\(\int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [1145]

Optimal result

Integrand size = 37, antiderivative size = 266 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (568 A+759 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (232 A+297 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d}+\frac {10 a A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d} \] Output:

4/693*a^3*(568*A+759*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/ 
2)+2/693*a^3*(568*A+759*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^ 
(1/2)+2/693*a^3*(232*A+297*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c 
))^(1/2)+2/231*a^2*(32*A+33*C)*cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(1/2)*sin 
(d*x+c)/d+10/99*a*A*cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2 
/11*A*cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.68 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.50 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sqrt {\cos (c+d x)} (22928 A+27456 C+2 (6989 A+6666 C) \cos (c+d x)+16 (325 A+198 C) \cos (2 (c+d x))+1735 A \cos (3 (c+d x))+396 C \cos (3 (c+d x))+448 A \cos (4 (c+d x))+63 A \cos (5 (c+d x))) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{5544 d (1+\cos (c+d x))} \] Input:

Integrate[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d* 
x]^2),x]
 

Output:

(a^2*Sqrt[Cos[c + d*x]]*(22928*A + 27456*C + 2*(6989*A + 6666*C)*Cos[c + d 
*x] + 16*(325*A + 198*C)*Cos[2*(c + d*x)] + 1735*A*Cos[3*(c + d*x)] + 396* 
C*Cos[3*(c + d*x)] + 448*A*Cos[4*(c + d*x)] + 63*A*Cos[5*(c + d*x)])*Sqrt[ 
a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(5544*d*(1 + Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 1.91 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.13, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.459, Rules used = {3042, 4753, 3042, 4575, 27, 3042, 4505, 27, 3042, 4505, 27, 3042, 4503, 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.

\(\displaystyle \int \cos ^{\frac {11}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^{11/2} (a \sec (c+d x)+a)^{5/2} \left (A+C \sec (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4753

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{5/2} \left (C \sec ^2(c+d x)+A\right )}{\sec ^{\frac {11}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\)

\(\Big \downarrow \) 4575

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\sec (c+d x) a+a)^{5/2} (5 a A+a (4 A+11 C) \sec (c+d x))}{2 \sec ^{\frac {9}{2}}(c+d x)}dx}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\sec (c+d x) a+a)^{5/2} (5 a A+a (4 A+11 C) \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)}dx}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a A+a (4 A+11 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4505

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{9} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 (32 A+33 C) a^2+(56 A+99 C) \sec (c+d x) a^2\right )}{2 \sec ^{\frac {7}{2}}(c+d x)}dx+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 (32 A+33 C) a^2+(56 A+99 C) \sec (c+d x) a^2\right )}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 (32 A+33 C) a^2+(56 A+99 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4505

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {2}{7} \int \frac {\sqrt {\sec (c+d x) a+a} \left (5 (232 A+297 C) a^3+(776 A+1089 C) \sec (c+d x) a^3\right )}{2 \sec ^{\frac {5}{2}}(c+d x)}dx+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \int \frac {\sqrt {\sec (c+d x) a+a} \left (5 (232 A+297 C) a^3+(776 A+1089 C) \sec (c+d x) a^3\right )}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (5 (232 A+297 C) a^3+(776 A+1089 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4503

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \left (3 a^3 (568 A+759 C) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \left (3 a^3 (568 A+759 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4292

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \left (3 a^3 (568 A+759 C) \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \left (3 a^3 (568 A+759 C) \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4291

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {10 a^2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{9} \left (\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+3 a^3 (568 A+759 C) \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

Input:

Int[Cos[c + d*x]^(11/2)*(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])/(11*d*Sec[c + d*x]^(9/2)) + ((10*a^2*A*(a + a*Sec[c + d*x])^(3/ 
2)*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + ((6*a^3*(32*A + 33*C)*Sqrt[a + 
 a*Sec[c + d*x]]*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ((2*a^4*(232*A + 
 297*C)*Sin[c + d*x])/(d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + 3* 
a^3*(568*A + 759*C)*((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*Se 
c[c + d*x]])))/7)/9)/(11*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 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]
 

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 = 1.77 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.49

\[\frac {2 \sin \left (d x +c \right ) \left (\left (63 \cos \left (d x +c \right )^{5}+224 \cos \left (d x +c \right )^{4}+355 \cos \left (d x +c \right )^{3}+426 \cos \left (d x +c \right )^{2}+568 \cos \left (d x +c \right )+1136\right ) A +\left (99 \cos \left (d x +c \right )^{3}+396 \cos \left (d x +c \right )^{2}+759 \cos \left (d x +c \right )+1518\right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, a^{2}}{693 d \left (\cos \left (d x +c \right )+1\right )}\]

Input:

int(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)
 

Output:

2/693/d*sin(d*x+c)*((63*cos(d*x+c)^5+224*cos(d*x+c)^4+355*cos(d*x+c)^3+426 
*cos(d*x+c)^2+568*cos(d*x+c)+1136)*A+(99*cos(d*x+c)^3+396*cos(d*x+c)^2+759 
*cos(d*x+c)+1518)*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 = 146, normalized size of antiderivative = 0.55 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (63 \, A a^{2} \cos \left (d x + c\right )^{5} + 224 \, A a^{2} \cos \left (d x + c\right )^{4} + {\left (355 \, A + 99 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 6 \, {\left (71 \, A + 66 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (568 \, A + 759 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (568 \, A + 759 \, C\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, a 
lgorithm="fricas")
 

Output:

2/693*(63*A*a^2*cos(d*x + c)^5 + 224*A*a^2*cos(d*x + c)^4 + (355*A + 99*C) 
*a^2*cos(d*x + c)^3 + 6*(71*A + 66*C)*a^2*cos(d*x + c)^2 + (568*A + 759*C) 
*a^2*cos(d*x + c) + 2*(568*A + 759*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. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(11/2)*(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. 695 vs. \(2 (230) = 460\).

Time = 0.28 (sec) , antiderivative size = 695, normalized size of antiderivative = 2.61 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, a 
lgorithm="maxima")
 

Output:

1/22176*(sqrt(2)*(31878*a^2*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) + 8778*a^2*cos(8/11*arctan2(si 
n(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 34 
65*a^2*cos(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*s 
in(11/2*d*x + 11/2*c) + 1287*a^2*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) + 385*a^2*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) - 
 31878*a^2*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))) - 8778*a^2*cos(11/2*d*x + 11/2*c)*sin(8/11*arct 
an2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 3465*a^2*cos(11/2*d 
*x + 11/2*c)*sin(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2* 
c))) - 1287*a^2*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))) - 385*a^2*cos(11/2*d*x + 11/2*c)*sin(2/11*a 
rctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 126*a^2*sin(11/2 
*d*x + 11/2*c) + 385*a^2*sin(9/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2 
*d*x + 11/2*c))) + 1287*a^2*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(1 
1/2*d*x + 11/2*c))) + 3465*a^2*sin(5/11*arctan2(sin(11/2*d*x + 11/2*c), co 
s(11/2*d*x + 11/2*c))) + 8778*a^2*sin(3/11*arctan2(sin(11/2*d*x + 11/2*c), 
 cos(11/2*d*x + 11/2*c))) + 31878*a^2*sin(1/11*arctan2(sin(11/2*d*x + 11/2 
*c), cos(11/2*d*x + 11/2*c))))*A*sqrt(a) - 132*sqrt(2)*(77*a^2*cos(7/4*...
 

Giac [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.09 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (693 \, \sqrt {2} A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 693 \, \sqrt {2} C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (1617 \, \sqrt {2} A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2541 \, \sqrt {2} C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (3003 \, \sqrt {2} A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3927 \, \sqrt {2} C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (2475 \, \sqrt {2} A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3267 \, \sqrt {2} C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 4 \, {\left (275 \, \sqrt {2} A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 363 \, \sqrt {2} C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, {\left (25 \, \sqrt {2} A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 33 \, \sqrt {2} C a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{693 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {11}{2}} d} \] Input:

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, a 
lgorithm="giac")
 

Output:

8/693*(693*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 693*sqrt(2)*C*a^8*sgn(cos(d*x 
 + c)) + (1617*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 2541*sqrt(2)*C*a^8*sgn(co 
s(d*x + c)) + (3003*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 3927*sqrt(2)*C*a^8*s 
gn(cos(d*x + c)) + (2475*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 3267*sqrt(2)*C* 
a^8*sgn(cos(d*x + c)) + 4*(275*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 363*sqrt( 
2)*C*a^8*sgn(cos(d*x + c)) + 2*(25*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 33*sq 
rt(2)*C*a^8*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c 
)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^ 
2)*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. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{11/2}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:

int(cos(c + d*x)^(11/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2), 
x)
 

Output:

int(cos(c + d*x)^(11/2)*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2), 
 x)
 

Reduce [F]

\[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )^{4}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )^{2}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )^{2}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5}d x \right ) a \right ) \] Input:

int(cos(d*x+c)^(11/2)*(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)** 
5*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)**5*sec(c + d*x)**3,x)*c + int(sqrt(sec(c + d*x) + 1)*sqrt(cos( 
c + d*x))*cos(c + d*x)**5*sec(c + d*x)**2,x)*a + int(sqrt(sec(c + d*x) + 1 
)*sqrt(cos(c + d*x))*cos(c + d*x)**5*sec(c + d*x)**2,x)*c + 2*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,x)*a)