\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [999]

Optimal result

Integrand size = 41, antiderivative size = 339 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (15 a^2 A b^4-6 A b^6+6 a^5 b B-5 a^3 b^3 B+2 a b^5 B-2 a^6 C-a^4 b^2 (12 A+C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}-\frac {(3 A b-a B) \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {\left (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

-(15*a^2*A*b^4-6*A*b^6+6*B*a^5*b-5*B*a^3*b^3+2*B*a*b^5-2*a^6*C-a^4*b^2*(12 
*A+C))*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^4/(a-b)^(5/2)/ 
(a+b)^(5/2)/d-(3*A*b-B*a)*arctanh(sin(d*x+c))/a^4/d-1/2*(11*A*a^2*b^2-6*A* 
b^4-5*B*a^3*b+2*B*a*b^3-a^4*(2*A-3*C))*tan(d*x+c)/a^3/(a^2-b^2)^2/d+1/2*(A 
*b^2-a*(B*b-C*a))*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2-1/2*(3*A*b^4 
+4*B*a^3*b-B*a*b^3-2*a^4*C-a^2*b^2*(6*A+C))*tan(d*x+c)/a^2/(a^2-b^2)^2/d/( 
a+b*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 4.77 (sec) , antiderivative size = 444, normalized size of antiderivative = 1.31 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\cos (c+d x) \left (C+B \sec (c+d x)+A \sec ^2(c+d x)\right ) \left (-\frac {8 \left (-15 a^2 A b^4+6 A b^6-6 a^5 b B+5 a^3 b^3 B-2 a b^5 B+2 a^6 C+a^4 b^2 (12 A+C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right ) \cos (c+d x)}{\left (-a^2+b^2\right )^{5/2}}+8 (3 A b-a B) \cos (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 (-3 A b+a B) \cos (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 a \left (4 a^6 A-6 a^4 A b^2-7 a^2 A b^4+6 A b^6+5 a^3 b^3 B-2 a b^5 B-3 a^4 b^2 C+2 a b \left (9 A b^4+6 a^3 b B-3 a b^3 B+4 a^4 (A-C)+a^2 b^2 (-16 A+C)\right ) \cos (c+d x)+b^2 \left (-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B+a^4 (2 A-3 C)\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}\right )}{4 a^4 d (2 A+C+2 B \cos (c+d x)+C \cos (2 (c+d x)))} \] Input:

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

Output:

(Cos[c + d*x]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*((-8*(-15*a^2*A*b^4 
+ 6*A*b^6 - 6*a^5*b*B + 5*a^3*b^3*B - 2*a*b^5*B + 2*a^6*C + a^4*b^2*(12*A 
+ C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]*Cos[c + d*x])/( 
-a^2 + b^2)^(5/2) + 8*(3*A*b - a*B)*Cos[c + d*x]*Log[Cos[(c + d*x)/2] - Si 
n[(c + d*x)/2]] + 8*(-3*A*b + a*B)*Cos[c + d*x]*Log[Cos[(c + d*x)/2] + Sin 
[(c + d*x)/2]] + (2*a*(4*a^6*A - 6*a^4*A*b^2 - 7*a^2*A*b^4 + 6*A*b^6 + 5*a 
^3*b^3*B - 2*a*b^5*B - 3*a^4*b^2*C + 2*a*b*(9*A*b^4 + 6*a^3*b*B - 3*a*b^3* 
B + 4*a^4*(A - C) + a^2*b^2*(-16*A + C))*Cos[c + d*x] + b^2*(-11*a^2*A*b^2 
 + 6*A*b^4 + 5*a^3*b*B - 2*a*b^3*B + a^4*(2*A - 3*C))*Cos[2*(c + d*x)])*Si 
n[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2)))/(4*a^4*d*(2*A + C + 2 
*B*Cos[c + d*x] + C*Cos[2*(c + d*x)]))
 

Rubi [A] (verified)

Time = 2.29 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 3534, 25, 3042, 3534, 3042, 3534, 3042, 3480, 3042, 3138, 218, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\int \frac {\left (-\left ((2 A-3 C) a^4\right )-5 b B a^3+11 A b^2 a^2+2 b^3 B a-\left (2 B a^3-b (4 A+3 C) a^2+b^2 B a+A b^3\right ) \cos (c+d x) a-6 A b^4+\left (-2 C a^4+4 b B a^3-b^2 (6 A+C) a^2-b^3 B a+3 A b^4\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\tan (c+d x) \left (-2 a^4 C+4 a^3 b B-a^2 b^2 (6 A+C)-a b^3 B+3 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\int \frac {-\left ((2 A-3 C) a^4\right )-5 b B a^3+11 A b^2 a^2+2 b^3 B a-\left (2 B a^3-b (4 A+3 C) a^2+b^2 B a+A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-6 A b^4+\left (-2 C a^4+4 b B a^3-b^2 (6 A+C) a^2-b^3 B a+3 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {\tan (c+d x) \left (-2 a^4 C+4 a^3 b B-a^2 b^2 (6 A+C)-a b^3 B+3 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\int \frac {\left (2 (3 A b-a B) \left (a^2-b^2\right )^2+a \left (-2 C a^4+4 b B a^3-b^2 (6 A+C) a^2-b^3 B a+3 A b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\tan (c+d x) \left (-\left (a^4 (2 A-3 C)\right )-5 a^3 b B+11 a^2 A b^2+2 a b^3 B-6 A b^4\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {\tan (c+d x) \left (-2 a^4 C+4 a^3 b B-a^2 b^2 (6 A+C)-a b^3 B+3 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\int \frac {2 (3 A b-a B) \left (a^2-b^2\right )^2+a \left (-2 C a^4+4 b B a^3-b^2 (6 A+C) a^2-b^3 B a+3 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {\tan (c+d x) \left (-\left (a^4 (2 A-3 C)\right )-5 a^3 b B+11 a^2 A b^2+2 a b^3 B-6 A b^4\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {\tan (c+d x) \left (-2 a^4 C+4 a^3 b B-a^2 b^2 (6 A+C)-a b^3 B+3 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3480

\(\displaystyle \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \int \sec (c+d x)dx}{a}+\frac {\left (-2 a^6 C+6 a^5 b B-a^4 b^2 (12 A+C)-5 a^3 b^3 B+15 a^2 A b^4+2 a b^5 B-6 A b^6\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}+\frac {\tan (c+d x) \left (-\left (a^4 (2 A-3 C)\right )-5 a^3 b B+11 a^2 A b^2+2 a b^3 B-6 A b^4\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {\tan (c+d x) \left (-2 a^4 C+4 a^3 b B-a^2 b^2 (6 A+C)-a b^3 B+3 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {\left (-2 a^6 C+6 a^5 b B-a^4 b^2 (12 A+C)-5 a^3 b^3 B+15 a^2 A b^4+2 a b^5 B-6 A b^6\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}+\frac {\tan (c+d x) \left (-\left (a^4 (2 A-3 C)\right )-5 a^3 b B+11 a^2 A b^2+2 a b^3 B-6 A b^4\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {\tan (c+d x) \left (-2 a^4 C+4 a^3 b B-a^2 b^2 (6 A+C)-a b^3 B+3 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 \left (-2 a^6 C+6 a^5 b B-a^4 b^2 (12 A+C)-5 a^3 b^3 B+15 a^2 A b^4+2 a b^5 B-6 A b^6\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}+\frac {\tan (c+d x) \left (-\left (a^4 (2 A-3 C)\right )-5 a^3 b B+11 a^2 A b^2+2 a b^3 B-6 A b^4\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {\tan (c+d x) \left (-2 a^4 C+4 a^3 b B-a^2 b^2 (6 A+C)-a b^3 B+3 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 \left (-2 a^6 C+6 a^5 b B-a^4 b^2 (12 A+C)-5 a^3 b^3 B+15 a^2 A b^4+2 a b^5 B-6 A b^6\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}+\frac {\tan (c+d x) \left (-\left (a^4 (2 A-3 C)\right )-5 a^3 b B+11 a^2 A b^2+2 a b^3 B-6 A b^4\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {\tan (c+d x) \left (-2 a^4 C+4 a^3 b B-a^2 b^2 (6 A+C)-a b^3 B+3 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\tan (c+d x) \left (-2 a^4 C+4 a^3 b B-a^2 b^2 (6 A+C)-a b^3 B+3 A b^4\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\tan (c+d x) \left (-\left (a^4 (2 A-3 C)\right )-5 a^3 b B+11 a^2 A b^2+2 a b^3 B-6 A b^4\right )}{a d}+\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \text {arctanh}(\sin (c+d x))}{a d}+\frac {2 \left (-2 a^6 C+6 a^5 b B-a^4 b^2 (12 A+C)-5 a^3 b^3 B+15 a^2 A b^4+2 a b^5 B-6 A b^6\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

Input:

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

Output:

((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d 
*x])^2) - (((3*A*b^4 + 4*a^3*b*B - a*b^3*B - 2*a^4*C - a^2*b^2*(6*A + C))* 
Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) + (((2*(15*a^2*A*b^4 
- 6*A*b^6 + 6*a^5*b*B - 5*a^3*b^3*B + 2*a*b^5*B - 2*a^6*C - a^4*b^2*(12*A 
+ C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*S 
qrt[a + b]*d) + (2*(a^2 - b^2)^2*(3*A*b - a*B)*ArcTanh[Sin[c + d*x]])/(a*d 
))/a + ((11*a^2*A*b^2 - 6*A*b^4 - 5*a^3*b*B + 2*a*b^3*B - a^4*(2*A - 3*C)) 
*Tan[c + d*x])/(a*d))/(a*(a^2 - b^2)))/(2*a*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 3480

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b 
- a*B)/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ 
(b*c - a*d)   Int[1/(c + d*Sin[e + f*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]
 

rule 3534

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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-3 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}+\frac {\frac {2 \left (-\frac {\left (8 A \,a^{2} b^{2}+a A \,b^{3}-4 A \,b^{4}-6 B \,a^{3} b -B \,a^{2} b^{2}+2 B a \,b^{3}+4 a^{4} C +a^{3} b C \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (8 A \,a^{2} b^{2}-a A \,b^{3}-4 A \,b^{4}-6 B \,a^{3} b +B \,a^{2} b^{2}+2 B a \,b^{3}+4 a^{4} C -a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (12 A \,a^{4} b^{2}-15 a^{2} A \,b^{4}+6 A \,b^{6}-6 B \,a^{5} b +5 B \,a^{3} b^{3}-2 B a \,b^{5}+2 a^{6} C +a^{4} b^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{a^{4}}-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (3 A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}}{d}\)	\(427\)
default	\(\frac {-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-3 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}+\frac {\frac {2 \left (-\frac {\left (8 A \,a^{2} b^{2}+a A \,b^{3}-4 A \,b^{4}-6 B \,a^{3} b -B \,a^{2} b^{2}+2 B a \,b^{3}+4 a^{4} C +a^{3} b C \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (8 A \,a^{2} b^{2}-a A \,b^{3}-4 A \,b^{4}-6 B \,a^{3} b +B \,a^{2} b^{2}+2 B a \,b^{3}+4 a^{4} C -a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (12 A \,a^{4} b^{2}-15 a^{2} A \,b^{4}+6 A \,b^{6}-6 B \,a^{5} b +5 B \,a^{3} b^{3}-2 B a \,b^{5}+2 a^{6} C +a^{4} b^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{a^{4}}-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (3 A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}}{d}\)	\(427\)
risch	\(\text {Expression too large to display}\)	\(2188\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x,meth 
od=_RETURNVERBOSE)
 

Output:

1/d*(-A/a^3/(tan(1/2*d*x+1/2*c)+1)+1/a^4*(-3*A*b+B*a)*ln(tan(1/2*d*x+1/2*c 
)+1)+2/a^4*((-1/2*(8*A*a^2*b^2+A*a*b^3-4*A*b^4-6*B*a^3*b-B*a^2*b^2+2*B*a*b 
^3+4*C*a^4+C*a^3*b)*a*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*b*a 
*(8*A*a^2*b^2-A*a*b^3-4*A*b^4-6*B*a^3*b+B*a^2*b^2+2*B*a*b^3+4*C*a^4-C*a^3* 
b)/(a+b)/(a-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+1/2*(12*A*a^4*b^2-15*A*a^2*b^4+6*A*b^6-6*B*a^5*b+5*B*a^3* 
b^3-2*B*a*b^5+2*C*a^6+C*a^4*b^2)/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*a 
rctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))-A/a^3/(tan(1/2*d*x+1/ 
2*c)-1)+(3*A*b-B*a)/a^4*ln(tan(1/2*d*x+1/2*c)-1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1071 vs. \(2 (316) = 632\).

Time = 62.87 (sec) , antiderivative size = 2213, normalized size of antiderivative = 6.53 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*(((2*C*a^6*b^2 - 6*B*a^5*b^3 + (12*A + C)*a^4*b^4 + 5*B*a^3*b^5 - 15 
*A*a^2*b^6 - 2*B*a*b^7 + 6*A*b^8)*cos(d*x + c)^3 + 2*(2*C*a^7*b - 6*B*a^6* 
b^2 + (12*A + C)*a^5*b^3 + 5*B*a^4*b^4 - 15*A*a^3*b^5 - 2*B*a^2*b^6 + 6*A* 
a*b^7)*cos(d*x + c)^2 + (2*C*a^8 - 6*B*a^7*b + (12*A + C)*a^6*b^2 + 5*B*a^ 
5*b^3 - 15*A*a^4*b^4 - 2*B*a^3*b^5 + 6*A*a^2*b^6)*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^7*b^2 - 3*A*a^6*b^3 - 3*B*a^5*b 
^4 + 9*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7 - B*a*b^8 + 3*A*b^9)*cos(d*x 
+ c)^3 + 2*(B*a^8*b - 3*A*a^7*b^2 - 3*B*a^6*b^3 + 9*A*a^5*b^4 + 3*B*a^4*b^ 
5 - 9*A*a^3*b^6 - B*a^2*b^7 + 3*A*a*b^8)*cos(d*x + c)^2 + (B*a^9 - 3*A*a^8 
*b - 3*B*a^7*b^2 + 9*A*a^6*b^3 + 3*B*a^5*b^4 - 9*A*a^4*b^5 - B*a^3*b^6 + 3 
*A*a^2*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + 2*((B*a^7*b^2 - 3*A*a^6* 
b^3 - 3*B*a^5*b^4 + 9*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7 - B*a*b^8 + 3* 
A*b^9)*cos(d*x + c)^3 + 2*(B*a^8*b - 3*A*a^7*b^2 - 3*B*a^6*b^3 + 9*A*a^5*b 
^4 + 3*B*a^4*b^5 - 9*A*a^3*b^6 - B*a^2*b^7 + 3*A*a*b^8)*cos(d*x + c)^2 + ( 
B*a^9 - 3*A*a^8*b - 3*B*a^7*b^2 + 9*A*a^6*b^3 + 3*B*a^5*b^4 - 9*A*a^4*b^5 
- B*a^3*b^6 + 3*A*a^2*b^7)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*A*a 
^9 - 6*A*a^7*b^2 + 6*A*a^5*b^4 - 2*A*a^3*b^6 + ((2*A - 3*C)*a^7*b^2 + 5*B* 
a^6*b^3 - (13*A - 3*C)*a^5*b^4 - 7*B*a^4*b^5 + 17*A*a^3*b^6 + 2*B*a^2*b...
 

Sympy [F]

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

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

Output:

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**2/(a + b*c 
os(c + d*x))**3, x)
 

Maxima [F(-2)]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3, 
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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 699 vs. \(2 (316) = 632\).

Time = 0.21 (sec) , antiderivative size = 699, normalized size of antiderivative = 2.06 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-((2*C*a^6 - 6*B*a^5*b + 12*A*a^4*b^2 + C*a^4*b^2 + 5*B*a^3*b^3 - 15*A*a^2 
*b^4 - 2*B*a*b^5 + 6*A*b^6)*(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^8 - 2*a^6*b^2 + a^4*b^4)*sqrt(a^2 - b^2)) + (4*C*a^5*b*tan(1/2 
*d*x + 1/2*c)^3 - 6*B*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^4*b^2*tan(1/2 
*d*x + 1/2*c)^3 + 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 5*B*a^3*b^3*tan(1/2 
*d*x + 1/2*c)^3 - C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 7*A*a^2*b^4*tan(1/2*d 
*x + 1/2*c)^3 + 3*B*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 5*A*a*b^5*tan(1/2*d*x 
 + 1/2*c)^3 - 2*B*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 4*A*b^6*tan(1/2*d*x + 1/2 
*c)^3 + 4*C*a^5*b*tan(1/2*d*x + 1/2*c) - 6*B*a^4*b^2*tan(1/2*d*x + 1/2*c) 
+ 3*C*a^4*b^2*tan(1/2*d*x + 1/2*c) + 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c) - 5* 
B*a^3*b^3*tan(1/2*d*x + 1/2*c) - C*a^3*b^3*tan(1/2*d*x + 1/2*c) + 7*A*a^2* 
b^4*tan(1/2*d*x + 1/2*c) + 3*B*a^2*b^4*tan(1/2*d*x + 1/2*c) - 5*A*a*b^5*ta 
n(1/2*d*x + 1/2*c) + 2*B*a*b^5*tan(1/2*d*x + 1/2*c) - 4*A*b^6*tan(1/2*d*x 
+ 1/2*c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1 
/2*d*x + 1/2*c)^2 + a + b)^2) - (B*a - 3*A*b)*log(abs(tan(1/2*d*x + 1/2*c) 
 + 1))/a^4 + (B*a - 3*A*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 2*A*ta 
n(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3))/d
 

Mupad [B] (verification not implemented)

Time = 8.08 (sec) , antiderivative size = 11417, normalized size of antiderivative = 33.68 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((((3*A*b - B*a)*(((3*A*b - B*a)*((8*(4*B*a^18 + 4*C*a^18 + 12*A*a^8* 
b^10 - 6*A*a^9*b^9 - 54*A*a^10*b^8 + 24*A*a^11*b^7 + 96*A*a^12*b^6 - 42*A* 
a^13*b^5 - 78*A*a^14*b^4 + 36*A*a^15*b^3 + 24*A*a^16*b^2 - 4*B*a^9*b^9 + 2 
*B*a^10*b^8 + 18*B*a^11*b^7 - 4*B*a^12*b^6 - 36*B*a^13*b^5 + 6*B*a^14*b^4 
+ 34*B*a^15*b^3 - 8*B*a^16*b^2 - 2*C*a^11*b^7 + 2*C*a^12*b^6 + 6*C*a^15*b^ 
3 - 6*C*a^16*b^2 - 12*A*a^17*b - 12*B*a^17*b - 4*C*a^17*b))/(a^15*b + a^16 
 - a^9*b^7 - a^10*b^6 + 3*a^11*b^5 + 3*a^12*b^4 - 3*a^13*b^3 - 3*a^14*b^2) 
 - (8*tan(c/2 + (d*x)/2)*(3*A*b - B*a)*(8*a^17*b - 8*a^8*b^10 + 8*a^9*b^9 
+ 32*a^10*b^8 - 32*a^11*b^7 - 48*a^12*b^6 + 48*a^13*b^5 + 32*a^14*b^4 - 32 
*a^15*b^3 - 8*a^16*b^2))/(a^4*(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b 
^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2))))/a^4 - (8*tan(c/2 + (d*x)/2)*( 
72*A^2*b^12 + 4*B^2*a^12 + 4*C^2*a^12 - 72*A^2*a*b^11 - 8*B^2*a^11*b - 288 
*A^2*a^2*b^10 + 288*A^2*a^3*b^9 + 441*A^2*a^4*b^8 - 432*A^2*a^5*b^7 - 288* 
A^2*a^6*b^6 + 288*A^2*a^7*b^5 + 36*A^2*a^8*b^4 - 72*A^2*a^9*b^3 + 36*A^2*a 
^10*b^2 + 8*B^2*a^2*b^10 - 8*B^2*a^3*b^9 - 32*B^2*a^4*b^8 + 32*B^2*a^5*b^7 
 + 57*B^2*a^6*b^6 - 48*B^2*a^7*b^5 - 52*B^2*a^8*b^4 + 32*B^2*a^9*b^3 + 24* 
B^2*a^10*b^2 + C^2*a^8*b^4 + 4*C^2*a^10*b^2 - 48*A*B*a*b^11 - 24*A*B*a^11* 
b - 24*B*C*a^11*b + 48*A*B*a^2*b^10 + 192*A*B*a^3*b^9 - 192*A*B*a^4*b^8 - 
318*A*B*a^5*b^7 + 288*A*B*a^6*b^6 + 252*A*B*a^7*b^5 - 192*A*B*a^8*b^4 - 72 
*A*B*a^9*b^3 + 48*A*B*a^10*b^2 + 12*A*C*a^4*b^8 - 6*A*C*a^6*b^6 - 36*A*...
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 2778, normalized size of antiderivative = 8.19 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(4*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)*sin(c + d*x)**2*a**5*b**2*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)*sin(c + d*x)**2*a**4*b**4 + 2*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)*sin(c + d*x)* 
*2*a**3*b**4*c - 20*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)*sin(c + d*x)**2*a**2*b**6 + 8*s 
qrt(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)*sin(c + d*x)**2*b**8 - 4*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**7 
*c - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/s 
qrt(a**2 - b**2))*cos(c + d*x)*a**6*b**2 - 6*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 
**2*c + 8*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**4 - 2*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**4*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**2*b**6 - 8*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)*b 
**8 + 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b...