\(\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1315]

Optimal result

Integrand size = 43, antiderivative size = 388 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a \left (5 a^3 B-105 a b^2 B+4 a^2 b (5 A-33 C)-6 b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 a^2 \left (50 a b B-a^2 (3 A-59 C)+3 b^2 (5 A+3 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (5 A b^2+15 a b B+16 a^2 C+3 b^2 C\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (5 b B+8 a C) (b+a \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:

2/5*(20*B*a^3*b-20*B*a*b^3+30*a^2*b^2*(A-C)-b^4*(5*A+3*C)+a^4*(3*A+5*C))*E 
llipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*(B*a^4+18*B*a^2*b^2+B*b^4+4*a*b 
^3*(3*A+C)+4*a^3*b*(A+3*C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/15* 
a*(5*B*a^3-105*B*a*b^2+4*a^2*b*(5*A-33*C)-6*b^3*(5*A+3*C))*cos(d*x+c)^(1/2 
)*sin(d*x+c)/d-2/15*a^2*(50*B*a*b-a^2*(3*A-59*C)+3*b^2*(5*A+3*C))*cos(d*x+ 
c)^(3/2)*sin(d*x+c)/d+2/5*(5*A*b^2+15*B*a*b+16*C*a^2+3*C*b^2)*(b+a*cos(d*x 
+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/15*(5*B*b+8*C*a)*(b+a*cos(d*x+c))^3 
*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*C*(b+a*cos(d*x+c))^4*sin(d*x+c)/d/cos(d 
*x+c)^(5/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 13.96 (sec) , antiderivative size = 4016, normalized size of antiderivative = 10.35 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \] Input:

Integrate[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + 
C*Sec[c + d*x]^2),x]
 

Output:

(Cos[c + d*x]^(13/2)*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c 
+ d*x]^2)*((-2*(3*a^4*A + 30*a^2*A*b^2 - 10*A*b^4 + 20*a^3*b*B - 40*a*b^3* 
B + 5*a^4*C - 60*a^2*b^2*C - 6*b^4*C + 3*a^4*A*Cos[2*c] + 30*a^2*A*b^2*Cos 
[2*c] + 20*a^3*b*B*Cos[2*c] + 5*a^4*C*Cos[2*c])*Csc[c]*Sec[c])/(5*d) + (4* 
a^3*(4*A*b + a*B)*Cos[d*x]*Sin[c])/(3*d) + (2*a^4*A*Cos[2*d*x]*Sin[2*c])/( 
5*d) + (4*a^3*(4*A*b + a*B)*Cos[c]*Sin[d*x])/(3*d) + (4*b^4*C*Sec[c]*Sec[c 
 + d*x]^3*Sin[d*x])/(5*d) + (4*Sec[c]*Sec[c + d*x]^2*(3*b^4*C*Sin[c] + 5*b 
^4*B*Sin[d*x] + 20*a*b^3*C*Sin[d*x]))/(15*d) + (4*Sec[c]*Sec[c + d*x]*(5*b 
^4*B*Sin[c] + 20*a*b^3*C*Sin[c] + 15*A*b^4*Sin[d*x] + 60*a*b^3*B*Sin[d*x] 
+ 90*a^2*b^2*C*Sin[d*x] + 9*b^4*C*Sin[d*x]))/(15*d) + (2*a^4*A*Cos[2*c]*Si 
n[2*d*x])/(5*d)))/((b + a*Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A* 
Cos[2*c + 2*d*x])) - (16*a^3*A*b*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{ 
1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^4*(A + 
 B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin 
[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[ 
Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(b + a*Cos[c + d*x])^ 
4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - 
(16*a*A*b^3*Cos[c + d*x]^6*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin 
[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*S 
ec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[...
 

Rubi [A] (verified)

Time = 2.64 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.512, Rules used = {3042, 4600, 3042, 3526, 27, 3042, 3526, 27, 3042, 3526, 27, 3042, 3512, 27, 3042, 3502, 27, 3042, 3227, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^{5/2} (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4600

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^4 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\)

\(\Big \downarrow \) 3526

\(\displaystyle \frac {2}{5} \int \frac {(b+a \cos (c+d x))^3 \left (5 a (A-C) \cos ^2(c+d x)+(5 A b+3 C b+5 a B) \cos (c+d x)+5 b B+8 a C\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {(b+a \cos (c+d x))^3 \left (5 a (A-C) \cos ^2(c+d x)+(5 A b+3 C b+5 a B) \cos (c+d x)+5 b B+8 a C\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (5 a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(5 A b+3 C b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+5 b B+8 a C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3526

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {(b+a \cos (c+d x))^2 \left (5 a (3 a A-5 b B-11 a C) \cos ^2(c+d x)+\left (15 B a^2+2 b (15 A+C) a+5 b^2 B\right ) \cos (c+d x)+3 \left (16 C a^2+15 b B a+5 A b^2+3 b^2 C\right )\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {(b+a \cos (c+d x))^2 \left (5 a (3 a A-5 b B-11 a C) \cos ^2(c+d x)+\left (15 B a^2+2 b (15 A+C) a+5 b^2 B\right ) \cos (c+d x)+3 \left (16 C a^2+15 b B a+5 A b^2+3 b^2 C\right )\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (5 a (3 a A-5 b B-11 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (15 B a^2+2 b (15 A+C) a+5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (16 C a^2+15 b B a+5 A b^2+3 b^2 C\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3526

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (2 \int \frac {(b+a \cos (c+d x)) \left (192 C a^3+195 b B a^2-5 \left (-\left ((3 A-59 C) a^2\right )+50 b B a+3 b^2 (5 A+3 C)\right ) \cos ^2(c+d x) a+2 b^2 (45 A+19 C) a+5 b^3 B+\left (15 B a^3+b (45 A-101 C) a^2-65 b^2 B a-3 b^3 (5 A+3 C)\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {(b+a \cos (c+d x)) \left (192 C a^3+195 b B a^2-5 \left (-\left ((3 A-59 C) a^2\right )+50 b B a+3 b^2 (5 A+3 C)\right ) \cos ^2(c+d x) a+2 b^2 (45 A+19 C) a+5 b^3 B+\left (15 B a^3+b (45 A-101 C) a^2-65 b^2 B a-3 b^3 (5 A+3 C)\right ) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (192 C a^3+195 b B a^2-5 \left (-\left ((3 A-59 C) a^2\right )+50 b B a+3 b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a+2 b^2 (45 A+19 C) a+5 b^3 B+\left (15 B a^3+b (45 A-101 C) a^2-65 b^2 B a-3 b^3 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3512

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2}{5} \int \frac {5 \left (3 a \left (5 B a^3+4 b (5 A-33 C) a^2-105 b^2 B a-6 b^3 (5 A+3 C)\right ) \cos ^2(c+d x)+3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \cos (c+d x)+b \left (192 C a^3+195 b B a^2+2 b^2 (45 A+19 C) a+5 b^3 B\right )\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a \left (5 B a^3+4 b (5 A-33 C) a^2-105 b^2 B a-6 b^3 (5 A+3 C)\right ) \cos ^2(c+d x)+3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \cos (c+d x)+b \left (192 C a^3+195 b B a^2+2 b^2 (45 A+19 C) a+5 b^3 B\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a \left (5 B a^3+4 b (5 A-33 C) a^2-105 b^2 B a-6 b^3 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+b \left (192 C a^3+195 b B a^2+2 b^2 (45 A+19 C) a+5 b^3 B\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2}{3} \int \frac {3 \left (5 \left (B a^4+4 b (A+3 C) a^3+18 b^2 B a^2+4 b^3 (3 A+C) a+b^4 B\right )+3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 B+4 a^2 b (5 A-33 C)-105 a b^2 B-6 b^3 (5 A+3 C)\right )}{d}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left (B a^4+4 b (A+3 C) a^3+18 b^2 B a^2+4 b^3 (3 A+C) a+b^4 B\right )+3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 B+4 a^2 b (5 A-33 C)-105 a b^2 B-6 b^3 (5 A+3 C)\right )}{d}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left (B a^4+4 b (A+3 C) a^3+18 b^2 B a^2+4 b^3 (3 A+C) a+b^4 B\right )+3 \left ((3 A+5 C) a^4+20 b B a^3+30 b^2 (A-C) a^2-20 b^3 B a-b^4 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 B+4 a^2 b (5 A-33 C)-105 a b^2 B-6 b^3 (5 A+3 C)\right )}{d}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 B+4 a^2 b (5 A-33 C)-105 a b^2 B-6 b^3 (5 A+3 C)\right )}{d}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 B+4 a^2 b (5 A-33 C)-105 a b^2 B-6 b^3 (5 A+3 C)\right )}{d}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 B+4 a^2 b (5 A-33 C)-105 a b^2 B-6 b^3 (5 A+3 C)\right )}{d}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right )}{d}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (-\left (a^2 (3 A-59 C)\right )+50 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (16 a^2 C+15 a b B+5 A b^2+3 b^2 C\right ) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 B+4 a^2 b (5 A-33 C)-105 a b^2 B-6 b^3 (5 A+3 C)\right )}{d}+\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right )}{d}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right )}{d}\right )+\frac {2 (8 a C+5 b B) \sin (c+d x) (a \cos (c+d x)+b)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

Input:

Int[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),x]
 

Output:

(2*C*(b + a*Cos[c + d*x])^4*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*( 
5*b*B + 8*a*C)*(b + a*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2 
)) + ((6*(20*a^3*b*B - 20*a*b^3*B + 30*a^2*b^2*(A - C) - b^4*(5*A + 3*C) + 
 a^4*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + (10*(a^4*B + 18*a^2*b^2*B 
 + b^4*B + 4*a*b^3*(3*A + C) + 4*a^3*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2 
])/d + (2*a*(5*a^3*B - 105*a*b^2*B + 4*a^2*b*(5*A - 33*C) - 6*b^3*(5*A + 3 
*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/d - (2*a^2*(50*a*b*B - a^2*(3*A - 59 
*C) + 3*b^2*(5*A + 3*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/d + (6*(5*A*b^2 
+ 15*a*b*B + 16*a^2*C + 3*b^2*C)*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*S 
qrt[Cos[c + d*x]]))/3)/5
 

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 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

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 4600

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) 
*(x_)]^2), x_Symbol] :> Simp[d^(m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[ 
e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]
 

Maple [F(-1)]

Timed out.

hanged

Input:

int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.17 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a^{4} + 4 i \, {\left (A + 3 \, C\right )} a^{3} b + 18 i \, B a^{2} b^{2} + 4 i \, {\left (3 \, A + C\right )} a b^{3} + i \, B b^{4}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{4} - 4 i \, {\left (A + 3 \, C\right )} a^{3} b - 18 i \, B a^{2} b^{2} - 4 i \, {\left (3 \, A + C\right )} a b^{3} - i \, B b^{4}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{4} - 20 i \, B a^{3} b - 30 i \, {\left (A - C\right )} a^{2} b^{2} + 20 i \, B a b^{3} + i \, {\left (5 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{4} + 20 i \, B a^{3} b + 30 i \, {\left (A - C\right )} a^{2} b^{2} - 20 i \, B a b^{3} - i \, {\left (5 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 3 \, C b^{4} + 5 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (30 \, C a^{2} b^{2} + 20 \, B a b^{3} + {\left (5 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="fricas")
 

Output:

-1/15*(5*sqrt(2)*(I*B*a^4 + 4*I*(A + 3*C)*a^3*b + 18*I*B*a^2*b^2 + 4*I*(3* 
A + C)*a*b^3 + I*B*b^4)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a^4 - 4*I*(A + 3*C)*a^3*b - 18*I* 
B*a^2*b^2 - 4*I*(3*A + C)*a*b^3 - I*B*b^4)*cos(d*x + c)^3*weierstrassPInve 
rse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(-I*(3*A + 5*C)*a^4 
- 20*I*B*a^3*b - 30*I*(A - C)*a^2*b^2 + 20*I*B*a*b^3 + I*(5*A + 3*C)*b^4)* 
cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + 
 c) + I*sin(d*x + c))) + 3*sqrt(2)*(I*(3*A + 5*C)*a^4 + 20*I*B*a^3*b + 30* 
I*(A - C)*a^2*b^2 - 20*I*B*a*b^3 - I*(5*A + 3*C)*b^4)*cos(d*x + c)^3*weier 
strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) 
)) - 2*(3*A*a^4*cos(d*x + c)^4 + 3*C*b^4 + 5*(B*a^4 + 4*A*a^3*b)*cos(d*x + 
 c)^3 + 3*(30*C*a^2*b^2 + 20*B*a*b^3 + (5*A + 3*C)*b^4)*cos(d*x + c)^2 + 5 
*(4*C*a*b^3 + B*b^4)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos 
(d*x + c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(5/2)*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+ 
c)**2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^4*c 
os(d*x + c)^(5/2), x)
 

Mupad [B] (verification not implemented)

Time = 17.62 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.35 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos( 
c + d*x)^2),x)
 

Output:

(2*(B*a^4*ellipticF(c/2 + (d*x)/2, 2) + 12*B*a^3*b*ellipticE(c/2 + (d*x)/2 
, 2) + B*a^4*cos(c + d*x)^(1/2)*sin(c + d*x) + 18*B*a^2*b^2*ellipticF(c/2 
+ (d*x)/2, 2)))/(3*d) + (2*C*a^4*ellipticE(c/2 + (d*x)/2, 2))/d + (8*A*a*b 
^3*ellipticF(c/2 + (d*x)/2, 2))/d + (8*C*a^3*b*ellipticF(c/2 + (d*x)/2, 2) 
)/d + (4*A*a^3*b*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 
 + (d*x)/2, 2))/3))/d + (12*A*a^2*b^2*ellipticE(c/2 + (d*x)/2, 2))/d - (2* 
A*a^4*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + 
d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) + (2*A*b^4*sin(c + d*x)*hypergeom([- 
1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1 
/2)) + (2*B*b^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/ 
(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*b^4*sin(c + d*x)*hy 
pergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c 
 + d*x)^2)^(1/2)) + (8*B*a*b^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, co 
s(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (8*C*a*b^3* 
sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x 
)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (12*C*a^2*b^2*sin(c + d*x)*hypergeom([-1 
/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/ 
2))
 

Reduce [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{6}d x \right ) b^{4} c +4 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{5}d x \right ) a \,b^{3} c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{5}d x \right ) b^{5}+6 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) a^{2} b^{2} c +5 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) a \,b^{4}+4 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a^{3} b c +10 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a^{2} b^{3}+\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) a^{4} c +10 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) a^{3} b^{2}+5 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) a^{4} b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) a^{5} \] Input:

int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**6,x)*b**4*c + 4*int(s 
qrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**5,x)*a*b**3*c + int(sqrt(c 
os(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**5,x)*b**5 + 6*int(sqrt(cos(c + 
d*x))*cos(c + d*x)**2*sec(c + d*x)**4,x)*a**2*b**2*c + 5*int(sqrt(cos(c + 
d*x))*cos(c + d*x)**2*sec(c + d*x)**4,x)*a*b**4 + 4*int(sqrt(cos(c + d*x)) 
*cos(c + d*x)**2*sec(c + d*x)**3,x)*a**3*b*c + 10*int(sqrt(cos(c + d*x))*c 
os(c + d*x)**2*sec(c + d*x)**3,x)*a**2*b**3 + int(sqrt(cos(c + d*x))*cos(c 
 + d*x)**2*sec(c + d*x)**2,x)*a**4*c + 10*int(sqrt(cos(c + d*x))*cos(c + d 
*x)**2*sec(c + d*x)**2,x)*a**3*b**2 + 5*int(sqrt(cos(c + d*x))*cos(c + d*x 
)**2*sec(c + d*x),x)*a**4*b + int(sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*a* 
*5