\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [260]

Optimal result

Integrand size = 29, antiderivative size = 122 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {B x}{b^2}-\frac {2 \left (A b^3+a^3 B-2 a b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^2 (a+b)^{3/2} d}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

B*x/b^2-2*(A*b^3+B*a^3-2*B*a*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a 
+b)^(1/2))/(a-b)^(3/2)/b^2/(a+b)^(3/2)/d+a*(A*b-B*a)*sin(d*x+c)/b/(a^2-b^2 
)/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 1.36 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {B (c+d x)-\frac {2 \left (A b^3+a \left (a^2-2 b^2\right ) B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+\frac {a b (A b-a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}}{b^2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3447, 3042, 3500, 3042, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )+B \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 1.39 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {a \left (A b -B a \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (A \,b^{3}+a^{3} B -2 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2}}+\frac {2 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(161\)
default	\(\frac {-\frac {2 \left (-\frac {a \left (A b -B a \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (A \,b^{3}+a^{3} B -2 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2}}+\frac {2 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(161\)
risch	\(\frac {B x}{b^{2}}-\frac {2 i a \left (A b -B a \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{2} \left (-a^{2}+b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B a}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B a}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\)	\(583\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (114) = 228\).

Time = 0.12 (sec) , antiderivative size = 551, normalized size of antiderivative = 4.52 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\left [\frac {2 \, {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4}\right )} d x + {\left (B a^{4} - 2 \, B a^{2} b^{2} + A a b^{3} + {\left (B a^{3} b - 2 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (B a^{4} b - A a^{3} b^{2} - B a^{2} b^{3} + A a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d\right )}}, \frac {{\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5}\right )} d x \cos \left (d x + c\right ) + {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4}\right )} d x - {\left (B a^{4} - 2 \, B a^{2} b^{2} + A a b^{3} + {\left (B a^{3} b - 2 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (B a^{4} b - A a^{3} b^{2} - B a^{2} b^{3} + A a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d}\right ] \] Input:

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorithm="fri 
cas")
 

Output:

[1/2*(2*(B*a^4*b - 2*B*a^2*b^3 + B*b^5)*d*x*cos(d*x + c) + 2*(B*a^5 - 2*B* 
a^3*b^2 + B*a*b^4)*d*x + (B*a^4 - 2*B*a^2*b^2 + A*a*b^3 + (B*a^3*b - 2*B*a 
*b^3 + A*b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2* 
a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d* 
x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - 2 
*(B*a^4*b - A*a^3*b^2 - B*a^2*b^3 + A*a*b^4)*sin(d*x + c))/((a^4*b^3 - 2*a 
^2*b^5 + b^7)*d*cos(d*x + c) + (a^5*b^2 - 2*a^3*b^4 + a*b^6)*d), ((B*a^4*b 
 - 2*B*a^2*b^3 + B*b^5)*d*x*cos(d*x + c) + (B*a^5 - 2*B*a^3*b^2 + B*a*b^4) 
*d*x - (B*a^4 - 2*B*a^2*b^2 + A*a*b^3 + (B*a^3*b - 2*B*a*b^3 + A*b^4)*cos( 
d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*si 
n(d*x + c))) - (B*a^4*b - A*a^3*b^2 - B*a^2*b^3 + A*a*b^4)*sin(d*x + c))/( 
(a^4*b^3 - 2*a^2*b^5 + b^7)*d*cos(d*x + c) + (a^5*b^2 - 2*a^3*b^4 + a*b^6) 
*d)]
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

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

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

Time = 0.15 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.63 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (B a^{3} - 2 \, B a b^{2} + A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {{\left (d x + c\right )} B}{b^{2}} - \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}}}{d} \] Input:

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorithm="gia 
c")
 

Output:

(2*(B*a^3 - 2*B*a*b^2 + A*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a 
+ 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^ 
2 - b^2)))/((a^2*b^2 - b^4)*sqrt(a^2 - b^2)) + (d*x + c)*B/b^2 - 2*(B*a^2* 
tan(1/2*d*x + 1/2*c) - A*a*b*tan(1/2*d*x + 1/2*c))/((a^2*b - b^3)*(a*tan(1 
/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)))/d
 

Mupad [B] (verification not implemented)

Time = 30.07 (sec) , antiderivative size = 3775, normalized size of antiderivative = 30.94 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^2,x)
 

Output:

(2*B*atan(((B*((B*((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 
- 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - 
 a^3*b^3) - (B*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4 
*b^6 + 2*a^5*b^5 - 2*a^6*b^4)*32i)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)) 
)*1i)/b^2 + (32*tan(c/2 + (d*x)/2)*(A^2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2* 
a*b^5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A* 
B*a*b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2 - (B*((B 
*((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 - 3*B*a^3*b^6 + B 
*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (B*ta 
n(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 
- 2*a^6*b^4)*32i)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))*1i)/b^2 - (32*t 
an(c/2 + (d*x)/2)*(A^2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5 
*b + 3*B^2*a^2*b^4 + 4*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a*b^5 + 2*A*B*a 
^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2)/((64*(B^3*a^5 - A*B^2*b^ 
5 + A^2*B*b^5 + 2*B^3*a*b^4 - B^3*a^4*b + 2*B^3*a^2*b^3 - 3*B^3*a^3*b^2 - 
3*A*B^2*a*b^4 + A*B^2*a^2*b^3 + A*B^2*a^3*b^2))/(a*b^5 + b^6 - a^2*b^4 - a 
^3*b^3) + (B*((B*((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 - 
 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - 
a^3*b^3) - (B*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4* 
b^6 + 2*a^5*b^5 - 2*a^6*b^4)*32i)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.68 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a +a^{2} d x -b^{2} d x}{b d \left (a^{2}-b^{2}\right )} \] Input:

int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x)
 

Output:

( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*a + a**2*d*x - b**2*d*x)/(b*d*(a**2 - b**2))