3.7.64 \(\int \sec (c+d x) (a+b \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\) [664]

3.7.64.1 Optimal result
3.7.64.2 Mathematica [A] (verified)
3.7.64.3 Rubi [A] (verified)
3.7.64.4 Maple [A] (verified)
3.7.64.5 Fricas [A] (verification not implemented)
3.7.64.6 Sympy [F]
3.7.64.7 Maxima [A] (verification not implemented)
3.7.64.8 Giac [B] (verification not implemented)
3.7.64.9 Mupad [B] (verification not implemented)

3.7.64.1 Optimal result

Integrand size = 31, antiderivative size = 310 \[ \int \sec (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \text {arctanh}(\sin (c+d x))}{16 d}-\frac {a \left (4 a^4 C-32 b^4 (5 A+4 C)-a^2 b^2 (190 A+121 C)\right ) \tan (c+d x)}{60 b d}-\frac {\left (8 a^4 C-15 b^4 (6 A+5 C)-2 a^2 b^2 (130 A+89 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {a \left (70 A b^2-4 a^2 C+53 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac {a C (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {C (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d} \]

output
1/16*(8*a^4*(2*A+C)+12*a^2*b^2*(4*A+3*C)+b^4*(6*A+5*C))*arctanh(sin(d*x+c) 
)/d-1/60*a*(4*a^4*C-32*b^4*(5*A+4*C)-a^2*b^2*(190*A+121*C))*tan(d*x+c)/b/d 
-1/240*(8*a^4*C-15*b^4*(6*A+5*C)-2*a^2*b^2*(130*A+89*C))*sec(d*x+c)*tan(d* 
x+c)/d+1/120*a*(70*A*b^2-4*C*a^2+53*C*b^2)*(a+b*sec(d*x+c))^2*tan(d*x+c)/b 
/d-1/120*(4*C*a^2-5*b^2*(6*A+5*C))*(a+b*sec(d*x+c))^3*tan(d*x+c)/b/d-1/30* 
a*C*(a+b*sec(d*x+c))^4*tan(d*x+c)/b/d+1/6*C*(a+b*sec(d*x+c))^5*tan(d*x+c)/ 
b/d
 
3.7.64.2 Mathematica [A] (verified)

Time = 8.88 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.67 \[ \int \sec (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 \left (8 a^4 C+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \sec (c+d x)+10 b^2 \left (6 A b^2+36 a^2 C+5 b^2 C\right ) \sec ^3(c+d x)+40 b^4 C \sec ^5(c+d x)+64 a b \left (15 \left (a^2+b^2\right ) (A+C)+5 \left (A b^2+a^2 C+2 b^2 C\right ) \tan ^2(c+d x)+3 b^2 C \tan ^4(c+d x)\right )\right )}{240 d} \]

input
Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]
 
output
(15*(8*a^4*(2*A + C) + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*ArcTanh[S 
in[c + d*x]] + Tan[c + d*x]*(15*(8*a^4*C + 12*a^2*b^2*(4*A + 3*C) + b^4*(6 
*A + 5*C))*Sec[c + d*x] + 10*b^2*(6*A*b^2 + 36*a^2*C + 5*b^2*C)*Sec[c + d* 
x]^3 + 40*b^4*C*Sec[c + d*x]^5 + 64*a*b*(15*(a^2 + b^2)*(A + C) + 5*(A*b^2 
 + a^2*C + 2*b^2*C)*Tan[c + d*x]^2 + 3*b^2*C*Tan[c + d*x]^4)))/(240*d)
 
3.7.64.3 Rubi [A] (verified)

Time = 1.84 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.05, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {3042, 4571, 3042, 4490, 3042, 4490, 27, 3042, 4490, 3042, 4485, 3042, 4274, 3042, 4254, 24, 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.

\(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4571

\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^4 (b (6 A+5 C)-a C \sec (c+d x))dx}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{5} \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (3 a b (10 A+7 C)-\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \sec (c+d x)\right )dx-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (3 a b (10 A+7 C)+\left (5 b^2 (6 A+5 C)-4 a^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{4} \int 3 \sec (c+d x) (a+b \sec (c+d x))^2 \left (b \left (8 (5 A+3 C) a^2+5 b^2 (6 A+5 C)\right )+a \left (-4 C a^2+70 A b^2+53 b^2 C\right ) \sec (c+d x)\right )dx-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (b \left (8 (5 A+3 C) a^2+5 b^2 (6 A+5 C)\right )+a \left (-4 C a^2+70 A b^2+53 b^2 C\right ) \sec (c+d x)\right )dx-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (b \left (8 (5 A+3 C) a^2+5 b^2 (6 A+5 C)\right )+a \left (-4 C a^2+70 A b^2+53 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \int \sec (c+d x) (a+b \sec (c+d x)) \left (a b \left (8 (15 A+8 C) a^2+b^2 (230 A+181 C)\right )-\left (8 C a^4-2 b^2 (130 A+89 C) a^2-15 b^4 (6 A+5 C)\right ) \sec (c+d x)\right )dx+\frac {a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a b \left (8 (15 A+8 C) a^2+b^2 (230 A+181 C)\right )+\left (-8 C a^4+2 b^2 (130 A+89 C) a^2+15 b^4 (6 A+5 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 4485

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \sec (c+d x) \left (15 b \left (8 (2 A+C) a^4+12 b^2 (4 A+3 C) a^2+b^4 (6 A+5 C)\right )-4 a \left (4 C a^4-b^2 (190 A+121 C) a^2-32 b^4 (5 A+4 C)\right ) \sec (c+d x)\right )dx-\frac {b \left (8 a^4 C-2 a^2 b^2 (130 A+89 C)-15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (15 b \left (8 (2 A+C) a^4+12 b^2 (4 A+3 C) a^2+b^4 (6 A+5 C)\right )-4 a \left (4 C a^4-b^2 (190 A+121 C) a^2-32 b^4 (5 A+4 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {b \left (8 a^4 C-2 a^2 b^2 (130 A+89 C)-15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 b \left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \int \sec (c+d x)dx-4 a \left (4 a^4 C-a^2 b^2 (190 A+121 C)-32 b^4 (5 A+4 C)\right ) \int \sec ^2(c+d x)dx\right )-\frac {b \left (8 a^4 C-2 a^2 b^2 (130 A+89 C)-15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 b \left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-4 a \left (4 a^4 C-a^2 b^2 (190 A+121 C)-32 b^4 (5 A+4 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )-\frac {b \left (8 a^4 C-2 a^2 b^2 (130 A+89 C)-15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {4 a \left (4 a^4 C-a^2 b^2 (190 A+121 C)-32 b^4 (5 A+4 C)\right ) \int 1d(-\tan (c+d x))}{d}+15 b \left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )-\frac {b \left (8 a^4 C-2 a^2 b^2 (130 A+89 C)-15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 b \left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {4 a \left (4 a^4 C-a^2 b^2 (190 A+121 C)-32 b^4 (5 A+4 C)\right ) \tan (c+d x)}{d}\right )-\frac {b \left (8 a^4 C-2 a^2 b^2 (130 A+89 C)-15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {a \left (-4 a^2 C+70 A b^2+53 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {1}{2} \left (\frac {15 b \left (8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {4 a \left (4 a^4 C-a^2 b^2 (190 A+121 C)-32 b^4 (5 A+4 C)\right ) \tan (c+d x)}{d}\right )-\frac {b \left (8 a^4 C-2 a^2 b^2 (130 A+89 C)-15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )\right )-\frac {\left (4 a^2 C-5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}}{6 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}\)

input
Int[Sec[c + d*x]*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]
 
output
(C*(a + b*Sec[c + d*x])^5*Tan[c + d*x])/(6*b*d) + (-1/5*(a*C*(a + b*Sec[c 
+ d*x])^4*Tan[c + d*x])/d + (-1/4*((4*a^2*C - 5*b^2*(6*A + 5*C))*(a + b*Se 
c[c + d*x])^3*Tan[c + d*x])/d + (3*((a*(70*A*b^2 - 4*a^2*C + 53*b^2*C)*(a 
+ b*Sec[c + d*x])^2*Tan[c + d*x])/(3*d) + (-1/2*(b*(8*a^4*C - 15*b^4*(6*A 
+ 5*C) - 2*a^2*b^2*(130*A + 89*C))*Sec[c + d*x]*Tan[c + d*x])/d + ((15*b*( 
8*a^4*(2*A + C) + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*ArcTanh[Sin[c 
+ d*x]])/d - (4*a*(4*a^4*C - 32*b^4*(5*A + 4*C) - a^2*b^2*(190*A + 121*C)) 
*Tan[c + d*x])/d)/2)/3))/4)/5)/(6*b)
 

3.7.64.3.1 Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4254
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4274
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4485
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ 
e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1)   Int[(d*Csc 
[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x 
], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[ 
n, -1]
 

rule 4490
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[Csc[e + f*x]* 
(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* 
B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
 

rule 4571
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[ 
(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x] 
*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   In 
t[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 
3.7.64.4 Maple [A] (verified)

Time = 1.57 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.94

method result size
parts \(\frac {\left (A \,b^{4}+6 C \,a^{2} b^{2}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (4 a A \,b^{3}+4 a^{3} b C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (6 A \,a^{2} b^{2}+a^{4} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {C \,b^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}+\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{4}}{d}+\frac {4 A \,a^{3} b \tan \left (d x +c \right )}{d}-\frac {4 C a \,b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) \(291\)
derivativedivides \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 A \,a^{3} b \tan \left (d x +c \right )-4 a^{3} b C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 A \,a^{2} b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 C \,a^{2} b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 a A \,b^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-4 C a \,b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+A \,b^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \,b^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) \(360\)
default \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 A \,a^{3} b \tan \left (d x +c \right )-4 a^{3} b C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 A \,a^{2} b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 C \,a^{2} b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 a A \,b^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-4 C a \,b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+A \,b^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \,b^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) \(360\)
parallelrisch \(\frac {-240 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (\left (\frac {3 A}{8}+\frac {5 C}{16}\right ) b^{4}+3 \left (A +\frac {3 C}{4}\right ) a^{2} b^{2}+a^{4} \left (A +\frac {C}{2}\right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+240 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (\left (\frac {3 A}{8}+\frac {5 C}{16}\right ) b^{4}+3 \left (A +\frac {3 C}{4}\right ) a^{2} b^{2}+a^{4} \left (A +\frac {C}{2}\right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (1020 A +850 C \right ) b^{4}+4320 a^{2} \left (A +\frac {17 C}{12}\right ) b^{2}+720 a^{4} C \right ) \sin \left (3 d x +3 c \right )+\left (\left (180 A +150 C \right ) b^{4}+1440 \left (A +\frac {3 C}{4}\right ) a^{2} b^{2}+240 a^{4} C \right ) \sin \left (5 d x +5 c \right )+4800 a b \left (\left (\frac {6 A}{5}+\frac {8 C}{5}\right ) b^{2}+a^{2} \left (A +\frac {6 C}{5}\right )\right ) \sin \left (2 d x +2 c \right )+3840 a b \left (b^{2} \left (A +\frac {4 C}{5}\right )+\left (A +C \right ) a^{2}\right ) \sin \left (4 d x +4 c \right )+960 a b \left (\left (\frac {2 A}{3}+\frac {8 C}{15}\right ) b^{2}+a^{2} \left (A +\frac {2 C}{3}\right )\right ) \sin \left (6 d x +6 c \right )+2880 \left (\left (\frac {7 A}{24}+\frac {11 C}{16}\right ) b^{4}+a^{2} \left (A +\frac {7 C}{4}\right ) b^{2}+\frac {a^{4} C}{6}\right ) \sin \left (d x +c \right )}{240 d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\) \(422\)
norman \(\frac {-\frac {\left (64 A \,a^{3} b -48 A \,a^{2} b^{2}+64 a A \,b^{3}-10 A \,b^{4}-8 a^{4} C +64 a^{3} b C -60 C \,a^{2} b^{2}+64 C a \,b^{3}-11 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {\left (64 A \,a^{3} b +48 A \,a^{2} b^{2}+64 a A \,b^{3}+10 A \,b^{4}+8 a^{4} C +64 a^{3} b C +60 C \,a^{2} b^{2}+64 C a \,b^{3}+11 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (960 A \,a^{3} b -432 A \,a^{2} b^{2}+704 a A \,b^{3}-42 A \,b^{4}-72 a^{4} C +704 a^{3} b C -252 C \,a^{2} b^{2}+448 C a \,b^{3}+5 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}-\frac {\left (960 A \,a^{3} b +432 A \,a^{2} b^{2}+704 a A \,b^{3}+42 A \,b^{4}+72 a^{4} C +704 a^{3} b C +252 C \,a^{2} b^{2}+448 C a \,b^{3}-5 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}-\frac {\left (1600 A \,a^{3} b -240 A \,a^{2} b^{2}+960 a A \,b^{3}-10 A \,b^{4}-40 a^{4} C +960 a^{3} b C -60 C \,a^{2} b^{2}+832 C a \,b^{3}-75 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 d}+\frac {\left (1600 A \,a^{3} b +240 A \,a^{2} b^{2}+960 a A \,b^{3}+10 A \,b^{4}+40 a^{4} C +960 a^{3} b C +60 C \,a^{2} b^{2}+832 C a \,b^{3}+75 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6}}-\frac {\left (16 a^{4} A +48 A \,a^{2} b^{2}+6 A \,b^{4}+8 a^{4} C +36 C \,a^{2} b^{2}+5 C \,b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {\left (16 a^{4} A +48 A \,a^{2} b^{2}+6 A \,b^{4}+8 a^{4} C +36 C \,a^{2} b^{2}+5 C \,b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\) \(623\)
risch \(\text {Expression too large to display}\) \(1067\)

input
int(sec(d*x+c)*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBO 
SE)
 
output
(A*b^4+6*C*a^2*b^2)/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8* 
ln(sec(d*x+c)+tan(d*x+c)))-(4*A*a*b^3+4*C*a^3*b)/d*(-2/3-1/3*sec(d*x+c)^2) 
*tan(d*x+c)+(6*A*a^2*b^2+C*a^4)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d* 
x+c)+tan(d*x+c)))+C*b^4/d*(-(-1/6*sec(d*x+c)^5-5/24*sec(d*x+c)^3-5/16*sec( 
d*x+c))*tan(d*x+c)+5/16*ln(sec(d*x+c)+tan(d*x+c)))+1/d*A*ln(sec(d*x+c)+tan 
(d*x+c))*a^4+4*A*a^3*b/d*tan(d*x+c)-4*C*a*b^3/d*(-8/15-1/5*sec(d*x+c)^4-4/ 
15*sec(d*x+c)^2)*tan(d*x+c)
 
3.7.64.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.96 \[ \int \sec (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (8 \, {\left (2 \, A + C\right )} a^{4} + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, {\left (2 \, A + C\right )} a^{4} + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (192 \, C a b^{3} \cos \left (d x + c\right ) + 64 \, {\left (5 \, {\left (3 \, A + 2 \, C\right )} a^{3} b + 2 \, {\left (5 \, A + 4 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} + 40 \, C b^{4} + 15 \, {\left (8 \, C a^{4} + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} + 64 \, {\left (5 \, C a^{3} b + {\left (5 \, A + 4 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (36 \, C a^{2} b^{2} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="f 
ricas")
 
output
1/480*(15*(8*(2*A + C)*a^4 + 12*(4*A + 3*C)*a^2*b^2 + (6*A + 5*C)*b^4)*cos 
(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(8*(2*A + C)*a^4 + 12*(4*A + 3*C)*a 
^2*b^2 + (6*A + 5*C)*b^4)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(192*C 
*a*b^3*cos(d*x + c) + 64*(5*(3*A + 2*C)*a^3*b + 2*(5*A + 4*C)*a*b^3)*cos(d 
*x + c)^5 + 40*C*b^4 + 15*(8*C*a^4 + 12*(4*A + 3*C)*a^2*b^2 + (6*A + 5*C)* 
b^4)*cos(d*x + c)^4 + 64*(5*C*a^3*b + (5*A + 4*C)*a*b^3)*cos(d*x + c)^3 + 
10*(36*C*a^2*b^2 + (6*A + 5*C)*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d 
*x + c)^6)
 
3.7.64.6 Sympy [F]

\[ \int \sec (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{4} \sec {\left (c + d x \right )}\, dx \]

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)
 
output
Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**4*sec(c + d*x), x)
 
3.7.64.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 459, normalized size of antiderivative = 1.48 \[ \int \sec (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} b + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{3} + 128 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a b^{3} - 5 \, C b^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, C a^{2} b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, A b^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 720 \, A a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 480 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 1920 \, A a^{3} b \tan \left (d x + c\right )}{480 \, d} \]

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="m 
axima")
 
output
1/480*(640*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3*b + 640*(tan(d*x + c)^3 
 + 3*tan(d*x + c))*A*a*b^3 + 128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 1 
5*tan(d*x + c))*C*a*b^3 - 5*C*b^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^ 
3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 
 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 180*C*a^2*b 
^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c) 
^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 30*A*b^4*(2 
*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 
1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 120*C*a^4*(2*sin 
(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) 
- 1)) - 720*A*a^2*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + 
 c) + 1) + log(sin(d*x + c) - 1)) + 480*A*a^4*log(sec(d*x + c) + tan(d*x + 
 c)) + 1920*A*a^3*b*tan(d*x + c))/d
 
3.7.64.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1100 vs. \(2 (296) = 592\).

Time = 0.41 (sec) , antiderivative size = 1100, normalized size of antiderivative = 3.55 \[ \int \sec (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="g 
iac")
 
output
1/240*(15*(16*A*a^4 + 8*C*a^4 + 48*A*a^2*b^2 + 36*C*a^2*b^2 + 6*A*b^4 + 5* 
C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(16*A*a^4 + 8*C*a^4 + 48*A* 
a^2*b^2 + 36*C*a^2*b^2 + 6*A*b^4 + 5*C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 
 1)) + 2*(120*C*a^4*tan(1/2*d*x + 1/2*c)^11 - 960*A*a^3*b*tan(1/2*d*x + 1/ 
2*c)^11 - 960*C*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 720*A*a^2*b^2*tan(1/2*d*x 
+ 1/2*c)^11 + 900*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 960*A*a*b^3*tan(1/2* 
d*x + 1/2*c)^11 - 960*C*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 150*A*b^4*tan(1/2* 
d*x + 1/2*c)^11 + 165*C*b^4*tan(1/2*d*x + 1/2*c)^11 - 360*C*a^4*tan(1/2*d* 
x + 1/2*c)^9 + 4800*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 3520*C*a^3*b*tan(1/2* 
d*x + 1/2*c)^9 - 2160*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 1260*C*a^2*b^2*ta 
n(1/2*d*x + 1/2*c)^9 + 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 2240*C*a*b^3* 
tan(1/2*d*x + 1/2*c)^9 - 210*A*b^4*tan(1/2*d*x + 1/2*c)^9 + 25*C*b^4*tan(1 
/2*d*x + 1/2*c)^9 + 240*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 9600*A*a^3*b*tan(1/ 
2*d*x + 1/2*c)^7 - 5760*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 1440*A*a^2*b^2*ta 
n(1/2*d*x + 1/2*c)^7 + 360*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 5760*A*a*b^3 
*tan(1/2*d*x + 1/2*c)^7 - 4992*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 60*A*b^4*t 
an(1/2*d*x + 1/2*c)^7 + 450*C*b^4*tan(1/2*d*x + 1/2*c)^7 + 240*C*a^4*tan(1 
/2*d*x + 1/2*c)^5 + 9600*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 5760*C*a^3*b*tan 
(1/2*d*x + 1/2*c)^5 + 1440*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 360*C*a^2*b^ 
2*tan(1/2*d*x + 1/2*c)^5 + 5760*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 4992*C...
 
3.7.64.9 Mupad [B] (verification not implemented)

Time = 19.12 (sec) , antiderivative size = 690, normalized size of antiderivative = 2.23 \[ \int \sec (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {5\,A\,b^4}{4}+C\,a^4+\frac {11\,C\,b^4}{8}+6\,A\,a^2\,b^2+\frac {15\,C\,a^2\,b^2}{2}-8\,A\,a\,b^3-8\,A\,a^3\,b-8\,C\,a\,b^3-8\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {5\,C\,b^4}{24}-3\,C\,a^4-\frac {7\,A\,b^4}{4}-18\,A\,a^2\,b^2-\frac {21\,C\,a^2\,b^2}{2}+\frac {88\,A\,a\,b^3}{3}+40\,A\,a^3\,b+\frac {56\,C\,a\,b^3}{3}+\frac {88\,C\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {A\,b^4}{2}+2\,C\,a^4+\frac {15\,C\,b^4}{4}+12\,A\,a^2\,b^2+3\,C\,a^2\,b^2-48\,A\,a\,b^3-80\,A\,a^3\,b-\frac {208\,C\,a\,b^3}{5}-48\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {A\,b^4}{2}+2\,C\,a^4+\frac {15\,C\,b^4}{4}+12\,A\,a^2\,b^2+3\,C\,a^2\,b^2+48\,A\,a\,b^3+80\,A\,a^3\,b+\frac {208\,C\,a\,b^3}{5}+48\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,C\,b^4}{24}-3\,C\,a^4-\frac {7\,A\,b^4}{4}-18\,A\,a^2\,b^2-\frac {21\,C\,a^2\,b^2}{2}-\frac {88\,A\,a\,b^3}{3}-40\,A\,a^3\,b-\frac {56\,C\,a\,b^3}{3}-\frac {88\,C\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,b^4}{4}+C\,a^4+\frac {11\,C\,b^4}{8}+6\,A\,a^2\,b^2+\frac {15\,C\,a^2\,b^2}{2}+8\,A\,a\,b^3+8\,A\,a^3\,b+8\,C\,a\,b^3+8\,C\,a^3\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a^4+\frac {3\,A\,b^4}{8}+\frac {C\,a^4}{2}+\frac {5\,C\,b^4}{16}+3\,A\,a^2\,b^2+\frac {9\,C\,a^2\,b^2}{4}\right )}{4\,A\,a^4+\frac {3\,A\,b^4}{2}+2\,C\,a^4+\frac {5\,C\,b^4}{4}+12\,A\,a^2\,b^2+9\,C\,a^2\,b^2}\right )\,\left (2\,A\,a^4+\frac {3\,A\,b^4}{4}+C\,a^4+\frac {5\,C\,b^4}{8}+6\,A\,a^2\,b^2+\frac {9\,C\,a^2\,b^2}{2}\right )}{d} \]

input
int(((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^4)/cos(c + d*x),x)
 
output
(tan(c/2 + (d*x)/2)*((5*A*b^4)/4 + C*a^4 + (11*C*b^4)/8 + 6*A*a^2*b^2 + (1 
5*C*a^2*b^2)/2 + 8*A*a*b^3 + 8*A*a^3*b + 8*C*a*b^3 + 8*C*a^3*b) + tan(c/2 
+ (d*x)/2)^11*((5*A*b^4)/4 + C*a^4 + (11*C*b^4)/8 + 6*A*a^2*b^2 + (15*C*a^ 
2*b^2)/2 - 8*A*a*b^3 - 8*A*a^3*b - 8*C*a*b^3 - 8*C*a^3*b) - tan(c/2 + (d*x 
)/2)^3*((7*A*b^4)/4 + 3*C*a^4 - (5*C*b^4)/24 + 18*A*a^2*b^2 + (21*C*a^2*b^ 
2)/2 + (88*A*a*b^3)/3 + 40*A*a^3*b + (56*C*a*b^3)/3 + (88*C*a^3*b)/3) + ta 
n(c/2 + (d*x)/2)^9*((5*C*b^4)/24 - 3*C*a^4 - (7*A*b^4)/4 - 18*A*a^2*b^2 - 
(21*C*a^2*b^2)/2 + (88*A*a*b^3)/3 + 40*A*a^3*b + (56*C*a*b^3)/3 + (88*C*a^ 
3*b)/3) + tan(c/2 + (d*x)/2)^5*((A*b^4)/2 + 2*C*a^4 + (15*C*b^4)/4 + 12*A* 
a^2*b^2 + 3*C*a^2*b^2 + 48*A*a*b^3 + 80*A*a^3*b + (208*C*a*b^3)/5 + 48*C*a 
^3*b) + tan(c/2 + (d*x)/2)^7*((A*b^4)/2 + 2*C*a^4 + (15*C*b^4)/4 + 12*A*a^ 
2*b^2 + 3*C*a^2*b^2 - 48*A*a*b^3 - 80*A*a^3*b - (208*C*a*b^3)/5 - 48*C*a^3 
*b))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + ( 
d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + 
(d*x)/2)^12 + 1)) + (atanh((4*tan(c/2 + (d*x)/2)*(A*a^4 + (3*A*b^4)/8 + (C 
*a^4)/2 + (5*C*b^4)/16 + 3*A*a^2*b^2 + (9*C*a^2*b^2)/4))/(4*A*a^4 + (3*A*b 
^4)/2 + 2*C*a^4 + (5*C*b^4)/4 + 12*A*a^2*b^2 + 9*C*a^2*b^2))*(2*A*a^4 + (3 
*A*b^4)/4 + C*a^4 + (5*C*b^4)/8 + 6*A*a^2*b^2 + (9*C*a^2*b^2)/2))/d