\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {15}{2}}(c+d x) \, dx\) [1327]

Optimal result

Integrand size = 45, antiderivative size = 334 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {16 a^3 (8368 A+9230 B+10439 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{45045 d \sqrt {a+a \cos (c+d x)}}+\frac {8 a^3 (8368 A+9230 B+10439 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45045 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (8368 A+9230 B+10439 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (2224 A+2522 B+2717 C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (136 A+182 B+143 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{1287 d}+\frac {2 a (5 A+13 B) (a+a \cos (c+d x))^{3/2} \sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {13}{2}}(c+d x) \sin (c+d x)}{13 d} \] Output:

16/45045*a^3*(8368*A+9230*B+10439*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*co 
s(d*x+c))^(1/2)+8/45045*a^3*(8368*A+9230*B+10439*C)*sec(d*x+c)^(3/2)*sin(d 
*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/15015*a^3*(8368*A+9230*B+10439*C)*sec(d*x 
+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/9009*a^3*(2224*A+2522*B+27 
17*C)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/1287*a^2*(136 
*A+182*B+143*C)*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^(9/2)*sin(d*x+c)/d+2/143 
*a*(5*A+13*B)*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(11/2)*sin(d*x+c)/d+2/13*A 
*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(13/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.00 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.67 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (343612 A+325910 B+322751 C+70 (5552 A+5083 B+4576 C) \cos (c+d x)+14 (30334 A+31850 B+32747 C) \cos (2 (c+d x))+125520 A \cos (3 (c+d x))+138450 B \cos (3 (c+d x))+141570 C \cos (3 (c+d x))+125520 A \cos (4 (c+d x))+138450 B \cos (4 (c+d x))+156585 C \cos (4 (c+d x))+16736 A \cos (5 (c+d x))+18460 B \cos (5 (c+d x))+20878 C \cos (5 (c+d x))+16736 A \cos (6 (c+d x))+18460 B \cos (6 (c+d x))+20878 C \cos (6 (c+d x))) \sec ^{\frac {13}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{180180 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 
2)*Sec[c + d*x]^(15/2),x]
 

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(343612*A + 325910*B + 322751*C + 70*(5552 
*A + 5083*B + 4576*C)*Cos[c + d*x] + 14*(30334*A + 31850*B + 32747*C)*Cos[ 
2*(c + d*x)] + 125520*A*Cos[3*(c + d*x)] + 138450*B*Cos[3*(c + d*x)] + 141 
570*C*Cos[3*(c + d*x)] + 125520*A*Cos[4*(c + d*x)] + 138450*B*Cos[4*(c + d 
*x)] + 156585*C*Cos[4*(c + d*x)] + 16736*A*Cos[5*(c + d*x)] + 18460*B*Cos[ 
5*(c + d*x)] + 20878*C*Cos[5*(c + d*x)] + 16736*A*Cos[6*(c + d*x)] + 18460 
*B*Cos[6*(c + d*x)] + 20878*C*Cos[6*(c + d*x)])*Sec[c + d*x]^(13/2)*Tan[(c 
 + d*x)/2])/(180180*d)
 

Rubi [A] (verified)

Time = 2.22 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.08, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.422, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3459, 3042, 3251, 3042, 3251, 3042, 3250}

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.

\(\displaystyle \int \sec ^{\frac {15}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (c+d x)^{15/2} (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{5/2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {15}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{15/2}}dx\)

\(\Big \downarrow \) 3522

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^{5/2} (a (5 A+13 B)+a (6 A+13 C) \cos (c+d x))}{2 \cos ^{\frac {13}{2}}(c+d x)}dx}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (a (5 A+13 B)+a (6 A+13 C) \cos (c+d x))}{\cos ^{\frac {13}{2}}(c+d x)}dx}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+13 B)+a (6 A+13 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{13/2}}dx}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3454

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{11} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((136 A+182 B+143 C) a^2+(96 A+78 B+143 C) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {11}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((136 A+182 B+143 C) a^2+(96 A+78 B+143 C) \cos (c+d x) a^2\right )}{\cos ^{\frac {11}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((136 A+182 B+143 C) a^2+(96 A+78 B+143 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3454

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {2}{9} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((2224 A+2522 B+2717 C) a^3+3 (560 A+598 B+715 C) \cos (c+d x) a^3\right )}{2 \cos ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((2224 A+2522 B+2717 C) a^3+3 (560 A+598 B+715 C) \cos (c+d x) a^3\right )}{\cos ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((2224 A+2522 B+2717 C) a^3+3 (560 A+598 B+715 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3459

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3251

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3251

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3250

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a^2 (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}+\frac {1}{11} \left (\frac {2 a^3 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {1}{9} \left (\frac {2 a^4 (2224 A+2522 B+2717 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {3}{7} a^3 (8368 A+9230 B+10439 C) \left (\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4}{5} \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\right )\right )\right )}{13 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac {13}{2}}(c+d x)}\right )\)

Input:

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec 
[c + d*x]^(15/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^(5/2)*Sin 
[c + d*x])/(13*d*Cos[c + d*x]^(13/2)) + ((2*a^2*(5*A + 13*B)*(a + a*Cos[c 
+ d*x])^(3/2)*Sin[c + d*x])/(11*d*Cos[c + d*x]^(11/2)) + ((2*a^3*(136*A + 
182*B + 143*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9 
/2)) + ((2*a^4*(2224*A + 2522*B + 2717*C)*Sin[c + d*x])/(7*d*Cos[c + d*x]^ 
(7/2)*Sqrt[a + a*Cos[c + d*x]]) + (3*a^3*(8368*A + 9230*B + 10439*C)*((2*a 
*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*((2* 
a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a*S 
in[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])))/5))/7)/9) 
/11)/(13*a))
 

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 3250

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq 
rt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, 
 e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3251

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e 
+ f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim 
p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2)))   Int[Sqrt[a + b*Sin[e 
+ f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
 

rule 3454

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

rule 3459

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

rule 3522

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

rule 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.66

\[\frac {2 a^{2} \sin \left (d x +c \right ) \left (\left (66944 \cos \left (d x +c \right )^{6}+33472 \cos \left (d x +c \right )^{5}+25104 \cos \left (d x +c \right )^{4}+20920 \cos \left (d x +c \right )^{3}+18305 \cos \left (d x +c \right )^{2}+11970 \cos \left (d x +c \right )+3465\right ) A +\cos \left (d x +c \right ) \left (73840 \cos \left (d x +c \right )^{5}+36920 \cos \left (d x +c \right )^{4}+27690 \cos \left (d x +c \right )^{3}+23075 \cos \left (d x +c \right )^{2}+14560 \cos \left (d x +c \right )+4095\right ) B +\cos \left (d x +c \right )^{2} \left (83512 \cos \left (d x +c \right )^{4}+41756 \cos \left (d x +c \right )^{3}+31317 \cos \left (d x +c \right )^{2}+18590 \cos \left (d x +c \right )+5005\right ) C \right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{\frac {15}{2}}}{45045 d \left (1+\cos \left (d x +c \right )\right )}\]

Input:

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(15/ 
2),x)
 

Output:

2/45045/d*a^2*sin(d*x+c)*((66944*cos(d*x+c)^6+33472*cos(d*x+c)^5+25104*cos 
(d*x+c)^4+20920*cos(d*x+c)^3+18305*cos(d*x+c)^2+11970*cos(d*x+c)+3465)*A+c 
os(d*x+c)*(73840*cos(d*x+c)^5+36920*cos(d*x+c)^4+27690*cos(d*x+c)^3+23075* 
cos(d*x+c)^2+14560*cos(d*x+c)+4095)*B+cos(d*x+c)^2*(83512*cos(d*x+c)^4+417 
56*cos(d*x+c)^3+31317*cos(d*x+c)^2+18590*cos(d*x+c)+5005)*C)*((1+cos(d*x+c 
))*a)^(1/2)*cos(d*x+c)*sec(d*x+c)^(15/2)/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.57 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {15}{2}}(c+d x) \, dx=\frac {2 \, {\left (8 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + 4 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 3 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (4184 \, A + 4615 \, B + 3718 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \, {\left (523 \, A + 416 \, B + 143 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 315 \, {\left (38 \, A + 13 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3465 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c 
)^(15/2),x, algorithm="fricas")
 

Output:

2/45045*(8*(8368*A + 9230*B + 10439*C)*a^2*cos(d*x + c)^6 + 4*(8368*A + 92 
30*B + 10439*C)*a^2*cos(d*x + c)^5 + 3*(8368*A + 9230*B + 10439*C)*a^2*cos 
(d*x + c)^4 + 5*(4184*A + 4615*B + 3718*C)*a^2*cos(d*x + c)^3 + 35*(523*A 
+ 416*B + 143*C)*a^2*cos(d*x + c)^2 + 315*(38*A + 13*B)*a^2*cos(d*x + c) + 
 3465*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^7 + d* 
cos(d*x + c)^6)*sqrt(cos(d*x + c)))
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x 
+c)**(15/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1142 vs. \(2 (292) = 584\).

Time = 0.24 (sec) , antiderivative size = 1142, normalized size of antiderivative = 3.42 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c 
)^(15/2),x, algorithm="maxima")
 

Output:

8/45045*((45045*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 165165*s 
qrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 414414*sqrt(2)*a^(5/2 
)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 604890*sqrt(2)*a^(5/2)*sin(d*x + c 
)^7/(cos(d*x + c) + 1)^7 + 522665*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x 
+ c) + 1)^9 - 289185*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 
 + 88980*sqrt(2)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 11864*sqr 
t(2)*a^(5/2)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15)*A*(sin(d*x + c)^2/(cos 
(d*x + c) + 1)^2 + 1)^5/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(-si 
n(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(5*sin(d*x + c)^2/(cos(d*x + c) 
+ 1)^2 + 10*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^6/(cos(d 
*x + c) + 1)^6 + 5*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + sin(d*x + c)^10/( 
cos(d*x + c) + 1)^10 + 1)) + 65*(693*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x 
 + c) + 1) - 3003*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 69 
30*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 10098*sqrt(2)*a^( 
5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 9053*sqrt(2)*a^(5/2)*sin(d*x + 
c)^9/(cos(d*x + c) + 1)^9 - 4875*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x 
+ c) + 1)^11 + 1500*sqrt(2)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 
- 200*sqrt(2)*a^(5/2)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15)*B*(sin(d*x + 
c)^2/(cos(d*x + c) + 1)^2 + 1)^5/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(1 
5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(5*sin(d*x + c)^2/(c...
 

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c 
)^(15/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 4.21 (sec) , antiderivative size = 927, normalized size of antiderivative = 2.78 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {15}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:

int((1/cos(c + d*x))^(15/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) 
 + C*cos(c + d*x)^2),x)
 

Output:

((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*(a + a*(e 
xp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(8368*A + 9230*B + 10 
439*C)*16i)/(45045*d) - (C*a^2*exp(c*3i + d*x*3i)*(a + a*(exp(- c*1i - d*x 
*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*8i)/(3*d) + (C*a^2*exp(c*10i + d*x*1 
0i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*8i)/(3*d 
) - (a^2*exp(c*5i + d*x*5i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d* 
x*1i)/2))^(1/2)*(6*A + 15*B + 23*C)*16i)/(15*d) + (a^2*exp(c*8i + d*x*8i)* 
(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(6*A + 15*B 
+ 23*C)*16i)/(15*d) + (a^2*exp(c*6i + d*x*6i)*(a + a*(exp(- c*1i - d*x*1i) 
/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(348*A + 345*B + 379*C)*16i)/(105*d) - ( 
a^2*exp(c*7i + d*x*7i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i) 
/2))^(1/2)*(348*A + 345*B + 379*C)*16i)/(105*d) + (a^2*exp(c*4i + d*x*4i)* 
(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(1046*A + 10 
75*B + 1108*C)*16i)/(315*d) - (a^2*exp(c*9i + d*x*9i)*(a + a*(exp(- c*1i - 
 d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(1046*A + 1075*B + 1108*C)*16i)/ 
(315*d) + (a^2*exp(c*2i + d*x*2i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1 
i + d*x*1i)/2))^(1/2)*(8368*A + 9230*B + 10439*C)*8i)/(3465*d) - (a^2*exp( 
c*11i + d*x*11i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^( 
1/2)*(8368*A + 9230*B + 10439*C)*8i)/(3465*d) - (a^2*exp(c*13i + d*x*13i)* 
(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(8368*A +...
 

Reduce [F]

\[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {15}{2}}(c+d x) \, dx=\sqrt {a}\, a^{2} \left (2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{7}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{7}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{7}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{7}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{7}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{7}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{7}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{7}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{7}d x \right ) a \right ) \] Input:

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(15/ 
2),x)
 

Output:

sqrt(a)*a**2*(2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x) 
*sec(c + d*x)**7,x)*a + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos( 
c + d*x)*sec(c + d*x)**7,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 
 1)*cos(c + d*x)**4*sec(c + d*x)**7,x)*c + int(sqrt(sec(c + d*x))*sqrt(cos 
(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**7,x)*b + 2*int(sqrt(sec(c + d 
*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**7,x)*c + int(sqr 
t(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**7,x)* 
a + 2*int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c 
+ d*x)**7,x)*b + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x 
)**2*sec(c + d*x)**7,x)*c + int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)* 
sec(c + d*x)**7,x)*a)