\(\int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [1146]

Optimal result

Integrand size = 37, antiderivative size = 216 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (13 A+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (13 A+21 C) \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (13 A+21 C) \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {10 A \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d} \] Output:

64/315*a^3*(13*A+21*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2 
)+16/315*a^2*(13*A+21*C)*cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c 
)/d+2/105*a*(13*A+21*C)*cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c) 
/d+10/63*A*cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+2/9*A*cos( 
d*x+c)^(7/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.51 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sqrt {\cos (c+d x)} (5653 A+7476 C+4 (779 A+588 C) \cos (c+d x)+4 (254 A+63 C) \cos (2 (c+d x))+260 A \cos (3 (c+d x))+35 A \cos (4 (c+d x))) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{1260 d (1+\cos (c+d x))} \] Input:

Integrate[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x 
]^2),x]
 

Output:

(a^2*Sqrt[Cos[c + d*x]]*(5653*A + 7476*C + 4*(779*A + 588*C)*Cos[c + d*x] 
+ 4*(254*A + 63*C)*Cos[2*(c + d*x)] + 260*A*Cos[3*(c + d*x)] + 35*A*Cos[4* 
(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(1260*d*(1 + Cos[c + 
d*x]))
 

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 4753, 3042, 4575, 27, 3042, 4501, 3042, 4296, 3042, 4296, 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.

\(\displaystyle \int \cos ^{\frac {9}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^{9/2} (a \sec (c+d x)+a)^{5/2} \left (A+C \sec (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4753

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{5/2} \left (C \sec ^2(c+d x)+A\right )}{\sec ^{\frac {9}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\)

\(\Big \downarrow \) 4575

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\sec (c+d x) a+a)^{5/2} (5 a A+a (2 A+9 C) \sec (c+d x))}{2 \sec ^{\frac {7}{2}}(c+d x)}dx}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (5 a A+a (2 A+9 C) \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)}dx}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a A+a (2 A+9 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4501

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{7} a (13 A+21 C) \int \frac {(\sec (c+d x) a+a)^{5/2}}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {10 a A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{7} a (13 A+21 C) \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {10 a A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4296

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{7} a (13 A+21 C) \left (\frac {8}{5} a \int \frac {(\sec (c+d x) a+a)^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {10 a A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{7} a (13 A+21 C) \left (\frac {8}{5} a \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {10 a A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4296

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{7} a (13 A+21 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {10 a A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{7} a (13 A+21 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {10 a A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4291

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{7} a (13 A+21 C) \left (\frac {8}{5} a \left (\frac {8 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {10 a A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

Input:

Int[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x 
]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Sec[c + d*x])^(5/2)*Sin 
[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + ((10*a*A*(a + a*Sec[c + d*x])^(5/2)* 
Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (3*a*(13*A + 21*C)*((2*a*(a + a*S 
ec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*a*((8*a^2*S 
qrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt 
[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])))/5))/7)/(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 3042

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

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 4296

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

rule 4501

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

rule 4575

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

rule 4753

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

Maple [A] (verified)

Time = 1.72 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.51

\[\frac {2 \sin \left (d x +c \right ) \left (\left (35 \cos \left (d x +c \right )^{4}+130 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+292 \cos \left (d x +c \right )+584\right ) A +\left (63 \cos \left (d x +c \right )^{2}+294 \cos \left (d x +c \right )+903\right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, a^{2}}{315 d \left (\cos \left (d x +c \right )+1\right )}\]

Input:

int(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)
 

Output:

2/315/d*sin(d*x+c)*((35*cos(d*x+c)^4+130*cos(d*x+c)^3+219*cos(d*x+c)^2+292 
*cos(d*x+c)+584)*A+(63*cos(d*x+c)^2+294*cos(d*x+c)+903)*C)*cos(d*x+c)^(1/2 
)*(a*(1+sec(d*x+c)))^(1/2)*a^2/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.59 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (35 \, A a^{2} \cos \left (d x + c\right )^{4} + 130 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (73 \, A + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (146 \, A + 147 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (584 \, A + 903 \, C\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, al 
gorithm="fricas")
                                                                                    
                                                                                    
 

Output:

2/315*(35*A*a^2*cos(d*x + c)^4 + 130*A*a^2*cos(d*x + c)^3 + 3*(73*A + 21*C 
)*a^2*cos(d*x + c)^2 + 2*(146*A + 147*C)*a^2*cos(d*x + c) + (584*A + 903*C 
)*a^2)*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. \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(9/2)*(a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (186) = 372\).

Time = 0.27 (sec) , antiderivative size = 580, normalized size of antiderivative = 2.69 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, al 
gorithm="maxima")
 

Output:

1/5040*(sqrt(2)*(8190*a^2*cos(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d* 
x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 2100*a^2*cos(2/3*arctan2(sin(9/2*d*x + 
 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 756*a^2*cos(4/9*arc 
tan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 2 
25*a^2*cos(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/ 
2*d*x + 9/2*c) - 8190*a^2*cos(9/2*d*x + 9/2*c)*sin(8/9*arctan2(sin(9/2*d*x 
 + 9/2*c), cos(9/2*d*x + 9/2*c))) - 2100*a^2*cos(9/2*d*x + 9/2*c)*sin(2/3* 
arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 756*a^2*cos(9/2*d*x 
 + 9/2*c)*sin(4/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 2 
25*a^2*cos(9/2*d*x + 9/2*c)*sin(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2* 
d*x + 9/2*c))) + 70*a^2*sin(9/2*d*x + 9/2*c) + 225*a^2*sin(7/9*arctan2(sin 
(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 756*a^2*sin(5/9*arctan2(sin(9/ 
2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 2100*a^2*sin(1/3*arctan2(sin(9/2* 
d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 8190*a^2*sin(1/9*arctan2(sin(9/2*d* 
x + 9/2*c), cos(9/2*d*x + 9/2*c))))*A*sqrt(a) - 168*(75*sqrt(2)*a^2*cos(5/ 
4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 25*sqrt( 
2)*a^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 75*sqrt(2)*a 
^2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(25*sqrt(2)*a^ 
2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2* 
d*x + 2*c))))*C*sqrt(a))/d
 

Giac [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.12 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (13 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 21 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, \sqrt {2} {\left (13 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 21 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 63 \, \sqrt {2} {\left (13 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 21 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 210 \, \sqrt {2} {\left (3 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, \sqrt {2} {\left (A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {9}{2}} d} \] Input:

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, al 
gorithm="giac")
 

Output:

8/315*(((4*(2*sqrt(2)*(13*A*a^7*sgn(cos(d*x + c)) + 21*C*a^7*sgn(cos(d*x + 
 c)))*tan(1/2*d*x + 1/2*c)^2 + 9*sqrt(2)*(13*A*a^7*sgn(cos(d*x + c)) + 21* 
C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 63*sqrt(2)*(13*A*a^7*sg 
n(cos(d*x + c)) + 21*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 21 
0*sqrt(2)*(3*A*a^7*sgn(cos(d*x + c)) + 5*C*a^7*sgn(cos(d*x + c))))*tan(1/2 
*d*x + 1/2*c)^2 + 315*sqrt(2)*(A*a^7*sgn(cos(d*x + c)) + C*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)^(9/2)*d)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{9/2}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:

int(cos(c + d*x)^(9/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2),x 
)
 

Output:

int(cos(c + d*x)^(9/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2), 
x)
 

Reduce [F]

\[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{4}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4}d x \right ) a \right ) \] Input:

int(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)
 

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)** 
4*sec(c + d*x)**4,x)*c + 2*int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*c 
os(c + d*x)**4*sec(c + d*x)**3,x)*c + int(sqrt(sec(c + d*x) + 1)*sqrt(cos( 
c + d*x))*cos(c + d*x)**4*sec(c + d*x)**2,x)*a + int(sqrt(sec(c + d*x) + 1 
)*sqrt(cos(c + d*x))*cos(c + d*x)**4*sec(c + d*x)**2,x)*c + 2*int(sqrt(sec 
(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**4*sec(c + d*x),x)*a + int( 
sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**4,x)*a)