-\frac {C \log (a+b \cos (d+e x))}{b e}+\frac {B x}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A b-a B) \int \frac {1}{a+b \sin \left (d+e x+\frac {\pi }{2}\right )}dx}{b}-\frac {C \log (a+b \cos (d+e x))}{b e}+\frac {B x}{b}\)

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{b e \sqrt {a-b} \sqrt {a+b}}-\frac {C \log (a+b \cos (d+e x))}{b e}+\frac {B x}{b}\)

Input:

Int[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + b*Cos[d + e*x]),x]
 

Output:

(B*x)/b + (2*(A*b - a*B)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b] 
])/(Sqrt[a - b]*b*Sqrt[a + b]*e) - (C*Log[a + b*Cos[d + e*x]])/(b*e)
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3147

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p 
 - 1)/2] && NeQ[a^2 - b^2, 0]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 4877

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] : 
> With[{e = FreeFactors[Cos[c*(a + b*x)], x]}, Int[ActivateTrig[u*v], x] + 
Simp[d   Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[Cos[ 
c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && Intege 
rQ[(n - 1)/2] && NonsumQ[u] && (EqQ[F, Sin] || EqQ[F, sin])
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.63

method	result	size
derivativedivides	\(\frac {\frac {C \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+1\right )+2 B \arctan \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{b}+\frac {\frac {2 \left (-C a +C b \right ) \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+a +b \right )}{2 a -2 b}+\frac {2 \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{b}}{e}\)	\(142\)
default	\(\frac {\frac {C \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+1\right )+2 B \arctan \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{b}+\frac {\frac {2 \left (-C a +C b \right ) \ln \left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+a +b \right )}{2 a -2 b}+\frac {2 \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{b}}{e}\)	\(142\)
risch	\(-\frac {i x C}{b}+\frac {B x}{b}+\frac {2 i C \,a^{2} b \,e^{2} x}{a^{2} b^{2} e^{2}-b^{4} e^{2}}-\frac {2 i C \,b^{3} e^{2} x}{a^{2} b^{2} e^{2}-b^{4} e^{2}}+\frac {2 i C \,a^{2} b d e}{a^{2} b^{2} e^{2}-b^{4} e^{2}}-\frac {2 i C \,b^{3} d e}{a^{2} b^{2} e^{2}-b^{4} e^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {A a b -B \,a^{2}-i \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}+2 A B \,a^{3} b -2 A B a \,b^{3}-B^{2} a^{4}+B^{2} a^{2} b^{2}}}{b \left (A b -B a \right )}\right ) C \,a^{2}}{\left (a^{2}-b^{2}\right ) e b}+\frac {b \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {A a b -B \,a^{2}-i \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}+2 A B \,a^{3} b -2 A B a \,b^{3}-B^{2} a^{4}+B^{2} a^{2} b^{2}}}{b \left (A b -B a \right )}\right ) C}{\left (a^{2}-b^{2}\right ) e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {A a b -B \,a^{2}-i \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}+2 A B \,a^{3} b -2 A B a \,b^{3}-B^{2} a^{4}+B^{2} a^{2} b^{2}}}{b \left (A b -B a \right )}\right ) \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}+2 A B \,a^{3} b -2 A B a \,b^{3}-B^{2} a^{4}+B^{2} a^{2} b^{2}}}{\left (a^{2}-b^{2}\right ) e b}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {A a b -B \,a^{2}+i \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}+2 A B \,a^{3} b -2 A B a \,b^{3}-B^{2} a^{4}+B^{2} a^{2} b^{2}}}{b \left (A b -B a \right )}\right ) C \,a^{2}}{\left (a^{2}-b^{2}\right ) e b}+\frac {b \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {A a b -B \,a^{2}+i \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}+2 A B \,a^{3} b -2 A B a \,b^{3}-B^{2} a^{4}+B^{2} a^{2} b^{2}}}{b \left (A b -B a \right )}\right ) C}{\left (a^{2}-b^{2}\right ) e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {A a b -B \,a^{2}+i \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}+2 A B \,a^{3} b -2 A B a \,b^{3}-B^{2} a^{4}+B^{2} a^{2} b^{2}}}{b \left (A b -B a \right )}\right ) \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}+2 A B \,a^{3} b -2 A B a \,b^{3}-B^{2} a^{4}+B^{2} a^{2} b^{2}}}{\left (a^{2}-b^{2}\right ) e b}\)	\(933\)

Input:

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x,method=_RETURNVERBOSE 
)
 

Output:

1/e*(2/b*(1/2*C*ln(tan(1/2*e*x+1/2*d)^2+1)+B*arctan(tan(1/2*e*x+1/2*d)))+2 
/b*(1/2*(-C*a+C*b)/(a-b)*ln(a*tan(1/2*e*x+1/2*d)^2-b*tan(1/2*e*x+1/2*d)^2+ 
a+b)+(A*b-B*a)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*e*x+1/2*d)/((a-b)* 
(a+b))^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 326, normalized size of antiderivative = 3.75 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx=\left [\frac {2 \, {\left (B a^{2} - B b^{2}\right )} e x + {\left (B a - A b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (e x + d\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (e x + d\right ) + b\right )} \sin \left (e x + d\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}}\right ) - {\left (C a^{2} - C b^{2}\right )} \log \left (b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} e}, \frac {2 \, {\left (B a^{2} - B b^{2}\right )} e x - 2 \, {\left (B a - A b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (e x + d\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (e x + d\right )}\right ) - {\left (C a^{2} - C b^{2}\right )} \log \left (b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} e}\right ] \] Input:

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x, algorithm="fri 
cas")
 

Output:

[1/2*(2*(B*a^2 - B*b^2)*e*x + (B*a - A*b)*sqrt(-a^2 + b^2)*log((2*a*b*cos( 
e*x + d) + (2*a^2 - b^2)*cos(e*x + d)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(e*x + 
d) + b)*sin(e*x + d) - a^2 + 2*b^2)/(b^2*cos(e*x + d)^2 + 2*a*b*cos(e*x + 
d) + a^2)) - (C*a^2 - C*b^2)*log(b^2*cos(e*x + d)^2 + 2*a*b*cos(e*x + d) + 
 a^2))/((a^2*b - b^3)*e), 1/2*(2*(B*a^2 - B*b^2)*e*x - 2*(B*a - A*b)*sqrt( 
a^2 - b^2)*arctan(-(a*cos(e*x + d) + b)/(sqrt(a^2 - b^2)*sin(e*x + d))) - 
(C*a^2 - C*b^2)*log(b^2*cos(e*x + d)^2 + 2*a*b*cos(e*x + d) + a^2))/((a^2* 
b - b^3)*e)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 672 vs. \(2 (73) = 146\).

Time = 13.95 (sec) , antiderivative size = 672, normalized size of antiderivative = 7.72 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx =\text {Too large to display} \] Input:

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x)
 

Output:

Piecewise((zoo*x*(A + B*cos(d) + C*sin(d))/cos(d), Eq(a, 0) & Eq(b, 0) & E 
q(e, 0)), (A*tan(d/2 + e*x/2)/(b*e) + B*x/b - B*tan(d/2 + e*x/2)/(b*e) + C 
*log(tan(d/2 + e*x/2)**2 + 1)/(b*e), Eq(a, b)), (A/(b*e*tan(d/2 + e*x/2)) 
+ B*x/b + B/(b*e*tan(d/2 + e*x/2)) + C*log(tan(d/2 + e*x/2)**2 + 1)/(b*e) 
- 2*C*log(tan(d/2 + e*x/2))/(b*e), Eq(a, -b)), ((A*x + B*sin(d + e*x)/e - 
C*cos(d + e*x)/e)/a, Eq(b, 0)), (x*(A + B*cos(d) + C*sin(d))/(a + b*cos(d) 
), Eq(e, 0)), (-A*b*sqrt(-a/(a - b) - b/(a - b))*log(-sqrt(-a/(a - b) - b/ 
(a - b)) + tan(d/2 + e*x/2))/(a*b*e + b**2*e) + A*b*sqrt(-a/(a - b) - b/(a 
 - b))*log(sqrt(-a/(a - b) - b/(a - b)) + tan(d/2 + e*x/2))/(a*b*e + b**2* 
e) + B*a*e*x/(a*b*e + b**2*e) + B*a*sqrt(-a/(a - b) - b/(a - b))*log(-sqrt 
(-a/(a - b) - b/(a - b)) + tan(d/2 + e*x/2))/(a*b*e + b**2*e) - B*a*sqrt(- 
a/(a - b) - b/(a - b))*log(sqrt(-a/(a - b) - b/(a - b)) + tan(d/2 + e*x/2) 
)/(a*b*e + b**2*e) + B*b*e*x/(a*b*e + b**2*e) - C*a*log(-sqrt(-a/(a - b) - 
 b/(a - b)) + tan(d/2 + e*x/2))/(a*b*e + b**2*e) - C*a*log(sqrt(-a/(a - b) 
 - b/(a - b)) + tan(d/2 + e*x/2))/(a*b*e + b**2*e) + C*a*log(tan(d/2 + e*x 
/2)**2 + 1)/(a*b*e + b**2*e) - C*b*log(-sqrt(-a/(a - b) - b/(a - b)) + tan 
(d/2 + e*x/2))/(a*b*e + b**2*e) - C*b*log(sqrt(-a/(a - b) - b/(a - b)) + t 
an(d/2 + e*x/2))/(a*b*e + b**2*e) + C*b*log(tan(d/2 + e*x/2)**2 + 1)/(a*b* 
e + b**2*e), True))
 

Maxima [F(-2)]

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

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x, algorithm="max 
ima")
 

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(4*b^2-4*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. 454 vs. \(2 (78) = 156\).

Time = 0.15 (sec) , antiderivative size = 454, normalized size of antiderivative = 5.22 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx=-\frac {\frac {C {\left (a + b\right )} {\left (a - b\right )}^{2} \log \left (\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + \frac {2 \, a + \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{2 \, {\left (a - b\right )}}\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left | b \right |}} + \frac {{\left (\sqrt {a^{2} - b^{2}} B {\left (2 \, a - b\right )} {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} A b {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} A {\left | a - b \right |} {\left | b \right |} + \sqrt {a^{2} - b^{2}} B {\left | a - b \right |} {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {e x + d}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{\sqrt {\frac {2 \, a + \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left | b \right |}} + \frac {{\left (2 \, B a - A b - B b + A {\left | b \right |} - B {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {e x + d}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{\sqrt {\frac {2 \, a - \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{b^{2} - a {\left | b \right |}} + \frac {{\left (C a - C b\right )} \log \left (\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + \frac {2 \, a - \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{2 \, {\left (a - b\right )}}\right )}{b^{2} - a {\left | b \right |}}}{e} \] Input:

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x, algorithm="gia 
c")
 

Output:

-(C*(a + b)*(a - b)^2*log(tan(1/2*e*x + 1/2*d)^2 + 1/2*(2*a + sqrt(-4*(a + 
 b)*(a - b) + 4*a^2))/(a - b))/((a^2 - 2*a*b + b^2)*b^2 + (a^3 - 2*a^2*b + 
 a*b^2)*abs(b)) + (sqrt(a^2 - b^2)*B*(2*a - b)*abs(a - b) - sqrt(a^2 - b^2 
)*A*b*abs(a - b) - sqrt(a^2 - b^2)*A*abs(a - b)*abs(b) + sqrt(a^2 - b^2)*B 
*abs(a - b)*abs(b))*(pi*floor(1/2*(e*x + d)/pi + 1/2) + arctan(2*sqrt(1/2) 
*tan(1/2*e*x + 1/2*d)/sqrt((2*a + sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a - b 
))))/((a^2 - 2*a*b + b^2)*b^2 + (a^3 - 2*a^2*b + a*b^2)*abs(b)) + (2*B*a - 
 A*b - B*b + A*abs(b) - B*abs(b))*(pi*floor(1/2*(e*x + d)/pi + 1/2) + arct 
an(2*sqrt(1/2)*tan(1/2*e*x + 1/2*d)/sqrt((2*a - sqrt(-4*(a + b)*(a - b) + 
4*a^2))/(a - b))))/(b^2 - a*abs(b)) + (C*a - C*b)*log(tan(1/2*e*x + 1/2*d) 
^2 + 1/2*(2*a - sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a - b))/(b^2 - a*abs(b) 
))/e
 

Mupad [B] (verification not implemented)

Time = 2.19 (sec) , antiderivative size = 886, normalized size of antiderivative = 10.18 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx =\text {Too large to display} \] Input:

int((A + B*cos(d + e*x) + C*sin(d + e*x))/(a + b*cos(d + e*x)),x)
 

Output:

(log(tan(d/2 + (e*x)/2) + 1i)*(B*1i + C))/(b*e) - (log(tan(d/2 + (e*x)/2) 
- 1i)*(B*1i - C))/(b*e) - (log(A^2*b^3 + B^2*b^3 - 4*C^2*a^3 + 4*C^2*b^3 + 
 A^2*a*b^2 + B^2*a*b^2 + 4*C^2*a*b^2 - 4*C^2*a^2*b - 2*A*B*a*b^2 - 2*A*B*a 
^2*b + A^2*b^2*tan(d/2 + (e*x)/2)*(b^2 - a^2)^(1/2) + B^2*b^2*tan(d/2 + (e 
*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*a^2*tan(d/2 + (e*x)/2)*(b^2 - a^2)^(1/2) 
+ 4*C^2*b^2*tan(d/2 + (e*x)/2)*(b^2 - a^2)^(1/2) - 4*A*C*b^2*(b^2 - a^2)^( 
1/2) + 4*B*C*a^2*(b^2 - a^2)^(1/2) - 4*A*C*b^3*tan(d/2 + (e*x)/2) - 4*B*C* 
a^3*tan(d/2 + (e*x)/2) - 4*A*C*a*b*(b^2 - a^2)^(1/2) + 4*B*C*a*b*(b^2 - a^ 
2)^(1/2) + 4*A*C*a^2*b*tan(d/2 + (e*x)/2) + 4*B*C*a*b^2*tan(d/2 + (e*x)/2) 
 - 2*A*B*a*b*tan(d/2 + (e*x)/2)*(b^2 - a^2)^(1/2))*(C*a^2 - C*b^2 + A*b*(b 
^2 - a^2)^(1/2) - B*a*(b^2 - a^2)^(1/2)))/(b*e*(a^2 - b^2)) - (log(A^2*b^3 
 + B^2*b^3 - 4*C^2*a^3 + 4*C^2*b^3 + A^2*a*b^2 + B^2*a*b^2 + 4*C^2*a*b^2 - 
 4*C^2*a^2*b - 2*A*B*a*b^2 - 2*A*B*a^2*b - A^2*b^2*tan(d/2 + (e*x)/2)*(b^2 
 - a^2)^(1/2) - B^2*b^2*tan(d/2 + (e*x)/2)*(b^2 - a^2)^(1/2) + 4*C^2*a^2*t 
an(d/2 + (e*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*b^2*tan(d/2 + (e*x)/2)*(b^2 - 
a^2)^(1/2) + 4*A*C*b^2*(b^2 - a^2)^(1/2) - 4*B*C*a^2*(b^2 - a^2)^(1/2) - 4 
*A*C*b^3*tan(d/2 + (e*x)/2) - 4*B*C*a^3*tan(d/2 + (e*x)/2) + 4*A*C*a*b*(b^ 
2 - a^2)^(1/2) - 4*B*C*a*b*(b^2 - a^2)^(1/2) + 4*A*C*a^2*b*tan(d/2 + (e*x) 
/2) + 4*B*C*a*b^2*tan(d/2 + (e*x)/2) + 2*A*B*a*b*tan(d/2 + (e*x)/2)*(b^2 - 
 a^2)^(1/2))*(C*a^2 - C*b^2 - A*b*(b^2 - a^2)^(1/2) + B*a*(b^2 - a^2)^(...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.30 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx=\frac {-\mathrm {log}\left (\cos \left (e x +d \right ) b +a \right ) c +b e x}{b e} \] Input:

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x)
 

Output:

( - log(cos(d + e*x)*b + a)*c + b*e*x)/(b*e)