\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [529]

Optimal result

Integrand size = 35, antiderivative size = 317 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(19 A-15 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(1201 A-1029 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{210 a d \sqrt {a+a \cos (c+d x)}}+\frac {(397 A-273 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{210 a d \sqrt {a+a \cos (c+d x)}}-\frac {(67 A-63 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(11 A-7 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}} \] Output:

1/4*(19*A-15*B)*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)*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)*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)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2) 
-1/2*(A-B)*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)*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 = 10.09 (sec) , antiderivative size = 2966, normalized size of antiderivative = 9.36 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[((A + B*Cos[c + d*x])*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]*(-1/28*((A - B)*(1 - 2*Sin[c/2 + (d*x)/2]))/((1 + S 
in[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)) + ((A - B)*(1 + 2*S 
in[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)*(315*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Si 
n[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*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/70 + ( 
(A - B)*(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*Sin[c/2 + (d*x)/2]^2])/(1 + Sin[c/2 + (d*x)/2])))/70 - ((-A - 3*B)*Csc 
[c/2 + (d*x)/2]^9*(363825*Sin[c/2 + (d*x)/2]^2 - 4729725*Sin[c/2 + (d*x)/2 
]^4 + 26785605*Sin[c/2 + (d*x)/2]^6 - 86790165*Sin[c/2 + (d*x)/2]^8 + 1776 
77808*Sin[c/2 + (d*x)/2]^10 - 239283044*Sin[c/2 + (d*x)/2]^12 + 52080*Hype 
rgeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*...
 

Rubi [A] (verified)

Time = 2.06 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.09, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3440, 3042, 3457, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3440

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (11 A-7 B)-8 a (A-B) \cos (c+d x)}{2 \cos ^{\frac {9}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (11 A-7 B)-8 a (A-B) \cos (c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (11 A-7 B)-8 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A - B)*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)*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)*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)*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)*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*(1 
201*A - 1029*B)*Sin[c + d*x])/(d*Sqrt[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 3440

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Sin[e + f*x])^m*((c + 
d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

rule 3457

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

Maple [A] (verified)

Time = 14.48 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (1995 \cos \left (d x +c \right )^{6}+3990 \cos \left (d x +c \right )^{5}+1995 \cos \left (d x +c \right )^{4}\right )+B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-1575 \cos \left (d x +c \right )^{6}-3150 \cos \left (d x +c \right )^{5}-1575 \cos \left (d x +c \right )^{4}\right )+\cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (1201 \cos \left (d x +c \right )^{4}+804 \cos \left (d x +c \right )^{3}-196 \cos \left (d x +c \right )^{2}+36 \cos \left (d x +c \right )-60\right ) \sqrt {2}\, A +\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \left (-1029 \cos \left (d x +c \right )^{3}-756 \cos \left (d x +c \right )^{2}+84 \cos \left (d x +c \right )-84\right ) \sqrt {2}\, B \right )}{420 d \left (\cos \left (d x +c \right )+1\right )^{2} a^{2}}\)	\(285\)
parts	\(-\frac {A \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (\cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (1201 \cos \left (d x +c \right )^{4}+804 \cos \left (d x +c \right )^{3}-196 \cos \left (d x +c \right )^{2}+36 \cos \left (d x +c \right )-60\right ) \sqrt {2}+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (1995 \cos \left (d x +c \right )^{6}+3990 \cos \left (d x +c \right )^{5}+1995 \cos \left (d x +c \right )^{4}\right )\right )}{420 d \left (\cos \left (d x +c \right )+1\right )^{2} a^{2}}+\frac {B \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \left (49 \cos \left (d x +c \right )^{3}+36 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+4\right ) \sqrt {2}+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (75 \cos \left (d x +c \right )^{6}+150 \cos \left (d x +c \right )^{5}+75 \cos \left (d x +c \right )^{4}\right )\right )}{20 d \left (\cos \left (d x +c \right )+1\right )^{2} a^{2}}\)	\(326\)

Input:

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

Output:

-1/420/d*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)*sec(d*x+c)^(9/2)/(cos(d*x+c)+1)^ 
2*(A*arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1995 
*cos(d*x+c)^6+3990*cos(d*x+c)^5+1995*cos(d*x+c)^4)+B*arcsin(cot(d*x+c)-csc 
(d*x+c))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-1575*cos(d*x+c)^6-3150*cos(d* 
x+c)^5-1575*cos(d*x+c)^4)+cos(d*x+c)*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)^2*sin(d* 
x+c)*(-1029*cos(d*x+c)^3-756*cos(d*x+c)^2+84*cos(d*x+c)-84)*2^(1/2)*B)/a^2
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.75 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {105 \, \sqrt {2} {\left ({\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (1201 \, A - 1029 \, B\right )} \cos \left (d x + c\right )^{4} + 12 \, {\left (67 \, A - 63 \, B\right )} \cos \left (d x + c\right )^{3} - 28 \, {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (3 \, A - 7 \, B\right )} \cos \left (d x + c\right ) - 60 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{420 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:

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

Output:

-1/420*(105*sqrt(2)*((19*A - 15*B)*cos(d*x + c)^5 + 2*(19*A - 15*B)*cos(d* 
x + c)^4 + (19*A - 15*B)*cos(d*x + c)^3)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos 
(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((1201*A - 1 
029*B)*cos(d*x + c)^4 + 12*(67*A - 63*B)*cos(d*x + c)^3 - 28*(7*A - 3*B)*c 
os(d*x + c)^2 + 12*(3*A - 7*B)*cos(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. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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