\(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [969]

Optimal result

Integrand size = 41, antiderivative size = 303 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {1}{2} b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) x+\frac {a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \sin (c+d x)}{6 d}-\frac {b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(4 A b+3 a B) (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:

1/2*b^2*(2*A*b^2+8*B*a*b+12*C*a^2+C*b^2)*x+1/2*a*(8*A*b^3+B*a^3+12*B*a*b^2 
+4*a^2*b*(A+2*C))*arctanh(sin(d*x+c))/d-1/6*b*(39*B*a^2*b-6*B*b^3+4*a*b^2* 
(11*A-6*C)+4*a^3*(2*A+3*C))*sin(d*x+c)/d-1/6*b^2*(18*B*a*b+3*b^2*(6*A-C)+a 
^2*(4*A+6*C))*cos(d*x+c)*sin(d*x+c)/d+1/6*(12*A*b^2+15*B*a*b+a^2*(4*A+6*C) 
)*(a+b*cos(d*x+c))^2*tan(d*x+c)/d+1/6*(4*A*b+3*B*a)*(a+b*cos(d*x+c))^3*sec 
(d*x+c)*tan(d*x+c)/d+1/3*A*(a+b*cos(d*x+c))^4*sec(d*x+c)^2*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 8.05 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.16 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\sec ^3(c+d x) \left (36 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) (c+d x) \cos (c+d x)+12 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) (c+d x) \cos (3 (c+d x))-48 a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) \cos ^3(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+2 \left (32 a^4 A+144 a^2 A b^2+96 a^3 b B+24 a^4 C+9 b^4 C+12 \left (8 a^3 A b+2 a^4 B+3 b^4 B+12 a b^3 C\right ) \cos (c+d x)+4 \left (36 a^2 A b^2+24 a^3 b B+3 b^4 C+a^4 (4 A+6 C)\right ) \cos (2 (c+d x))+12 b^4 B \cos (3 (c+d x))+48 a b^3 C \cos (3 (c+d x))+3 b^4 C \cos (4 (c+d x))\right ) \sin (c+d x)\right )}{96 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S 
ec[c + d*x]^4,x]
 

Output:

(Sec[c + d*x]^3*(36*b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*(c + d*x)*C 
os[c + d*x] + 12*b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*(c + d*x)*Cos[ 
3*(c + d*x)] - 48*a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*Cos 
[c + d*x]^3*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/ 
2] + Sin[(c + d*x)/2]]) + 2*(32*a^4*A + 144*a^2*A*b^2 + 96*a^3*b*B + 24*a^ 
4*C + 9*b^4*C + 12*(8*a^3*A*b + 2*a^4*B + 3*b^4*B + 12*a*b^3*C)*Cos[c + d* 
x] + 4*(36*a^2*A*b^2 + 24*a^3*b*B + 3*b^4*C + a^4*(4*A + 6*C))*Cos[2*(c + 
d*x)] + 12*b^4*B*Cos[3*(c + d*x)] + 48*a*b^3*C*Cos[3*(c + d*x)] + 3*b^4*C* 
Cos[4*(c + d*x)])*Sin[c + d*x]))/(96*d)
 

Rubi [A] (verified)

Time = 2.27 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.390, Rules used = {3042, 3526, 3042, 3526, 3042, 3526, 3042, 3512, 27, 3042, 3502, 27, 3042, 3214, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int (a+b \cos (c+d x)) \left (-2 b \left ((4 A+6 C) a^2+18 b B a+3 b^2 (6 A-C)\right ) \cos ^2(c+d x)-b \left (3 B a^2+2 b (4 A-9 C) a-6 b^2 B\right ) \cos (c+d x)+3 \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )\right ) \sec (c+d x)dx+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-2 b \left ((4 A+6 C) a^2+18 b B a+3 b^2 (6 A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (3 B a^2+2 b (4 A-9 C) a-6 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 3512

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {1}{2} \int 2 \left (3 \left (12 C a^2+8 b B a+2 A b^2+b^2 C\right ) \cos (c+d x) b^2-\left (4 (2 A+3 C) a^3+39 b B a^2+4 b^2 (11 A-6 C) a-6 b^3 B\right ) \cos ^2(c+d x) b+3 a \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )\right ) \sec (c+d x)dx-\frac {b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \left (3 \left (12 C a^2+8 b B a+2 A b^2+b^2 C\right ) \cos (c+d x) b^2-\left (4 (2 A+3 C) a^3+39 b B a^2+4 b^2 (11 A-6 C) a-6 b^3 B\right ) \cos ^2(c+d x) b+3 a \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )\right ) \sec (c+d x)dx-\frac {b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \frac {3 \left (12 C a^2+8 b B a+2 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2-\left (4 (2 A+3 C) a^3+39 b B a^2+4 b^2 (11 A-6 C) a-6 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b+3 a \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int 3 \left (\left (12 C a^2+8 b B a+2 A b^2+b^2 C\right ) \cos (c+d x) b^2+a \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )\right ) \sec (c+d x)dx-\frac {b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}-\frac {b \sin (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \left (\left (12 C a^2+8 b B a+2 A b^2+b^2 C\right ) \cos (c+d x) b^2+a \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )\right ) \sec (c+d x)dx-\frac {b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}-\frac {b \sin (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \frac {\left (12 C a^2+8 b B a+2 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}-\frac {b \sin (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (a \left (a^3 B+4 a^2 b (A+2 C)+12 a b^2 B+8 A b^3\right ) \int \sec (c+d x)dx+b^2 x \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right )\right )-\frac {b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}-\frac {b \sin (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (a \left (a^3 B+4 a^2 b (A+2 C)+12 a b^2 B+8 A b^3\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+b^2 x \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right )\right )-\frac {b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}-\frac {b \sin (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (-\frac {b^2 \sin (c+d x) \cos (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {\tan (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \cos (c+d x))^2}{d}+3 \left (b^2 x \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right )+\frac {a \left (a^3 B+4 a^2 b (A+2 C)+12 a b^2 B+8 A b^3\right ) \text {arctanh}(\sin (c+d x))}{d}\right )-\frac {b \sin (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{d}\right )+\frac {(3 a B+4 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}\)

Input:

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + 
 d*x]^4,x]
 

Output:

(A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (((4*A*b + 
3*a*B)*(a + b*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (3*(b^2*( 
2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*x + (a*(8*A*b^3 + a^3*B + 12*a*b^2*B 
 + 4*a^2*b*(A + 2*C))*ArcTanh[Sin[c + d*x]])/d) - (b*(39*a^2*b*B - 6*b^3*B 
 + 4*a*b^2*(11*A - 6*C) + 4*a^3*(2*A + 3*C))*Sin[c + d*x])/d - (b^2*(18*a* 
b*B + 3*b^2*(6*A - C) + a^2*(4*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/d + (( 
12*A*b^2 + 15*a*b*B + a^2*(4*A + 6*C))*(a + b*Cos[c + d*x])^2*Tan[c + d*x] 
)/d)/2)/3
 

Defintions of rubi rules used

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 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3502

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

rule 3512

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 3526

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

rule 4257

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

Maple [A] (verified)

Time = 13.06 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.78

method	result	size
parts	\(-\frac {A \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 A \,a^{3} b +B \,a^{4}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (B \,b^{4}+4 C a \,b^{3}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (A \,b^{4}+4 B a \,b^{3}+6 C \,a^{2} b^{2}\right ) \left (d x +c \right )}{d}+\frac {\left (4 a A \,b^{3}+6 B \,a^{2} b^{2}+4 a^{3} b C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (6 A \,a^{2} b^{2}+4 B \,a^{3} b +a^{4} C \right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(235\)
derivativedivides	\(\frac {-A \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} C \tan \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 B \,a^{3} b \tan \left (d x +c \right )+4 a^{3} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 A \,a^{2} b^{2} \tan \left (d x +c \right )+6 B \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \,a^{2} b^{2} \left (d x +c \right )+4 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B a \,b^{3} \left (d x +c \right )+4 C \sin \left (d x +c \right ) a \,b^{3}+A \,b^{4} \left (d x +c \right )+B \sin \left (d x +c \right ) b^{4}+C \,b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(295\)
default	\(\frac {-A \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} C \tan \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 B \,a^{3} b \tan \left (d x +c \right )+4 a^{3} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 A \,a^{2} b^{2} \tan \left (d x +c \right )+6 B \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \,a^{2} b^{2} \left (d x +c \right )+4 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B a \,b^{3} \left (d x +c \right )+4 C \sin \left (d x +c \right ) a \,b^{3}+A \,b^{4} \left (d x +c \right )+B \sin \left (d x +c \right ) b^{4}+C \,b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(295\)
parallelrisch	\(\frac {-48 \left (\frac {B \,a^{3}}{4}+a^{2} b \left (A +2 C \right )+3 B a \,b^{2}+2 A \,b^{3}\right ) \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+48 \left (\frac {B \,a^{3}}{4}+a^{2} b \left (A +2 C \right )+3 B a \,b^{2}+2 A \,b^{3}\right ) \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+24 b^{2} x \left (6 a^{2} C +4 B a b +\left (A +\frac {C}{2}\right ) b^{2}\right ) d \cos \left (3 d x +3 c \right )+\left (8 a^{4} \left (2 A +3 C \right )+96 B \,a^{3} b +144 A \,a^{2} b^{2}+9 C \,b^{4}\right ) \sin \left (3 d x +3 c \right )+24 \left (4 A \,a^{3} b +B \,a^{4}+B \,b^{4}+4 C a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+12 \left (B \,b^{4}+4 C a \,b^{3}\right ) \sin \left (4 d x +4 c \right )+3 C \sin \left (5 d x +5 c \right ) b^{4}+72 b^{2} x \left (6 a^{2} C +4 B a b +\left (A +\frac {C}{2}\right ) b^{2}\right ) d \cos \left (d x +c \right )+48 \left (a^{4} \left (A +\frac {C}{2}\right )+2 B \,a^{3} b +3 A \,a^{2} b^{2}+\frac {C \,b^{4}}{8}\right ) \sin \left (d x +c \right )}{24 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(384\)
risch	\(x A \,b^{4}+4 x B a \,b^{3}+6 x C \,a^{2} b^{2}+\frac {b^{4} C x}{2}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,b^{4}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{4}}{2 d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} C a \,b^{3}}{d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} C \,b^{4}}{8 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} C a \,b^{3}}{d}-\frac {i a^{2} \left (12 A a b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-36 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-72 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-48 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 A a b \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-4 A \,a^{2}-36 A \,b^{2}-24 B a b -6 a^{2} C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} C \,b^{4}}{8 d}-\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {4 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{3}}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b C}{d}+\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}+\frac {4 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{3}}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b C}{d}\)	\(588\)

Input:

int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,meth 
od=_RETURNVERBOSE)
 

Output:

-A*a^4/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(4*A*a^3*b+B*a^4)/d*(1/2*sec(d 
*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(B*b^4+4*C*a*b^3)/d*sin(d* 
x+c)+(A*b^4+4*B*a*b^3+6*C*a^2*b^2)/d*(d*x+c)+(4*A*a*b^3+6*B*a^2*b^2+4*C*a^ 
3*b)/d*ln(sec(d*x+c)+tan(d*x+c))+(6*A*a^2*b^2+4*B*a^3*b+C*a^4)/d*tan(d*x+c 
)+C*b^4/d*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.89 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {6 \, {\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} + {\left (2 \, A + C\right )} b^{4}\right )} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (B a^{4} + 4 \, {\left (A + 2 \, C\right )} a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a^{4} + 4 \, {\left (A + 2 \, C\right )} a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, C b^{4} \cos \left (d x + c\right )^{4} + 2 \, A a^{4} + 6 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{4} + 12 \, B a^{3} b + 18 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, 
x, algorithm="fricas")
 

Output:

1/12*(6*(12*C*a^2*b^2 + 8*B*a*b^3 + (2*A + C)*b^4)*d*x*cos(d*x + c)^3 + 3* 
(B*a^4 + 4*(A + 2*C)*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*cos(d*x + c)^3*log( 
sin(d*x + c) + 1) - 3*(B*a^4 + 4*(A + 2*C)*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^ 
3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(3*C*b^4*cos(d*x + c)^4 + 2*A 
*a^4 + 6*(4*C*a*b^3 + B*b^4)*cos(d*x + c)^3 + 2*((2*A + 3*C)*a^4 + 12*B*a^ 
3*b + 18*A*a^2*b^2)*cos(d*x + c)^2 + 3*(B*a^4 + 4*A*a^3*b)*cos(d*x + c))*s 
in(d*x + c))/(d*cos(d*x + c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)* 
*4,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.11 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 72 \, {\left (d x + c\right )} C a^{2} b^{2} + 48 \, {\left (d x + c\right )} B a b^{3} + 12 \, {\left (d x + c\right )} A b^{4} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{4} - 3 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{3} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{3} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C a b^{3} \sin \left (d x + c\right ) + 12 \, B b^{4} \sin \left (d x + c\right ) + 12 \, C a^{4} \tan \left (d x + c\right ) + 48 \, B a^{3} b \tan \left (d x + c\right ) + 72 \, A a^{2} b^{2} \tan \left (d x + c\right )}{12 \, d} \] Input:

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, 
x, algorithm="maxima")
 

Output:

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 72*(d*x + c)*C*a^2*b^2 + 
 48*(d*x + c)*B*a*b^3 + 12*(d*x + c)*A*b^4 + 3*(2*d*x + 2*c + sin(2*d*x + 
2*c))*C*b^4 - 3*B*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)) - 12*A*a^3*b*(2*sin(d*x + c)/(sin(d*x + 
c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*C*a^3*b*(l 
og(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*B*a^2*b^2*(log(sin(d*x 
+ c) + 1) - log(sin(d*x + c) - 1)) + 24*A*a*b^3*(log(sin(d*x + c) + 1) - l 
og(sin(d*x + c) - 1)) + 48*C*a*b^3*sin(d*x + c) + 12*B*b^4*sin(d*x + c) + 
12*C*a^4*tan(d*x + c) + 48*B*a^3*b*tan(d*x + c) + 72*A*a^2*b^2*tan(d*x + c 
))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 550, normalized size of antiderivative = 1.82 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx =\text {Too large to display} \] Input:

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, 
x, algorithm="giac")
 

Output:

1/6*(3*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*A*b^4 + C*b^4)*(d*x + c) + 3*(B*a^4 + 
 4*A*a^3*b + 8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*log(abs(tan(1/2*d*x + 1 
/2*c) + 1)) - 3*(B*a^4 + 4*A*a^3*b + 8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3) 
*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(8*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 
+ 2*B*b^4*tan(1/2*d*x + 1/2*c)^3 - C*b^4*tan(1/2*d*x + 1/2*c)^3 + 8*C*a*b^ 
3*tan(1/2*d*x + 1/2*c) + 2*B*b^4*tan(1/2*d*x + 1/2*c) + C*b^4*tan(1/2*d*x 
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 2*(6*A*a^4*tan(1/2*d*x + 1/2*c) 
^5 - 3*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 12* 
A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 24*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 36*A* 
a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^4 
*tan(1/2*d*x + 1/2*c)^3 - 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 72*A*a^2*b^2 
*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^4*tan(1/2*d*x + 1/2*c) + 3*B*a^4*tan(1/2*d 
*x + 1/2*c) + 6*C*a^4*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b*tan(1/2*d*x + 1/2* 
c) + 24*B*a^3*b*tan(1/2*d*x + 1/2*c) + 36*A*a^2*b^2*tan(1/2*d*x + 1/2*c))/ 
(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 3.06 (sec) , antiderivative size = 4849, normalized size of antiderivative = 16.00 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Too large to display} \] Input:

int(((a + b*cos(c + d*x))^4*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^4,x)
                                                                                    
                                                                                    
 

Output:

(atan(((((B*a^4)/2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 2*A*a^3*b + 4*C*a^3*b)*(32* 
A*b^4 + 16*B*a^4 + 16*C*b^4 + 192*B*a^2*b^2 + 192*C*a^2*b^2 + 128*A*a*b^3 
+ 64*A*a^3*b + 128*B*a*b^3 + 128*C*a^3*b) + tan(c/2 + (d*x)/2)*(32*A^2*b^8 
 + 8*B^2*a^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6 
*b^2 + 512*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 192*C^2*a^2* 
b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 256*A*B*a*b^7 + 64 
*A*B*a^7*b + 128*B*C*a*b^7 + 128*B*C*a^7*b + 1536*A*B*a^3*b^5 + 896*A*B*a^ 
5*b^3 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2 + 1536*B*C*a^ 
3*b^5 + 1536*B*C*a^5*b^3))*((B*a^4)/2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 2*A*a^3* 
b + 4*C*a^3*b)*1i - (((B*a^4)/2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 2*A*a^3*b + 4* 
C*a^3*b)*(32*A*b^4 + 16*B*a^4 + 16*C*b^4 + 192*B*a^2*b^2 + 192*C*a^2*b^2 + 
 128*A*a*b^3 + 64*A*a^3*b + 128*B*a*b^3 + 128*C*a^3*b) - tan(c/2 + (d*x)/2 
)*(32*A^2*b^8 + 8*B^2*a^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 
+ 128*A^2*a^6*b^2 + 512*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 
 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 256*A 
*B*a*b^7 + 64*A*B*a^7*b + 128*B*C*a*b^7 + 128*B*C*a^7*b + 1536*A*B*a^3*b^5 
 + 896*A*B*a^5*b^3 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2 
+ 1536*B*C*a^3*b^5 + 1536*B*C*a^5*b^3))*((B*a^4)/2 + 6*B*a^2*b^2 + 4*A*a*b 
^3 + 2*A*a^3*b + 4*C*a^3*b)*1i)/(1024*A^3*a^2*b^10 - (((B*a^4)/2 + 6*B*a^2 
*b^2 + 4*A*a*b^3 + 2*A*a^3*b + 4*C*a^3*b)*(32*A*b^4 + 16*B*a^4 + 16*C*b...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 704, normalized size of antiderivative = 2.32 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx =\text {Too large to display} \] Input:

int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
 

Output:

(3*cos(c + d*x)**2*sin(c + d*x)**3*b**4*c - 3*cos(c + d*x)**2*sin(c + d*x) 
*b**4*c - 15*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b 
 - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b*c - 60 
*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**3 + 15*cos 
(c + d*x)*log(tan((c + d*x)/2) - 1)*a**4*b + 24*cos(c + d*x)*log(tan((c + 
d*x)/2) - 1)*a**3*b*c + 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b** 
3 + 15*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4*b + 24* 
cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b*c + 60*cos(c 
 + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**3 - 15*cos(c + d 
*x)*log(tan((c + d*x)/2) + 1)*a**4*b - 24*cos(c + d*x)*log(tan((c + d*x)/2 
) + 1)*a**3*b*c - 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b**3 + 24 
*cos(c + d*x)*sin(c + d*x)**3*a*b**3*c + 6*cos(c + d*x)*sin(c + d*x)**3*b* 
*5 + 36*cos(c + d*x)*sin(c + d*x)**2*a**2*b**2*c*d*x + 30*cos(c + d*x)*sin 
(c + d*x)**2*a*b**4*d*x + 3*cos(c + d*x)*sin(c + d*x)**2*b**4*c*d*x - 15*c 
os(c + d*x)*sin(c + d*x)*a**4*b - 24*cos(c + d*x)*sin(c + d*x)*a*b**3*c - 
6*cos(c + d*x)*sin(c + d*x)*b**5 - 36*cos(c + d*x)*a**2*b**2*c*d*x - 30*co 
s(c + d*x)*a*b**4*d*x - 3*cos(c + d*x)*b**4*c*d*x + 4*sin(c + d*x)**3*a**5 
 + 6*sin(c + d*x)**3*a**4*c + 60*sin(c + d*x)**3*a**3*b**2 - 6*sin(c + d*x 
)*a**5 - 6*sin(c + d*x)*a**4*c - 60*sin(c + d*x)*a**3*b**2)/(6*cos(c + d*x 
)*d*(sin(c + d*x)**2 - 1))