\(\int \frac {\cos ^3(c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [570]

Optimal result

Integrand size = 33, antiderivative size = 332 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {a \left (2 A b^2+\left (4 a^2+b^2\right ) C\right ) x}{b^5}+\frac {2 a^2 \left (2 a^2 A b^2-3 A b^4+4 a^4 C-5 a^2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^5 (a+b)^{3/2} d}+\frac {\left (a^2 b^2 (6 A-7 C)+12 a^4 C-b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac {a \left (A b^2+2 a^2 C-b^2 C\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

-a*(2*A*b^2+(4*a^2+b^2)*C)*x/b^5+2*a^2*(2*A*a^2*b^2-3*A*b^4+4*C*a^4-5*C*a^ 
2*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^5/ 
(a+b)^(3/2)/d+1/3*(a^2*b^2*(6*A-7*C)+12*a^4*C-b^4*(3*A+2*C))*sin(d*x+c)/b^ 
4/(a^2-b^2)/d-a*(A*b^2+2*C*a^2-C*b^2)*cos(d*x+c)*sin(d*x+c)/b^3/(a^2-b^2)/ 
d+1/3*(3*A*b^2+4*C*a^2-C*b^2)*cos(d*x+c)^2*sin(d*x+c)/b^2/(a^2-b^2)/d-(A*b 
^2+C*a^2)*cos(d*x+c)^3*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 2.95 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.65 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {-12 a \left (2 A b^2+\left (4 a^2+b^2\right ) C\right ) (c+d x)+\frac {24 a^2 \left (-3 A b^4+a^2 b^2 (2 A-5 C)+4 a^4 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+3 b \left (4 A b^2+3 \left (4 a^2+b^2\right ) C\right ) \sin (c+d x)+\frac {12 a^3 b \left (A b^2+a^2 C\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-6 a b^2 C \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^5 d} \] Input:

Integrate[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^2,x 
]
 

Output:

(-12*a*(2*A*b^2 + (4*a^2 + b^2)*C)*(c + d*x) + (24*a^2*(-3*A*b^4 + a^2*b^2 
*(2*A - 5*C) + 4*a^4*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2 
]])/(-a^2 + b^2)^(3/2) + 3*b*(4*A*b^2 + 3*(4*a^2 + b^2)*C)*Sin[c + d*x] + 
(12*a^3*b*(A*b^2 + a^2*C)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d* 
x])) - 6*a*b^2*C*Sin[2*(c + d*x)] + b^3*C*Sin[3*(c + d*x)])/(12*b^5*d)
 

Rubi [A] (verified)

Time = 1.96 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 3527, 3042, 3528, 25, 3042, 3528, 27, 3042, 3502, 27, 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.

\(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

\(\displaystyle -\frac {\int \frac {\cos ^2(c+d x) \left (-\left (\left (3 A b^2+\left (4 a^2-b^2\right ) C\right ) \cos ^2(c+d x)\right )-a b (A+C) \cos (c+d x)+3 \left (C a^2+A b^2\right )\right )}{a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (-3 A b^2-\left (4 a^2-b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (C a^2+A b^2\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3528

\(\displaystyle -\frac {\frac {\int -\frac {\cos (c+d x) \left (-6 a \left (2 C a^2+A b^2-b^2 C\right ) \cos ^2(c+d x)-b \left (3 A b^2+\left (a^2+2 b^2\right ) C\right ) \cos (c+d x)+2 a \left (4 C a^2+b^2 (3 A-C)\right )\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {-\frac {\int \frac {\cos (c+d x) \left (-6 a \left (2 C a^2+A b^2-b^2 C\right ) \cos ^2(c+d x)-b \left (3 A b^2+\left (a^2+2 b^2\right ) C\right ) \cos (c+d x)+2 a \left (4 C a^2+b^2 (3 A-C)\right )\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (-6 a \left (2 C a^2+A b^2-b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (3 A b^2+\left (a^2+2 b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a \left (4 C a^2+b^2 (3 A-C)\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3528

\(\displaystyle -\frac {-\frac {\frac {\int -\frac {2 \left (3 \left (2 C a^2+A b^2-b^2 C\right ) a^2-b \left (3 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x) a-\left (12 C a^4+b^2 (6 A-7 C) a^2-b^4 (3 A+2 C)\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)}dx}{2 b}-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {3 \left (2 C a^2+A b^2-b^2 C\right ) a^2-b \left (3 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x) a-\left (12 C a^4+b^2 (6 A-7 C) a^2-b^4 (3 A+2 C)\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)}dx}{b}-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {3 \left (2 C a^2+A b^2-b^2 C\right ) a^2-b \left (3 A b^2+\left (2 a^2+b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+\left (-12 C a^4-b^2 (6 A-7 C) a^2+b^4 (3 A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {-\frac {-\frac {\frac {\int \frac {3 \left (b \left (2 C a^2+A b^2-b^2 C\right ) a^2+\left (a^2-b^2\right ) \left (4 C a^2+2 A b^2+b^2 C\right ) \cos (c+d x) a\right )}{a+b \cos (c+d x)}dx}{b}-\frac {\left (12 a^4 C+a^2 b^2 (6 A-7 C)-b^4 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \int \frac {b \left (2 C a^2+A b^2-b^2 C\right ) a^2+\left (a^2-b^2\right ) \left (4 C a^2+2 A b^2+b^2 C\right ) \cos (c+d x) a}{a+b \cos (c+d x)}dx}{b}-\frac {\left (12 a^4 C+a^2 b^2 (6 A-7 C)-b^4 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \int \frac {b \left (2 C a^2+A b^2-b^2 C\right ) a^2+\left (a^2-b^2\right ) \left (4 C a^2+2 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (12 a^4 C+a^2 b^2 (6 A-7 C)-b^4 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \left (\frac {a^2 \left (-4 a^4 C-a^2 b^2 (2 A-5 C)+3 A b^4\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}+\frac {a x \left (a^2-b^2\right ) \left (4 a^2 C+2 A b^2+b^2 C\right )}{b}\right )}{b}-\frac {\left (12 a^4 C+a^2 b^2 (6 A-7 C)-b^4 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \left (\frac {a^2 \left (-4 a^4 C-a^2 b^2 (2 A-5 C)+3 A b^4\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {a x \left (a^2-b^2\right ) \left (4 a^2 C+2 A b^2+b^2 C\right )}{b}\right )}{b}-\frac {\left (12 a^4 C+a^2 b^2 (6 A-7 C)-b^4 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {-\frac {-\frac {\frac {3 \left (\frac {2 a^2 \left (-4 a^4 C-a^2 b^2 (2 A-5 C)+3 A b^4\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}+\frac {a x \left (a^2-b^2\right ) \left (4 a^2 C+2 A b^2+b^2 C\right )}{b}\right )}{b}-\frac {\left (12 a^4 C+a^2 b^2 (6 A-7 C)-b^4 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{b}-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}}{3 b}-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {-\frac {\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d}-\frac {-\frac {3 a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b d}-\frac {\frac {3 \left (\frac {a x \left (a^2-b^2\right ) \left (4 a^2 C+2 A b^2+b^2 C\right )}{b}+\frac {2 a^2 \left (-4 a^4 C-a^2 b^2 (2 A-5 C)+3 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {\left (12 a^4 C+a^2 b^2 (6 A-7 C)-b^4 (3 A+2 C)\right ) \sin (c+d x)}{b d}}{b}}{3 b}}{b \left (a^2-b^2\right )}\)

Input:

Int[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^2,x]
 

Output:

-(((A*b^2 + a^2*C)*Cos[c + d*x]^3*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Co 
s[c + d*x]))) - (-1/3*((3*A*b^2 + 4*a^2*C - b^2*C)*Cos[c + d*x]^2*Sin[c + 
d*x])/(b*d) - ((-3*a*(A*b^2 + 2*a^2*C - b^2*C)*Cos[c + d*x]*Sin[c + d*x])/ 
(b*d) - ((3*((a*(a^2 - b^2)*(2*A*b^2 + 4*a^2*C + b^2*C)*x)/b + (2*a^2*(3*A 
*b^4 - a^2*b^2*(2*A - 5*C) - 4*a^4*C)*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*(6*A - 7*C) 
+ 12*a^4*C - b^4*(3*A + 2*C))*Sin[c + d*x])/(b*d))/b)/(3*b))/(b*(a^2 - b^2 
))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 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 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]
 

rule 3527

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

rule 3528

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

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 323, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {\left (-A \,b^{3}-3 a^{2} b C -C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 A \,b^{3}-6 a^{2} b C -\frac {2}{3} C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A \,b^{3}-3 a^{2} b C +C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+a \left (2 A \,b^{2}+4 a^{2} C +C \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{5}}+\frac {2 a^{2} \left (\frac {a b \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 A \,a^{2} b^{2}-3 A \,b^{4}+4 a^{4} C -5 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{5}}}{d}\)	\(323\)
default	\(\frac {-\frac {2 \left (\frac {\left (-A \,b^{3}-3 a^{2} b C -C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 A \,b^{3}-6 a^{2} b C -\frac {2}{3} C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A \,b^{3}-3 a^{2} b C +C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+a \left (2 A \,b^{2}+4 a^{2} C +C \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{5}}+\frac {2 a^{2} \left (\frac {a b \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 A \,a^{2} b^{2}-3 A \,b^{4}+4 a^{4} C -5 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{5}}}{d}\)	\(323\)
risch	\(-\frac {2 a x A}{b^{3}}-\frac {4 a^{3} x C}{b^{5}}-\frac {a C x}{b^{3}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{2 b^{4} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 b^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 b^{2} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C}{8 b^{2} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{2 b^{4} d}-\frac {i C a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 b^{3} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C}{8 b^{2} d}+\frac {i C a \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 b^{3} d}+\frac {2 i a^{3} \left (A \,b^{2}+a^{2} C \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{5} \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 {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\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}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {4 a^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{5}}+\frac {5 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\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}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}+\frac {4 a^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{5}}-\frac {5 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {C \sin \left (3 d x +3 c \right )}{12 b^{2} d}\)	\(984\)

Input:

int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x,method=_RETURNVER 
BOSE)
 

Output:

1/d*(-2/b^5*(((-A*b^3-3*C*a^2*b-C*a*b^2-C*b^3)*tan(1/2*d*x+1/2*c)^5+(-2*A* 
b^3-6*a^2*b*C-2/3*C*b^3)*tan(1/2*d*x+1/2*c)^3+(-A*b^3-3*C*a^2*b+C*a*b^2-C* 
b^3)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^3+a*(2*A*b^2+4*C*a^2+C*b 
^2)*arctan(tan(1/2*d*x+1/2*c)))+2*a^2/b^5*(a*b*(A*b^2+C*a^2)/(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)+(2*A*a 
^2*b^2-3*A*b^4+4*C*a^4-5*C*a^2*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))))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 985, normalized size of antiderivative = 2.97 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm= 
"fricas")
 

Output:

[-1/6*(6*(4*C*a^7*b + (2*A - 7*C)*a^5*b^3 - 2*(2*A - C)*a^3*b^5 + (2*A + C 
)*a*b^7)*d*x*cos(d*x + c) + 6*(4*C*a^8 + (2*A - 7*C)*a^6*b^2 - 2*(2*A - C) 
*a^4*b^4 + (2*A + C)*a^2*b^6)*d*x + 3*(4*C*a^7 + (2*A - 5*C)*a^5*b^2 - 3*A 
*a^3*b^4 + (4*C*a^6*b + (2*A - 5*C)*a^4*b^3 - 3*A*a^2*b^5)*cos(d*x + c))*s 
qrt(-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*co 
s(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - 2*(12*C*a^7*b + (6*A - 19*C)*a 
^5*b^3 - (9*A - 5*C)*a^3*b^5 + (3*A + 2*C)*a*b^7 + (C*a^4*b^4 - 2*C*a^2*b^ 
6 + C*b^8)*cos(d*x + c)^3 - 2*(C*a^5*b^3 - 2*C*a^3*b^5 + C*a*b^7)*cos(d*x 
+ c)^2 + (6*C*a^6*b^2 + (3*A - 10*C)*a^4*b^4 - 2*(3*A - C)*a^2*b^6 + (3*A 
+ 2*C)*b^8)*cos(d*x + c))*sin(d*x + c))/((a^4*b^6 - 2*a^2*b^8 + b^10)*d*co 
s(d*x + c) + (a^5*b^5 - 2*a^3*b^7 + a*b^9)*d), -1/3*(3*(4*C*a^7*b + (2*A - 
 7*C)*a^5*b^3 - 2*(2*A - C)*a^3*b^5 + (2*A + C)*a*b^7)*d*x*cos(d*x + c) + 
3*(4*C*a^8 + (2*A - 7*C)*a^6*b^2 - 2*(2*A - C)*a^4*b^4 + (2*A + C)*a^2*b^6 
)*d*x - 3*(4*C*a^7 + (2*A - 5*C)*a^5*b^2 - 3*A*a^3*b^4 + (4*C*a^6*b + (2*A 
 - 5*C)*a^4*b^3 - 3*A*a^2*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*co 
s(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (12*C*a^7*b + (6*A - 19* 
C)*a^5*b^3 - (9*A - 5*C)*a^3*b^5 + (3*A + 2*C)*a*b^7 + (C*a^4*b^4 - 2*C*a^ 
2*b^6 + C*b^8)*cos(d*x + c)^3 - 2*(C*a^5*b^3 - 2*C*a^3*b^5 + C*a*b^7)*cos( 
d*x + c)^2 + (6*C*a^6*b^2 + (3*A - 10*C)*a^4*b^4 - 2*(3*A - C)*a^2*b^6 ...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**3*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm= 
"maxima")
 

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 = 0.17 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {6 \, {\left (4 \, C a^{6} + 2 \, A a^{4} b^{2} - 5 \, C a^{4} b^{2} - 3 \, A a^{2} b^{4}\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^{5} - b^{7}\right )} \sqrt {a^{2} - b^{2}}} - \frac {6 \, {\left (C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\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 )}} + \frac {3 \, {\left (4 \, C a^{3} + 2 \, A a b^{2} + C a b^{2}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {2 \, {\left (9 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{4}}}{3 \, d} \] Input:

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm= 
"giac")
 

Output:

-1/3*(6*(4*C*a^6 + 2*A*a^4*b^2 - 5*C*a^4*b^2 - 3*A*a^2*b^4)*(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^5 - b^7)*sqrt(a^2 - b^2)) 
- 6*(C*a^5*tan(1/2*d*x + 1/2*c) + A*a^3*b^2*tan(1/2*d*x + 1/2*c))/((a^2*b^ 
4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) + 
3*(4*C*a^3 + 2*A*a*b^2 + C*a*b^2)*(d*x + c)/b^5 - 2*(9*C*a^2*tan(1/2*d*x + 
 1/2*c)^5 + 3*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*A*b^2*tan(1/2*d*x + 1/2*c)^ 
5 + 3*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 18*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 6*A 
*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 9*C*a^2*tan 
(1/2*d*x + 1/2*c) - 3*C*a*b*tan(1/2*d*x + 1/2*c) + 3*A*b^2*tan(1/2*d*x + 1 
/2*c) + 3*C*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^4) 
)/d
 

Mupad [B] (verification not implemented)

Time = 5.93 (sec) , antiderivative size = 6989, normalized size of antiderivative = 21.05 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^3*(A + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^2,x)
 

Output:

(atan(((((((32*(5*A*a^4*b^14 - 3*A*a^3*b^15 - 3*A*a^2*b^16 + A*a^5*b^13 - 
2*A*a^6*b^12 + C*a^3*b^15 - 5*C*a^4*b^14 - 4*C*a^5*b^13 + 9*C*a^6*b^12 + 2 
*C*a^7*b^11 - 4*C*a^8*b^10 + 2*A*a*b^17 + C*a*b^17))/(a*b^14 + b^15 - a^2* 
b^13 - a^3*b^12) - (32*tan(c/2 + (d*x)/2)*(C*a^3*4i + a*b^2*(2*A + C)*1i)* 
(2*a*b^15 - 2*a^2*b^14 - 4*a^3*b^13 + 4*a^4*b^12 + 2*a^5*b^11 - 2*a^6*b^10 
))/(b^5*(a*b^10 + b^11 - a^2*b^9 - a^3*b^8)))*(C*a^3*4i + a*b^2*(2*A + C)* 
1i))/b^5 + (32*tan(c/2 + (d*x)/2)*(32*C^2*a^12 - 32*C^2*a^11*b + 4*A^2*a^2 
*b^10 - 8*A^2*a^3*b^9 + 5*A^2*a^4*b^8 + 16*A^2*a^5*b^7 - 16*A^2*a^6*b^6 - 
8*A^2*a^7*b^5 + 8*A^2*a^8*b^4 + C^2*a^2*b^10 - 2*C^2*a^3*b^9 + 7*C^2*a^4*b 
^8 - 12*C^2*a^5*b^7 + 7*C^2*a^6*b^6 - 2*C^2*a^7*b^5 + 2*C^2*a^8*b^4 + 48*C 
^2*a^9*b^3 - 48*C^2*a^10*b^2 + 4*A*C*a^2*b^10 - 8*A*C*a^3*b^9 + 12*A*C*a^4 
*b^8 - 16*A*C*a^5*b^7 + 10*A*C*a^6*b^6 + 56*A*C*a^7*b^5 - 56*A*C*a^8*b^4 - 
 32*A*C*a^9*b^3 + 32*A*C*a^10*b^2))/(a*b^10 + b^11 - a^2*b^9 - a^3*b^8))*( 
C*a^3*4i + a*b^2*(2*A + C)*1i)*1i)/b^5 - (((((32*(5*A*a^4*b^14 - 3*A*a^3*b 
^15 - 3*A*a^2*b^16 + A*a^5*b^13 - 2*A*a^6*b^12 + C*a^3*b^15 - 5*C*a^4*b^14 
 - 4*C*a^5*b^13 + 9*C*a^6*b^12 + 2*C*a^7*b^11 - 4*C*a^8*b^10 + 2*A*a*b^17 
+ C*a*b^17))/(a*b^14 + b^15 - a^2*b^13 - a^3*b^12) + (32*tan(c/2 + (d*x)/2 
)*(C*a^3*4i + a*b^2*(2*A + C)*1i)*(2*a*b^15 - 2*a^2*b^14 - 4*a^3*b^13 + 4* 
a^4*b^12 + 2*a^5*b^11 - 2*a^6*b^10))/(b^5*(a*b^10 + b^11 - a^2*b^9 - a^3*b 
^8)))*(C*a^3*4i + a*b^2*(2*A + C)*1i))/b^5 - (32*tan(c/2 + (d*x)/2)*(32...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1212, normalized size of antiderivative = 3.65 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x)
 

Output:

(24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( 
a**2 - b**2))*cos(c + d*x)*a**6*b*c + 12*sqrt(a**2 - b**2)*atan((tan((c + 
d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**5*b**3 
- 30*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt 
(a**2 - b**2))*cos(c + d*x)*a**4*b**3*c - 18*sqrt(a**2 - b**2)*atan((tan(( 
c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**3*b 
**5 + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ 
sqrt(a**2 - b**2))*a**7*c + 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a 
- tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**6*b**2 - 30*sqrt(a**2 - b**2)* 
atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**5*b** 
2*c - 18*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ 
sqrt(a**2 - b**2))*a**4*b**4 - cos(c + d*x)*sin(c + d*x)**3*a**4*b**4*c + 
2*cos(c + d*x)*sin(c + d*x)**3*a**2*b**6*c - cos(c + d*x)*sin(c + d*x)**3* 
b**8*c + 6*cos(c + d*x)*sin(c + d*x)*a**6*b**2*c + 3*cos(c + d*x)*sin(c + 
d*x)*a**5*b**4 - 9*cos(c + d*x)*sin(c + d*x)*a**4*b**4*c - 6*cos(c + d*x)* 
sin(c + d*x)*a**3*b**6 + 3*cos(c + d*x)*sin(c + d*x)*a*b**8 + 3*cos(c + d* 
x)*sin(c + d*x)*b**8*c - 12*cos(c + d*x)*a**7*b*c**2 - 12*cos(c + d*x)*a** 
7*b*c*d*x - 6*cos(c + d*x)*a**6*b**3*c - 6*cos(c + d*x)*a**6*b**3*d*x + 21 
*cos(c + d*x)*a**5*b**3*c**2 + 21*cos(c + d*x)*a**5*b**3*c*d*x + 12*cos(c 
+ d*x)*a**4*b**5*c + 12*cos(c + d*x)*a**4*b**5*d*x - 6*cos(c + d*x)*a**...