\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [909]

Optimal result

Integrand size = 41, antiderivative size = 312 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (2 A b^2-4 a b B+6 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 b^4 d}-\frac {2 a \left (a^2 A b^2-2 A b^4-2 a^3 b B+3 a b^3 B+3 a^4 C-4 a^2 b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^4 (a+b)^{3/2} d}+\frac {\left (2 a^2 b B-b^3 B-a b^2 (A-2 C)-3 a^3 C\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (2 A b^2-2 a b B+3 a^2 C-b^2 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:

1/2*(2*A*b^2-4*B*a*b+6*C*a^2+C*b^2)*arctanh(sin(d*x+c))/b^4/d-2*a*(A*a^2*b 
^2-2*A*b^4-2*B*a^3*b+3*B*a*b^3+3*C*a^4-4*C*a^2*b^2)*arctanh((a-b)^(1/2)*ta 
n(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^4/(a+b)^(3/2)/d+(2*B*a^2*b-B*b 
^3-a*b^2*(A-2*C)-3*a^3*C)*tan(d*x+c)/b^3/(a^2-b^2)/d+1/2*(2*A*b^2-2*B*a*b+ 
3*C*a^2-C*b^2)*sec(d*x+c)*tan(d*x+c)/b^2/(a^2-b^2)/d-(A*b^2-a*(B*b-C*a))*s 
ec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 4.93 (sec) , antiderivative size = 519, normalized size of antiderivative = 1.66 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {(b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {8 a \left (-2 A b^4-2 a^3 b B+3 a b^3 B+a^2 b^2 (A-4 C)+3 a^4 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))}{\left (a^2-b^2\right )^{3/2}}-2 \left (2 A b^2-4 a b B+6 a^2 C+b^2 C\right ) (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (2 A b^2-4 a b B+6 a^2 C+b^2 C\right ) (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^2 C (b+a \cos (c+d x))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b (b B-2 a C) (b+a \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {b^2 C (b+a \cos (c+d x))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b (b B-2 a C) (b+a \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 a^2 b \left (A b^2+a (-b B+a C)\right ) \sin (c+d x)}{(-a+b) (a+b)}\right )}{2 b^4 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^2} \] Input:

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

Output:

((b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((8*a*(-2*A* 
b^4 - 2*a^3*b*B + 3*a*b^3*B + a^2*b^2*(A - 4*C) + 3*a^4*C)*ArcTanh[((-a + 
b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x]))/(a^2 - b^2)^(3 
/2) - 2*(2*A*b^2 - 4*a*b*B + 6*a^2*C + b^2*C)*(b + a*Cos[c + d*x])*Log[Cos 
[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(2*A*b^2 - 4*a*b*B + 6*a^2*C + b^2*C 
)*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^2*C*( 
b + a*Cos[c + d*x]))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*b*(b*B - 
 2*a*C)*(b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c 
+ d*x)/2]) - (b^2*C*(b + a*Cos[c + d*x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x 
)/2])^2 + (4*b*(b*B - 2*a*C)*(b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(Cos[( 
c + d*x)/2] + Sin[(c + d*x)/2]) + (4*a^2*b*(A*b^2 + a*(-(b*B) + a*C))*Sin[ 
c + d*x])/((-a + b)*(a + b))))/(2*b^4*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Co 
s[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)
 

Rubi [A] (verified)

Time = 2.09 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.04, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 4586, 3042, 4580, 25, 3042, 4570, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4586

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4580

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4486

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {2 a \left (-3 a^4 C+2 a^3 b B-a^2 b^2 (A-4 C)-3 a b^3 B+2 A b^4\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}+\frac {\left (a^2-b^2\right ) \left (6 a^2 C-4 a b B+2 A b^2+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {2 \tan (c+d x) \left (-3 a^3 C+2 a^2 b B-a b^2 (A-2 C)-b^3 B\right )}{b d}}{2 b}-\frac {\tan (c+d x) \sec (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right )}{2 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {-\frac {\frac {\frac {4 a \left (-3 a^4 C+2 a^3 b B-a^2 b^2 (A-4 C)-3 a b^3 B+2 A b^4\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}+\frac {\left (a^2-b^2\right ) \left (6 a^2 C-4 a b B+2 A b^2+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {2 \tan (c+d x) \left (-3 a^3 C+2 a^2 b B-a b^2 (A-2 C)-b^3 B\right )}{b d}}{2 b}-\frac {\tan (c+d x) \sec (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right )}{2 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

-(((A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*( 
a + b*Sec[c + d*x]))) - (-1/2*((2*A*b^2 - 2*a*b*B + 3*a^2*C - b^2*C)*Sec[c 
 + d*x]*Tan[c + d*x])/(b*d) - ((((a^2 - b^2)*(2*A*b^2 - 4*a*b*B + 6*a^2*C 
+ b^2*C)*ArcTanh[Sin[c + d*x]])/(b*d) + (4*a*(2*A*b^4 + 2*a^3*b*B - 3*a*b^ 
3*B - a^2*b^2*(A - 4*C) - 3*a^4*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/ 
Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b + (2*(2*a^2*b*B - b^3*B - a 
*b^2*(A - 2*C) - 3*a^3*C)*Tan[c + d*x])/(b*d))/(2*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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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 4257

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

rule 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4486

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( 
e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b   Int[Csc[e + f*x], 
 x], x] + Simp[(A*b - a*B)/b   Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x 
] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
 

rule 4570

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e 
_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S 
ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) 
)), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ 
b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; 
 FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]
 

rule 4580

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ 
(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x 
_Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 
1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[Csc[e + f*x]*(a + b*Csc[e 
+ f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* 
(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & 
& NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 4586

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[(-d)*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + 
b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 1)/(b*f*(a^2 - b^2)*(m + 1)) 
), x] + Simp[d/(b*(a^2 - b^2)*(m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*( 
d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) - a*(b*B - a*C)*(n - 1) + b*(a*A 
 - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(m + n + 1) + C*(a^2*n 
+ b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C 
}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]
 

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {-\frac {C}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 B b -4 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 A \,b^{2}-4 B a b +6 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}+\frac {2 a \left (\frac {a b \left (A \,b^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (A \,a^{2} b^{2}-2 A \,b^{4}-2 B \,a^{3} b +3 B a \,b^{3}+3 a^{4} C -4 C \,a^{2} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}+\frac {C}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B b -4 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 A \,b^{2}+4 B a b -6 C \,a^{2}-C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}}{d}\)	\(365\)
default	\(\frac {-\frac {C}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 B b -4 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 A \,b^{2}-4 B a b +6 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}+\frac {2 a \left (\frac {a b \left (A \,b^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (A \,a^{2} b^{2}-2 A \,b^{4}-2 B \,a^{3} b +3 B a \,b^{3}+3 a^{4} C -4 C \,a^{2} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}+\frac {C}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B b -4 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 A \,b^{2}+4 B a b -6 C \,a^{2}-C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}}{d}\)	\(365\)
risch	\(\text {Expression too large to display}\)	\(1753\)

Input:

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

Output:

1/d*(-1/2*C/b^2/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(2*B*b-4*C*a-C*b)/b^3/(tan(1/ 
2*d*x+1/2*c)+1)+1/2*(2*A*b^2-4*B*a*b+6*C*a^2+C*b^2)/b^4*ln(tan(1/2*d*x+1/2 
*c)+1)+2*a/b^4*(a*b*(A*b^2-B*a*b+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)-(A*a^2*b^2-2*A*b^4-2*B*a^3* 
b+3*B*a*b^3+3*C*a^4-4*C*a^2*b^2)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh(( 
a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))+1/2*C/b^2/(tan(1/2*d*x+1/2*c 
)-1)^2-1/2*(2*B*b-4*C*a-C*b)/b^3/(tan(1/2*d*x+1/2*c)-1)+1/2/b^4*(-2*A*b^2+ 
4*B*a*b-6*C*a^2-C*b^2)*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. 724 vs. \(2 (297) = 594\).

Time = 33.30 (sec) , antiderivative size = 1505, normalized size of antiderivative = 4.82 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(2*((3*C*a^6 - 2*B*a^5*b + (A - 4*C)*a^4*b^2 + 3*B*a^3*b^3 - 2*A*a^2* 
b^4)*cos(d*x + c)^3 + (3*C*a^5*b - 2*B*a^4*b^2 + (A - 4*C)*a^3*b^3 + 3*B*a 
^2*b^4 - 2*A*a*b^5)*cos(d*x + c)^2)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c 
) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)* 
sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2 
)) + ((6*C*a^7 - 4*B*a^6*b + (2*A - 11*C)*a^5*b^2 + 8*B*a^4*b^3 - 4*(A - C 
)*a^3*b^4 - 4*B*a^2*b^5 + (2*A + C)*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*b - 4 
*B*a^5*b^2 + (2*A - 11*C)*a^4*b^3 + 8*B*a^3*b^4 - 4*(A - C)*a^2*b^5 - 4*B* 
a*b^6 + (2*A + C)*b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((6*C*a^7 - 
 4*B*a^6*b + (2*A - 11*C)*a^5*b^2 + 8*B*a^4*b^3 - 4*(A - C)*a^3*b^4 - 4*B* 
a^2*b^5 + (2*A + C)*a*b^6)*cos(d*x + c)^3 + (6*C*a^6*b - 4*B*a^5*b^2 + (2* 
A - 11*C)*a^4*b^3 + 8*B*a^3*b^4 - 4*(A - C)*a^2*b^5 - 4*B*a*b^6 + (2*A + C 
)*b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(C*a^4*b^3 - 2*C*a^2*b^5 
 + C*b^7 - 2*(3*C*a^6*b - 2*B*a^5*b^2 + (A - 5*C)*a^4*b^3 + 3*B*a^3*b^4 - 
(A - 2*C)*a^2*b^5 - B*a*b^6)*cos(d*x + c)^2 - (3*C*a^5*b^2 - 2*B*a^4*b^3 - 
 6*C*a^3*b^4 + 4*B*a^2*b^5 + 3*C*a*b^6 - 2*B*b^7)*cos(d*x + c))*sin(d*x + 
c))/((a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3 + (a^4*b^5 - 2*a^2*b^7 
 + b^9)*d*cos(d*x + c)^2), -1/4*(4*((3*C*a^6 - 2*B*a^5*b + (A - 4*C)*a^4*b 
^2 + 3*B*a^3*b^3 - 2*A*a^2*b^4)*cos(d*x + c)^3 + (3*C*a^5*b - 2*B*a^4*b^2 
+ (A - 4*C)*a^3*b^3 + 3*B*a^2*b^4 - 2*A*a*b^5)*cos(d*x + c)^2)*sqrt(-a^...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(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*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.37 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, C a^{5} - 2 \, B a^{4} b + A a^{3} b^{2} - 4 \, C a^{3} b^{2} + 3 \, B a^{2} b^{3} - 2 \, A a 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^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {4 \, {\left (C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\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 {{\left (6 \, C a^{2} - 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} + \frac {{\left (6 \, C a^{2} - 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac {2 \, {\left (4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b \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 )}^{2} b^{3}}}{2 \, d} \] Input:

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

Output:

-1/2*(4*(3*C*a^5 - 2*B*a^4*b + A*a^3*b^2 - 4*C*a^3*b^2 + 3*B*a^2*b^3 - 2*A 
*a*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^4 - 
 b^6)*sqrt(-a^2 + b^2)) - 4*(C*a^4*tan(1/2*d*x + 1/2*c) - B*a^3*b*tan(1/2* 
d*x + 1/2*c) + A*a^2*b^2*tan(1/2*d*x + 1/2*c))/((a^2*b^3 - b^5)*(a*tan(1/2 
*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)) - (6*C*a^2 - 4*B*a*b 
+ 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + (6*C*a^2 - 4*B 
*a*b + 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 2*(4*C*a* 
tan(1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*x + 1/2*c)^3 + C*b*tan(1/2*d*x + 
1/2*c)^3 - 4*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c) + C*b*t 
an(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^3))/d
 

Mupad [B] (verification not implemented)

Time = 23.91 (sec) , antiderivative size = 9931, normalized size of antiderivative = 31.83 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((((8*(4*A*b^15 + 2*C*b^15 - 4*A*a^2*b^13 + 12*A*a^3*b^12 - 4*A*a^ 
5*b^10 + 12*B*a^2*b^13 + 12*B*a^3*b^12 - 20*B*a^4*b^11 - 4*B*a^5*b^10 + 8* 
B*a^6*b^9 + 6*C*a^2*b^13 - 16*C*a^3*b^12 - 14*C*a^4*b^11 + 28*C*a^5*b^10 + 
 6*C*a^6*b^9 - 12*C*a^7*b^8 - 8*A*a*b^14 - 8*B*a*b^14))/(a*b^11 + b^12 - a 
^2*b^10 - a^3*b^9) - (8*tan(c/2 + (d*x)/2)*(3*C*a^2 + b^2*(A + C/2) - 2*B* 
a*b)*(8*a*b^13 - 8*a^2*b^12 - 16*a^3*b^11 + 16*a^4*b^10 + 8*a^5*b^9 - 8*a^ 
6*b^8))/(b^4*(a*b^8 + b^9 - a^2*b^7 - a^3*b^6)))*(3*C*a^2 + b^2*(A + C/2) 
- 2*B*a*b))/b^4 - (8*tan(c/2 + (d*x)/2)*(4*A^2*b^10 + 72*C^2*a^10 + C^2*b^ 
10 - 8*A^2*a*b^9 - 2*C^2*a*b^9 - 72*C^2*a^9*b + 12*A^2*a^2*b^8 + 16*A^2*a^ 
3*b^7 - 20*A^2*a^4*b^6 - 8*A^2*a^5*b^5 + 8*A^2*a^6*b^4 + 16*B^2*a^2*b^8 - 
32*B^2*a^3*b^7 + 20*B^2*a^4*b^6 + 64*B^2*a^5*b^5 - 64*B^2*a^6*b^4 - 32*B^2 
*a^7*b^3 + 32*B^2*a^8*b^2 + 11*C^2*a^2*b^8 - 20*C^2*a^3*b^7 + 23*C^2*a^4*b 
^6 - 26*C^2*a^5*b^5 + 17*C^2*a^6*b^4 + 120*C^2*a^7*b^3 - 120*C^2*a^8*b^2 + 
 4*A*C*b^10 - 16*A*B*a*b^9 - 8*A*C*a*b^9 - 8*B*C*a*b^9 - 96*B*C*a^9*b + 32 
*A*B*a^2*b^8 - 32*A*B*a^3*b^7 - 64*A*B*a^4*b^6 + 72*A*B*a^5*b^5 + 32*A*B*a 
^6*b^4 - 32*A*B*a^7*b^3 + 20*A*C*a^2*b^8 - 32*A*C*a^3*b^7 + 36*A*C*a^4*b^6 
 + 88*A*C*a^5*b^5 - 100*A*C*a^6*b^4 - 48*A*C*a^7*b^3 + 48*A*C*a^8*b^2 + 16 
*B*C*a^2*b^8 - 40*B*C*a^3*b^7 + 64*B*C*a^4*b^6 - 40*B*C*a^5*b^5 - 176*B*C* 
a^6*b^4 + 176*B*C*a^7*b^3 + 96*B*C*a^8*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3* 
b^6))*(3*C*a^2 + b^2*(A + C/2) - 2*B*a*b)*1i)/b^4 - (((((8*(4*A*b^15 + ...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 2974, normalized size of antiderivative = 9.53 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 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**6*c + 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 + 16*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**2*c - 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**3*b**4 + 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**6*c - 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**5*b**2 - 16*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**2*c + 
 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t( - a**2 + b**2))*cos(c + d*x)*a**3*b**4 - 12*sqrt( - a**2 + b**2)*atan(( 
tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x 
)**2*a**5*b*c + 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + 
 d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**4*b**3 + 16*sqrt( - a 
**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + 
b**2))*sin(c + d*x)**2*a**3*b**3*c - 4*sqrt( - a**2 + b**2)*atan((tan((c + 
 d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*...