\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [1346]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F(-1)]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 45, antiderivative size = 333 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(19 A-15 B+11 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(1201 A-1029 B+665 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt {a+a \cos (c+d x)}}+\frac {(397 A-273 B+245 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{210 a d \sqrt {a+a \cos (c+d x)}}-\frac {(67 A-63 B+35 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(11 A-7 B+7 C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}} \] Output:

1/4*(19*A-15*B+11*C)*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2 
)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/a^(3/2 
)/d-1/210*(1201*A-1029*B+665*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*cos(d 
*x+c))^(1/2)+1/210*(397*A-273*B+245*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+ 
a*cos(d*x+c))^(1/2)-1/70*(67*A-63*B+35*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/ 
(a+a*cos(d*x+c))^(1/2)-1/2*(A-B+C)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos( 
d*x+c))^(3/2)+1/14*(11*A-7*B+7*C)*sec(d*x+c)^(7/2)*sin(d*x+c)/a/d/(a+a*cos 
(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 9.79 (sec) , antiderivative size = 3137, normalized size of antiderivative = 9.42 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2))/(a 
+ a*Cos[c + d*x])^(3/2),x]
 

Output:

(2*Cos[c/2 + (d*x)/2]^3*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2 
*Sin[c/2 + (d*x)/2]^2]*((4*C*Sin[c/2 + (d*x)/2])/(7*(1 - 2*Sin[c/2 + (d*x) 
/2]^2)^(7/2)) - ((A - B + C)*(1 - 2*Sin[c/2 + (d*x)/2]))/(28*(1 + Sin[c/2 
+ (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)) + ((A - B + C)*(1 + 2*Sin[ 
c/2 + (d*x)/2]))/(28*(1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2) 
^(7/2)) + (24*C*Sin[c/2 + (d*x)/2])/(35*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2) 
) + (32*C*Sin[c/2 + (d*x)/2])/(35*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + (6 
4*C*Sin[c/2 + (d*x)/2])/(35*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) - ((A - B + 
C)*(315*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2] 
] + (5 + 3*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + 
(d*x)/2]^2)^(5/2)) - (11 + 17*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2] 
)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + (61 + 71*Sin[c/2 + (d*x)/2])/((1 - 
 Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (193*Sqrt[1 - 2*S 
in[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/70 + ((A - B + C)*(315*Ar 
cTan[(1 + 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (5 - 3 
*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2 
)^(5/2)) - (11 - 17*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*(1 - 2*S 
in[c/2 + (d*x)/2]^2)^(3/2)) + (61 - 71*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + 
 (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (193*Sqrt[1 - 2*Sin[c/2 + ( 
d*x)/2]^2])/(1 + Sin[c/2 + (d*x)/2])))/70 - ((-A - 3*B + 7*C)*Csc[c/2 +...
 

Rubi [A] (verified)

Time = 2.18 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.08, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4709, 3042, 3520, 27, 3042, 3463, 27, 3042, 3463, 27, 3042, 3463, 27, 3042, 3463, 27, 3042, 3261, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sec ^{\frac {9}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^{9/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{(a \cos (c+d x)+a)^{3/2}}dx\)

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\cos ^{\frac {9}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 3520

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (11 A-7 B+7 C)-4 a (2 A-2 B+C) \cos (c+d x)}{2 \cos ^{\frac {9}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (11 A-7 B+7 C)-4 a (2 A-2 B+C) \cos (c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (11 A-7 B+7 C)-4 a (2 A-2 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {a^2 (67 A-63 B+35 C)-6 a^2 (11 A-7 B+7 C) \cos (c+d x)}{2 \cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{7 a}+\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^2 (67 A-63 B+35 C)-6 a^2 (11 A-7 B+7 C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^2 (67 A-63 B+35 C)-6 a^2 (11 A-7 B+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^3 (397 A-273 B+245 C)-4 a^3 (67 A-63 B+35 C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{5 a}+\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (397 A-273 B+245 C)-4 a^3 (67 A-63 B+35 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (397 A-273 B+245 C)-4 a^3 (67 A-63 B+35 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^4 (1201 A-1029 B+665 C)-2 a^4 (397 A-273 B+245 C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 a^3 (397 A-273 B+245 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (397 A-273 B+245 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^4 (1201 A-1029 B+665 C)-2 a^4 (397 A-273 B+245 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (397 A-273 B+245 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^4 (1201 A-1029 B+665 C)-2 a^4 (397 A-273 B+245 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (397 A-273 B+245 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {105 a^5 (19 A-15 B+11 C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^4 (1201 A-1029 B+665 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (397 A-273 B+245 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^4 (1201 A-1029 B+665 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-105 a^4 (19 A-15 B+11 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (397 A-273 B+245 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^4 (1201 A-1029 B+665 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-105 a^4 (19 A-15 B+11 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3261

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (397 A-273 B+245 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {210 a^5 (19 A-15 B+11 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^4 (1201 A-1029 B+665 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (11 A-7 B+7 C) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (67 A-63 B+35 C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (397 A-273 B+245 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^4 (1201 A-1029 B+665 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {105 \sqrt {2} a^{7/2} (19 A-15 B+11 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}}{5 a}}{7 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

Input:

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2))/(a + a*Co 
s[c + d*x])^(3/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A - B + C)*Sin[c + d*x])/(d* 
Cos[c + d*x]^(7/2)*(a + a*Cos[c + d*x])^(3/2)) + ((2*a*(11*A - 7*B + 7*C)* 
Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d*x]]) - ((2*a^2* 
(67*A - 63*B + 35*C)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[ 
c + d*x]]) - ((2*a^3*(397*A - 273*B + 245*C)*Sin[c + d*x])/(3*d*Cos[c + d* 
x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-105*Sqrt[2]*a^(7/2)*(19*A - 15*B + 
 11*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + 
a*Cos[c + d*x]])])/d + (2*a^4*(1201*A - 1029*B + 665*C)*Sin[c + d*x])/(d*S 
qrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(5*a))/(7*a))/(4*a^2))
 

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 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3261
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[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 3463
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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
+ 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2))   Int[(a 
 + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 
1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], 
 x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq 
Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m 
 + 1/2, 0])
 

rule 3520
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[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* 
Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x 
] + Simp[1/(b*(b*c - a*d)*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c 
+ d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a 
*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c 
*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c 
^2 - d^2, 0] && LtQ[m, -2^(-1)]
 

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 = 2.83 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.18

method result size
default \(-\frac {\sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1995 \cos \left (d x +c \right )^{6}+3990 \cos \left (d x +c \right )^{5}+1995 \cos \left (d x +c \right )^{4}\right )+B \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (-1575 \cos \left (d x +c \right )^{6}-3150 \cos \left (d x +c \right )^{5}-1575 \cos \left (d x +c \right )^{4}\right )+C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1155 \cos \left (d x +c \right )^{6}+2310 \cos \left (d x +c \right )^{5}+1155 \cos \left (d x +c \right )^{4}\right )+\sin \left (d x +c \right ) \left (1201 \cos \left (d x +c \right )^{4}+804 \cos \left (d x +c \right )^{3}-196 \cos \left (d x +c \right )^{2}+36 \cos \left (d x +c \right )-60\right ) \sqrt {2}\, A \cos \left (d x +c \right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \left (-1029 \cos \left (d x +c \right )^{3}-756 \cos \left (d x +c \right )^{2}+84 \cos \left (d x +c \right )-84\right ) \sqrt {2}\, B +\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \left (665 \cos \left (d x +c \right )^{2}+420 \cos \left (d x +c \right )-140\right ) \sqrt {2}\, C \right )}{420 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}\) \(392\)
parts \(-\frac {A \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (\sin \left (d x +c \right ) \left (1201 \cos \left (d x +c \right )^{4}+804 \cos \left (d x +c \right )^{3}-196 \cos \left (d x +c \right )^{2}+36 \cos \left (d x +c \right )-60\right ) \sqrt {2}\, \cos \left (d x +c \right )+\arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1995 \cos \left (d x +c \right )^{6}+3990 \cos \left (d x +c \right )^{5}+1995 \cos \left (d x +c \right )^{4}\right )\right )}{420 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {B \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (\sin \left (d x +c \right ) \left (49 \cos \left (d x +c \right )^{3}+36 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+4\right ) \sqrt {2}\, \cos \left (d x +c \right )^{2}+\arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (75 \cos \left (d x +c \right )^{6}+150 \cos \left (d x +c \right )^{5}+75 \cos \left (d x +c \right )^{4}\right )\right )}{20 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {C \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (\sin \left (d x +c \right ) \left (19 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )-4\right ) \sqrt {2}\, \cos \left (d x +c \right )^{3}+\arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (33 \cos \left (d x +c \right )^{6}+66 \cos \left (d x +c \right )^{5}+33 \cos \left (d x +c \right )^{4}\right )\right )}{12 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}\) \(474\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(3/2 
),x,method=_RETURNVERBOSE)
 

Output:

-1/420/d/a^2*2^(1/2)*((1+cos(d*x+c))*a)^(1/2)*sec(d*x+c)^(9/2)/(1+cos(d*x+ 
c))^2*(A*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)* 
(1995*cos(d*x+c)^6+3990*cos(d*x+c)^5+1995*cos(d*x+c)^4)+B*arcsin(-csc(d*x+ 
c)+cot(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-1575*cos(d*x+c)^6-3150* 
cos(d*x+c)^5-1575*cos(d*x+c)^4)+C*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+ 
c)/(1+cos(d*x+c)))^(1/2)*(1155*cos(d*x+c)^6+2310*cos(d*x+c)^5+1155*cos(d*x 
+c)^4)+sin(d*x+c)*(1201*cos(d*x+c)^4+804*cos(d*x+c)^3-196*cos(d*x+c)^2+36* 
cos(d*x+c)-60)*2^(1/2)*A*cos(d*x+c)+sin(d*x+c)*cos(d*x+c)^2*(-1029*cos(d*x 
+c)^3-756*cos(d*x+c)^2+84*cos(d*x+c)-84)*2^(1/2)*B+sin(d*x+c)*cos(d*x+c)^3 
*(665*cos(d*x+c)^2+420*cos(d*x+c)-140)*2^(1/2)*C)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.77 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {105 \, \sqrt {2} {\left ({\left (19 \, A - 15 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (19 \, A - 15 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (19 \, A - 15 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (1201 \, A - 1029 \, B + 665 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \, {\left (67 \, A - 63 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \, {\left (7 \, A - 3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (3 \, A - 7 \, B\right )} \cos \left (d x + c\right ) - 60 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{420 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c) 
)^(3/2),x, algorithm="fricas")
 

Output:

-1/420*(105*sqrt(2)*((19*A - 15*B + 11*C)*cos(d*x + c)^5 + 2*(19*A - 15*B 
+ 11*C)*cos(d*x + c)^4 + (19*A - 15*B + 11*C)*cos(d*x + c)^3)*sqrt(a)*arct 
an(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + 
c))) + 2*((1201*A - 1029*B + 665*C)*cos(d*x + c)^4 + 12*(67*A - 63*B + 35* 
C)*cos(d*x + c)^3 - 28*(7*A - 3*B + 5*C)*cos(d*x + c)^2 + 12*(3*A - 7*B)*c 
os(d*x + c) - 60*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c 
)))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(9/2)/(a+a*cos(d*x+ 
c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c) 
)^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {9}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c) 
)^(3/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(9/2)/(a*co 
s(d*x + c) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int(((1/cos(c + d*x))^(9/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + 
a*cos(c + d*x))^(3/2),x)
 

Output:

int(((1/cos(c + d*x))^(9/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + 
a*cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{2}} \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(3/2 
),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec( 
c + d*x)**4)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*b + int((sqrt(sec(c 
 + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**4)/(cos(c + 
d*x)**2 + 2*cos(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x))*sqrt(cos(c + 
d*x) + 1)*sec(c + d*x)**4)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*a))/a 
**2