\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [520]

Optimal result

Integrand size = 35, antiderivative size = 295 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} (A-B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 (257 A-129 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (29 A-93 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (19 A-3 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (A-9 B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}} \] Output:

-2^(1/2)*(A-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)/a^(1/2)/d+2/315*(257 
*A-129*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-2/315*(29*A 
-93*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/105*(19*A-3* 
B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-2/63*(A-9*B)*sec(d 
*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/9*A*sec(d*x+c)^(9/2)*sin 
(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.77 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 e^{-\frac {1}{2} i (c+d x)} \cos \left (\frac {1}{2} (c+d x)\right ) \left (-315 i (A-B) \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-\frac {1}{4} (-1279 A+423 B+(214 A-918 B) \cos (c+d x)-8 (157 A-69 B) \cos (2 (c+d x))+58 A \cos (3 (c+d x))-186 B \cos (3 (c+d x))-257 A \cos (4 (c+d x))+129 B \cos (4 (c+d x))) \sec ^{\frac {9}{2}}(c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right )+i \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{315 d \sqrt {a (1+\cos (c+d x))}} \] Input:

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(11/2))/Sqrt[a + a*Cos[c + d* 
x]],x]
 

Output:

(2*Cos[(c + d*x)/2]*((-315*I)*(A - B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*( 
c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(S 
qrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - ((-1279*A + 423*B + (214*A - 918* 
B)*Cos[c + d*x] - 8*(157*A - 69*B)*Cos[2*(c + d*x)] + 58*A*Cos[3*(c + d*x) 
] - 186*B*Cos[3*(c + d*x)] - 257*A*Cos[4*(c + d*x)] + 129*B*Cos[4*(c + d*x 
)])*Sec[c + d*x]^(9/2)*(Cos[(c + d*x)/2] + I*Sin[(c + d*x)/2])*Sin[(c + d* 
x)/2])/4))/(315*d*E^((I/2)*(c + d*x))*Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 2.00 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.14, 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, 3463, 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.

\(\displaystyle \int \frac {\sec ^{\frac {11}{2}}(c+d x) (A+B \cos (c+d x))}{\sqrt {a \cos (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{11/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 3440

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {11}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {3 \left (a^2 (19 A-3 B)-2 a^2 (A-9 B) \cos (c+d x)\right )}{2 \cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{7 a}+\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{9 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {a^2 (19 A-3 B)-2 a^2 (A-9 B) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {a^2 (19 A-3 B)-2 a^2 (A-9 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 \int -\frac {a^3 (29 A-93 B)-4 a^3 (19 A-3 B) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{5 a}+\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (29 A-93 B)-4 a^3 (19 A-3 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{5 a}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (29 A-93 B)-4 a^3 (19 A-3 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^4 (257 A-129 B)-2 a^4 (29 A-93 B) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 a^3 (29 A-93 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{5 a}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (29 A-93 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^4 (257 A-129 B)-2 a^4 (29 A-93 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{5 a}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (29 A-93 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^4 (257 A-129 B)-2 a^4 (29 A-93 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 3463

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (29 A-93 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {315 a^5 (A-B)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^4 (257 A-129 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{5 a}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (29 A-93 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^4 (257 A-129 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-315 a^4 (A-B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{5 a}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (29 A-93 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^4 (257 A-129 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-315 a^4 (A-B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 3261

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (29 A-93 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {630 a^5 (A-B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^4 (257 A-129 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{5 a}\right )}{7 a}}{9 a}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a (A-9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {2 a^2 (19 A-3 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (29 A-93 B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^4 (257 A-129 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {315 \sqrt {2} a^{7/2} (A-B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}}{5 a}\right )}{7 a}}{9 a}\right )\)

Input:

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(11/2))/Sqrt[a + a*Cos[c + d*x]],x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sin[c + d*x])/(9*d*Cos[c + d*x 
]^(9/2)*Sqrt[a + a*Cos[c + d*x]]) - ((2*a*(A - 9*B)*Sin[c + d*x])/(7*d*Cos 
[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d*x]]) - (3*((2*a^2*(19*A - 3*B)*Sin[c 
+ d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) - ((2*a^3*(29*A 
- 93*B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - 
((-315*Sqrt[2]*a^(7/2)*(A - 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*(257*A - 129*B)*Sin[ 
c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(5*a)))/ 
(7*a))/(9*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 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 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 = 14.14 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {11}{2}} \left (A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (315 \cos \left (d x +c \right )^{6}+315 \cos \left (d x +c \right )^{5}\right )+B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-315 \cos \left (d x +c \right )^{6}-315 \cos \left (d x +c \right )^{5}\right )+\cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (257 \cos \left (d x +c \right )^{4}-29 \cos \left (d x +c \right )^{3}+57 \cos \left (d x +c \right )^{2}-5 \cos \left (d x +c \right )+35\right ) \sqrt {2}\, A +\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \left (-129 \cos \left (d x +c \right )^{3}+93 \cos \left (d x +c \right )^{2}-9 \cos \left (d x +c \right )+45\right ) \sqrt {2}\, B \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) a}\)	\(265\)
parts	\(\frac {A \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {11}{2}} \left (\cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (257 \cos \left (d x +c \right )^{4}-29 \cos \left (d x +c \right )^{3}+57 \cos \left (d x +c \right )^{2}-5 \cos \left (d x +c \right )+35\right ) \sqrt {2}+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (315 \cos \left (d x +c \right )^{6}+315 \cos \left (d x +c \right )^{5}\right )\right )}{315 d \left (\cos \left (d x +c \right )+1\right ) a}-\frac {B \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {11}{2}} \left (\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \left (43 \cos \left (d x +c \right )^{3}-31 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )-15\right ) \sqrt {2}+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (105 \cos \left (d x +c \right )^{6}+105 \cos \left (d x +c \right )^{5}\right )\right )}{105 d \left (\cos \left (d x +c \right )+1\right ) a}\)	\(306\)

Input:

int((A+B*cos(d*x+c))*sec(d*x+c)^(11/2)/(a+a*cos(d*x+c))^(1/2),x,method=_RE 
TURNVERBOSE)
 

Output:

1/315/d*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)*sec(d*x+c)^(11/2)/(cos(d*x+c)+1)* 
(A*arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(315*co 
s(d*x+c)^6+315*cos(d*x+c)^5)+B*arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d*x+c)/( 
cos(d*x+c)+1))^(1/2)*(-315*cos(d*x+c)^6-315*cos(d*x+c)^5)+cos(d*x+c)*sin(d 
*x+c)*(257*cos(d*x+c)^4-29*cos(d*x+c)^3+57*cos(d*x+c)^2-5*cos(d*x+c)+35)*2 
^(1/2)*A+cos(d*x+c)^2*sin(d*x+c)*(-129*cos(d*x+c)^3+93*cos(d*x+c)^2-9*cos( 
d*x+c)+45)*2^(1/2)*B)/a
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.67 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\frac {315 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{5} + {\left (A - B\right )} a \cos \left (d x + c\right )^{4}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} + \frac {2 \, {\left ({\left (257 \, A - 129 \, B\right )} \cos \left (d x + c\right )^{4} - {\left (29 \, A - 93 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (A - 9 \, B\right )} \cos \left (d x + c\right ) + 35 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{315 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(11/2)/(a+a*cos(d*x+c))^(1/2),x, alg 
orithm="fricas")
 

Output:

1/315*(315*sqrt(2)*((A - B)*a*cos(d*x + c)^5 + (A - B)*a*cos(d*x + c)^4)*a 
rctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x 
 + c)))/sqrt(a) + 2*((257*A - 129*B)*cos(d*x + c)^4 - (29*A - 93*B)*cos(d* 
x + c)^3 + 3*(19*A - 3*B)*cos(d*x + c)^2 - 5*(A - 9*B)*cos(d*x + c) + 35*A 
)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + 
 c)^5 + a*d*cos(d*x + c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)**(11/2)/(a+a*cos(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {11}{2}}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(11/2)/(a+a*cos(d*x+c))^(1/2),x, alg 
orithm="maxima")
 

Output:

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(11/2)/sqrt(a*cos(d*x + c) + a 
), x)
 

Giac [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {11}{2}}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(11/2)/(a+a*cos(d*x+c))^(1/2),x, alg 
orithm="giac")
 

Output:

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(11/2)/sqrt(a*cos(d*x + c) + a 
), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(11/2))/(a + a*cos(c + d*x))^(1 
/2),x)
 

Output:

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(11/2))/(a + a*cos(c + d*x))^(1 
/2), x)
 

Reduce [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{5}}{\cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}}{\cos \left (d x +c \right )+1}d x \right ) a \right )}{a} \] Input:

int((A+B*cos(d*x+c))*sec(d*x+c)^(11/2)/(a+a*cos(d*x+c))^(1/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec( 
c + d*x)**5)/(cos(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x))*sqrt(cos(c 
+ d*x) + 1)*sec(c + d*x)**5)/(cos(c + d*x) + 1),x)*a))/a