\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [259]

Optimal result

Integrand size = 31, antiderivative size = 155 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {(A b-2 a B) x}{b^3}-\frac {2 a \left (a^2 A b-2 A b^3-2 a^3 B+3 a b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac {B \sin (c+d x)}{b^2 d}-\frac {a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 1.99 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {(A b-2 a B) (c+d x)+\frac {2 a \left (-a^2 A b+2 A b^3+2 a^3 B-3 a b^2 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+b B \sin (c+d x)+\frac {a^2 b (-A b+a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}}{b^3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 3467, 3042, 3502, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3467

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

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 3467

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f 
_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ 
(B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 
*(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2))   Int[(c + d* 
Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c 
- 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* 
a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 
1)*(c^2 - d^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[ 
n, -1]
 

rule 3502

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

Maple [A] (verified)

Time = 1.71 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {-\frac {2 a \left (\frac {a \left (A b -B a \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (A \,a^{2} b -2 A \,b^{3}-2 a^{3} B +3 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}+\frac {\frac {2 B b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \left (A b -2 B a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\)	\(205\)
default	\(\frac {-\frac {2 a \left (\frac {a \left (A b -B a \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (A \,a^{2} b -2 A \,b^{3}-2 a^{3} B +3 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}+\frac {\frac {2 B b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \left (A b -2 B a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\)	\(205\)
risch	\(\frac {x A}{b^{2}}-\frac {2 x B a}{b^{3}}-\frac {i B \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {i B \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {2 i a^{2} \left (-A b +B a \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{3} \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 {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 ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 a \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 {2 a^{4} \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^{3}}-\frac {3 a^{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}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}+\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 ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 a \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 {2 a^{4} \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^{3}}+\frac {3 a^{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}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}\)	\(818\)

Input:

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

Output:

1/d*(-2*a/b^3*(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*a^2*b-2*A*b^3-2*B*a^3+3*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^3*(B*b*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+(A*b-2*B* 
a)*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. 359 vs. \(2 (148) = 296\).

Time = 0.13 (sec) , antiderivative size = 789, normalized size of antiderivative = 5.09 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/2*(2*(2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A*a^2*b^4 + 2*B*a*b^5 - 
A*b^6)*d*x*cos(d*x + c) + 2*(2*B*a^6 - A*a^5*b - 4*B*a^4*b^2 + 2*A*a^3*b^3 
 + 2*B*a^2*b^4 - A*a*b^5)*d*x - (2*B*a^5 - A*a^4*b - 3*B*a^3*b^2 + 2*A*a^2 
*b^3 + (2*B*a^4*b - A*a^3*b^2 - 3*B*a^2*b^3 + 2*A*a*b^4)*cos(d*x + c))*sqr 
t(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*s 
qrt(-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*(2*B*a^5*b - A*a^4*b^2 - 3*B*a 
^3*b^3 + A*a^2*b^4 + B*a*b^5 + (B*a^4*b^2 - 2*B*a^2*b^4 + B*b^6)*cos(d*x + 
 c))*sin(d*x + c))/((a^4*b^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c) + (a^5*b^3 
- 2*a^3*b^5 + a*b^7)*d), -((2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A*a^2* 
b^4 + 2*B*a*b^5 - A*b^6)*d*x*cos(d*x + c) + (2*B*a^6 - A*a^5*b - 4*B*a^4*b 
^2 + 2*A*a^3*b^3 + 2*B*a^2*b^4 - A*a*b^5)*d*x - (2*B*a^5 - A*a^4*b - 3*B*a 
^3*b^2 + 2*A*a^2*b^3 + (2*B*a^4*b - A*a^3*b^2 - 3*B*a^2*b^3 + 2*A*a*b^4)*c 
os(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2) 
*sin(d*x + c))) - (2*B*a^5*b - A*a^4*b^2 - 3*B*a^3*b^3 + A*a^2*b^4 + B*a*b 
^5 + (B*a^4*b^2 - 2*B*a^2*b^4 + B*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b 
^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c) + (a^5*b^3 - 2*a^3*b^5 + a*b^7)*d)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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. 1116 vs. \(2 (148) = 296\).

Time = 0.25 (sec) , antiderivative size = 1116, normalized size of antiderivative = 7.20 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((4*B*a^6*b^2 - 2*A*a^5*b^3 - 2*B*a^5*b^3 + A*a^4*b^4 - 9*B*a^4*b^4 + 5*A* 
a^3*b^5 + 4*B*a^3*b^5 - 2*A*a^2*b^6 + 5*B*a^2*b^6 - 3*A*a*b^7 - 2*B*a*b^7 
+ A*b^8 + 2*B*a^3*abs(-a^2*b^3 + b^5) - A*a^2*b*abs(-a^2*b^3 + b^5) - B*a^ 
2*b*abs(-a^2*b^3 + b^5) + A*a*b^2*abs(-a^2*b^3 + b^5) - 2*B*a*b^2*abs(-a^2 
*b^3 + b^5) + A*b^3*abs(-a^2*b^3 + b^5))*(pi*floor(1/2*(d*x + c)/pi + 1/2) 
 + arctan(2*sqrt(1/2)*tan(1/2*d*x + 1/2*c)/sqrt((2*a^3*b^2 - 2*a*b^4 + sqr 
t(-4*(a^3*b^2 + a^2*b^3 - a*b^4 - b^5)*(a^3*b^2 - a^2*b^3 - a*b^4 + b^5) + 
 4*(a^3*b^2 - a*b^4)^2))/(a^3*b^2 - a^2*b^3 - a*b^4 + b^5))))/(a^3*b^2*abs 
(-a^2*b^3 + b^5) - a*b^4*abs(-a^2*b^3 + b^5) + (a^2*b^3 - b^5)^2) + ((a^2* 
b - a*b^2 - b^3)*sqrt(a^2 - b^2)*A*abs(-a^2*b^3 + b^5)*abs(-a + b) - (2*a^ 
3 - a^2*b - 2*a*b^2)*sqrt(a^2 - b^2)*B*abs(-a^2*b^3 + b^5)*abs(-a + b) - ( 
2*a^5*b^3 - a^4*b^4 - 5*a^3*b^5 + 2*a^2*b^6 + 3*a*b^7 - b^8)*sqrt(a^2 - b^ 
2)*A*abs(-a + b) + (4*a^6*b^2 - 2*a^5*b^3 - 9*a^4*b^4 + 4*a^3*b^5 + 5*a^2* 
b^6 - 2*a*b^7)*sqrt(a^2 - b^2)*B*abs(-a + b))*(pi*floor(1/2*(d*x + c)/pi + 
 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d*x + 1/2*c)/sqrt((2*a^3*b^2 - 2*a*b^4 
- sqrt(-4*(a^3*b^2 + a^2*b^3 - a*b^4 - b^5)*(a^3*b^2 - a^2*b^3 - a*b^4 + b 
^5) + 4*(a^3*b^2 - a*b^4)^2))/(a^3*b^2 - a^2*b^3 - a*b^4 + b^5))))/((a^2*b 
^3 - b^5)^2*(a^2 - 2*a*b + b^2) - (a^5*b^2 - 2*a^4*b^3 + 2*a^2*b^5 - a*b^6 
)*abs(-a^2*b^3 + b^5)) + 2*(2*B*a^3*tan(1/2*d*x + 1/2*c)^3 - A*a^2*b*tan(1 
/2*d*x + 1/2*c)^3 - B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - B*a*b^2*tan(1/2*d*...
 

Mupad [B] (verification not implemented)

Time = 28.14 (sec) , antiderivative size = 3276, normalized size of antiderivative = 21.14 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(log(tan(c/2 + (d*x)/2) + 1i)*(A*b - 2*B*a)*1i)/(b^3*d) - ((2*tan(c/2 + (d 
*x)/2)^3*(A*a^2*b - B*b^3 - 2*B*a^3 + B*a*b^2 + B*a^2*b))/(b^2*(a + b)*(a 
- b)) + (2*tan(c/2 + (d*x)/2)*(B*b^3 - 2*B*a^3 + A*a^2*b + B*a*b^2 - B*a^2 
*b))/(b^2*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d*x)/2)^4*(a - b) + 2*a 
*tan(c/2 + (d*x)/2)^2)) - (log(tan(c/2 + (d*x)/2) - 1i)*(A*b*1i - B*a*2i)) 
/(b^3*d) - (a*atan(((a*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*B^2*a^8 - 2*A^ 
2*a*b^7 - 8*B^2*a^7*b + 3*A^2*a^2*b^6 + 4*A^2*a^3*b^5 - 5*A^2*a^4*b^4 - 2* 
A^2*a^5*b^3 + 2*A^2*a^6*b^2 + 4*B^2*a^2*b^6 - 8*B^2*a^3*b^5 + 5*B^2*a^4*b^ 
4 + 16*B^2*a^5*b^3 - 16*B^2*a^6*b^2 - 4*A*B*a*b^7 - 8*A*B*a^7*b + 8*A*B*a^ 
2*b^6 - 8*A*B*a^3*b^5 - 16*A*B*a^4*b^4 + 18*A*B*a^5*b^3 + 8*A*B*a^6*b^2))/ 
(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (a*((32*(A*a^2*b^10 - A*b^12 - 3*A*a^3 
*b^9 + A*a^5*b^7 - 3*B*a^2*b^10 - 3*B*a^3*b^9 + 5*B*a^4*b^8 + B*a^5*b^7 - 
2*B*a^6*b^6 + 2*A*a*b^11 + 2*B*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) 
- (32*a*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 
 - A*a^2*b - 3*B*a*b^2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2 
*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 
 + 3*a^4*b^5 - a^6*b^3)))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 
- A*a^2*b - 3*B*a*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*(-(a + b) 
^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*b - 3*B*a*b^2)*1i)/(b^9 - 3 
*a^2*b^7 + 3*a^4*b^5 - a^6*b^3) + (a*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 +...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}+\sin \left (d x +c \right ) a^{2} b -\sin \left (d x +c \right ) b^{3}-a^{3} d x +a \,b^{2} d x}{b^{2} d \left (a^{2}-b^{2}\right )} \] Input:

int(cos(d*x+c)^2*(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)/sqrt(a 
**2 - b**2))*a**2 + sin(c + d*x)*a**2*b - sin(c + d*x)*b**3 - a**3*d*x + a 
*b**2*d*x)/(b**2*d*(a**2 - b**2))