\(\int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [523]

Optimal result

Integrand size = 35, antiderivative size = 275 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {32 a^2 (168 A+187 B) \sin (c+d x)}{3465 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (168 A+187 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {4 a^2 (168 A+187 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (168 A+187 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (12 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d} \] Output:

32/3465*a^2*(168*A+187*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^( 
1/2)+16/3465*a^2*(168*A+187*B)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+ 
c))^(1/2)+4/1155*a^2*(168*A+187*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec( 
d*x+c))^(1/2)+2/693*a^2*(168*A+187*B)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*s 
ec(d*x+c))^(1/2)+2/99*a^2*(12*A+11*B)*cos(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*s 
ec(d*x+c))^(1/2)+2/11*a*A*cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+ 
c)/d
 

Mathematica [A] (verified)

Time = 7.23 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.49 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a \sqrt {\cos (c+d x)} (55482 A+59158 B+(34734 A+35156 B) \cos (c+d x)+8 (1743 A+1507 B) \cos (2 (c+d x))+4935 A \cos (3 (c+d x))+3740 B \cos (3 (c+d x))+1470 A \cos (4 (c+d x))+770 B \cos (4 (c+d x))+315 A \cos (5 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{27720 d} \] Input:

Integrate[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d* 
x]),x]
 

Output:

(a*Sqrt[Cos[c + d*x]]*(55482*A + 59158*B + (34734*A + 35156*B)*Cos[c + d*x 
] + 8*(1743*A + 1507*B)*Cos[2*(c + d*x)] + 4935*A*Cos[3*(c + d*x)] + 3740* 
B*Cos[3*(c + d*x)] + 1470*A*Cos[4*(c + d*x)] + 770*B*Cos[4*(c + d*x)] + 31 
5*A*Cos[5*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(27720* 
d)
 

Rubi [A] (verified)

Time = 1.59 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3434, 3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4292, 3042, 4292, 3042, 4291}

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

\(\displaystyle \int \cos ^{\frac {11}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2} (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{11/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3434

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\)

\(\Big \downarrow \) 4505

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{11} \int \frac {\sqrt {\sec (c+d x) a+a} (a (12 A+11 B)+a (8 A+11 B) \sec (c+d x))}{2 \sec ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \int \frac {\sqrt {\sec (c+d x) a+a} (a (12 A+11 B)+a (8 A+11 B) \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (12 A+11 B)+a (8 A+11 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4503

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \left (\frac {1}{9} a (168 A+187 B) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 (12 A+11 B) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \left (\frac {1}{9} a (168 A+187 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^2 (12 A+11 B) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4292

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \left (\frac {1}{9} a (168 A+187 B) \left (\frac {6}{7} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (12 A+11 B) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \left (\frac {1}{9} a (168 A+187 B) \left (\frac {6}{7} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (12 A+11 B) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4292

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \left (\frac {1}{9} a (168 A+187 B) \left (\frac {6}{7} \left (\frac {4}{5} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (12 A+11 B) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \left (\frac {1}{9} a (168 A+187 B) \left (\frac {6}{7} \left (\frac {4}{5} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (12 A+11 B) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4292

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \left (\frac {1}{9} a (168 A+187 B) \left (\frac {6}{7} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (12 A+11 B) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \left (\frac {1}{9} a (168 A+187 B) \left (\frac {6}{7} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (12 A+11 B) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4291

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{11} \left (\frac {2 a^2 (12 A+11 B) \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {1}{9} a (168 A+187 B) \left (\frac {2 a \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {6}{7} \left (\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4}{5} \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

Input:

Int[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*A*Sqrt[a + a*Sec[c + d*x]]*Sin 
[c + d*x])/(11*d*Sec[c + d*x]^(9/2)) + ((2*a^2*(12*A + 11*B)*Sin[c + d*x]) 
/(9*d*Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]) + (a*(168*A + 187*B)*(( 
2*a*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (6*( 
(2*a*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (4* 
((2*a*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4 
*a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])))/5))/7 
))/9)/11)
 

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 3434

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Csc[e + f*x])^m*((c + 
d*Csc[e + f*x])^n/(g*Csc[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 4291

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] 
*(d_.)], x_Symbol] :> Simp[-2*a*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*S 
qrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4292

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a 
+ b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n))   Int[Sqrt[a + b*Csc 
[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && 
 EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
 

rule 4503

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

rule 4505

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

Maple [A] (verified)

Time = 1.62 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.50

method	result	size
default	\(\frac {2 \sin \left (d x +c \right ) \left (\left (315 \cos \left (d x +c \right )^{5}+735 \cos \left (d x +c \right )^{4}+840 \cos \left (d x +c \right )^{3}+1008 \cos \left (d x +c \right )^{2}+1344 \cos \left (d x +c \right )+2688\right ) A +\left (385 \cos \left (d x +c \right )^{4}+935 \cos \left (d x +c \right )^{3}+1122 \cos \left (d x +c \right )^{2}+1496 \cos \left (d x +c \right )+2992\right ) B \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, a}{3465 d \left (1+\cos \left (d x +c \right )\right )}\)	\(138\)

Input:

int(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x,method=_RE 
TURNVERBOSE)
 

Output:

2/3465/d*sin(d*x+c)*((315*cos(d*x+c)^5+735*cos(d*x+c)^4+840*cos(d*x+c)^3+1 
008*cos(d*x+c)^2+1344*cos(d*x+c)+2688)*A+(385*cos(d*x+c)^4+935*cos(d*x+c)^ 
3+1122*cos(d*x+c)^2+1496*cos(d*x+c)+2992)*B)*cos(d*x+c)^(1/2)*(a*(1+sec(d* 
x+c)))^(1/2)*a/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.52 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (315 \, A a \cos \left (d x + c\right )^{5} + 35 \, {\left (21 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{4} + 5 \, {\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right )^{3} + 6 \, {\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right ) + 16 \, {\left (168 \, A + 187 \, B\right )} a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, alg 
orithm="fricas")
 

Output:

2/3465*(315*A*a*cos(d*x + c)^5 + 35*(21*A + 11*B)*a*cos(d*x + c)^4 + 5*(16 
8*A + 187*B)*a*cos(d*x + c)^3 + 6*(168*A + 187*B)*a*cos(d*x + c)^2 + 8*(16 
8*A + 187*B)*a*cos(d*x + c) + 16*(168*A + 187*B)*a)*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(11/2)*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (239) = 478\).

Time = 0.28 (sec) , antiderivative size = 703, normalized size of antiderivative = 2.56 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, alg 
orithm="maxima")
 

Output:

1/110880*(21*sqrt(2)*(3630*a*cos(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos 
(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 990*a*cos(8/11*arctan2(sin( 
11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 429* 
a*cos(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11 
/2*d*x + 11/2*c) + 165*a*cos(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2 
*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 55*a*cos(2/11*arctan2(sin(11/2*d 
*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) - 3630*a*cos 
(11/2*d*x + 11/2*c)*sin(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x 
 + 11/2*c))) - 990*a*cos(11/2*d*x + 11/2*c)*sin(8/11*arctan2(sin(11/2*d*x 
+ 11/2*c), cos(11/2*d*x + 11/2*c))) - 429*a*cos(11/2*d*x + 11/2*c)*sin(6/1 
1*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 165*a*cos(11/ 
2*d*x + 11/2*c)*sin(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11 
/2*c))) - 55*a*cos(11/2*d*x + 11/2*c)*sin(2/11*arctan2(sin(11/2*d*x + 11/2 
*c), cos(11/2*d*x + 11/2*c))) + 30*a*sin(11/2*d*x + 11/2*c) + 55*a*sin(9/1 
1*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 165*a*sin(7/1 
1*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 429*a*sin(5/1 
1*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 990*a*sin(3/1 
1*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 3630*a*sin(1/ 
11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))*A*sqrt(a) - 4 
4*sqrt(2)*(189*(10*a*sin(4*d*x + 4*c) + a*sin(2*d*x + 2*c))*cos(9/4*arc...
 

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.04 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {4 \, {\left ({\left ({\left ({\left ({\left (2 \, \sqrt {2} {\left (483 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 517 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 11 \, \sqrt {2} {\left (483 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 517 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 198 \, \sqrt {2} {\left (56 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 69 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 462 \, \sqrt {2} {\left (33 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 32 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2310 \, \sqrt {2} {\left (3 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 4 \, B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3465 \, \sqrt {2} {\left (A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {11}{2}} d} \] Input:

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, alg 
orithm="giac")
 

Output:

4/3465*(((((2*sqrt(2)*(483*A*a^7*sgn(cos(d*x + c)) + 517*B*a^7*sgn(cos(d*x 
 + c)))*tan(1/2*d*x + 1/2*c)^2 + 11*sqrt(2)*(483*A*a^7*sgn(cos(d*x + c)) + 
 517*B*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 198*sqrt(2)*(56*A* 
a^7*sgn(cos(d*x + c)) + 69*B*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^ 
2 + 462*sqrt(2)*(33*A*a^7*sgn(cos(d*x + c)) + 32*B*a^7*sgn(cos(d*x + c)))) 
*tan(1/2*d*x + 1/2*c)^2 + 2310*sqrt(2)*(3*A*a^7*sgn(cos(d*x + c)) + 4*B*a^ 
7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 3465*sqrt(2)*(A*a^7*sgn(cos 
(d*x + c)) + B*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d* 
x + 1/2*c)^2 + a)^(11/2)*d)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^{11/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:

int(cos(c + d*x)^(11/2)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2),x)
 

Output:

int(cos(c + d*x)^(11/2)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2), x 
)
 

Reduce [F]

\[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5}d x \right ) a \right ) \] Input:

int(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x)
 

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**5*s 
ec(c + d*x)**2,x)*b + int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c 
+ d*x)**5*sec(c + d*x),x)*a + int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x) 
)*cos(c + d*x)**5*sec(c + d*x),x)*b + int(sqrt(sec(c + d*x) + 1)*sqrt(cos( 
c + d*x))*cos(c + d*x)**5,x)*a)