3.13.63 \(\int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1263]

3.13.63.1 Optimal result

Integrand size = 45, antiderivative size = 284 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^3 (2840 A+3212 B+3795 C) \sin (c+d x)}{3465 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2840 A+3212 B+3795 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (1160 A+1364 B+1485 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+44 B+33 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d}+\frac {2 a (5 A+11 B) \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d} \]

output

2/99*a*(5*A+11*B)*cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/1 
1*A*cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+2/3465*a^3*(1160* 
A+1364*B+1485*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+4/34 
65*a^3*(2840*A+3212*B+3795*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c 
))^(1/2)+2/3465*a^3*(2840*A+3212*B+3795*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/( 
a+a*sec(d*x+c))^(1/2)+2/231*a^2*(32*A+44*B+33*C)*cos(d*x+c)^(5/2)*sin(d*x+ 
c)*(a+a*sec(d*x+c))^(1/2)/d
 

3.13.63.2 Mathematica [A] (verified)

Time = 6.34 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.55 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sqrt {\cos (c+d x)} (114640 A+124366 B+137280 C+(69890 A+68552 B+66660 C) \cos (c+d x)+16 (1625 A+1397 B+990 C) \cos (2 (c+d x))+8675 A \cos (3 (c+d x))+5720 B \cos (3 (c+d x))+1980 C \cos (3 (c+d x))+2240 A \cos (4 (c+d x))+770 B \cos (4 (c+d x))+315 A \cos (5 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{27720 d} \]

input

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

output

(a^2*Sqrt[Cos[c + d*x]]*(114640*A + 124366*B + 137280*C + (69890*A + 68552 
*B + 66660*C)*Cos[c + d*x] + 16*(1625*A + 1397*B + 990*C)*Cos[2*(c + d*x)] 
 + 8675*A*Cos[3*(c + d*x)] + 5720*B*Cos[3*(c + d*x)] + 1980*C*Cos[3*(c + d 
*x)] + 2240*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[(c + d*x)/2])/(27720*d)
 

3.13.63.3 Rubi [A] (verified)

Time = 1.94 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.12, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {3042, 4753, 3042, 4574, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{9} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 (32 A+44 B+33 C) a^2+(56 A+44 B+99 C) \sec (c+d x) a^2\right )}{2 \sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (3 (32 A+44 B+33 C) a^2+(56 A+44 B+99 C) \sec (c+d x) a^2\right )}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 (32 A+44 B+33 C) a^2+(56 A+44 B+99 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4505

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {2}{7} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((1160 A+1364 B+1485 C) a^3+(776 A+836 B+1089 C) \sec (c+d x) a^3\right )}{2 \sec ^{\frac {5}{2}}(c+d x)}dx+\frac {6 a^3 (32 A+44 B+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((1160 A+1364 B+1485 C) a^3+(776 A+836 B+1089 C) \sec (c+d x) a^3\right )}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {6 a^3 (32 A+44 B+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((1160 A+1364 B+1485 C) a^3+(776 A+836 B+1089 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {6 a^3 (32 A+44 B+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4503

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} a^3 (2840 A+3212 B+3795 C) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^4 (1160 A+1364 B+1485 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^3 (32 A+44 B+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} a^3 (2840 A+3212 B+3795 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^4 (1160 A+1364 B+1485 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^3 (32 A+44 B+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4292

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} a^3 (2840 A+3212 B+3795 C) \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (1160 A+1364 B+1485 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^3 (32 A+44 B+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} a^3 (2840 A+3212 B+3795 C) \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^4 (1160 A+1364 B+1485 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^3 (32 A+44 B+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4291

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a^2 (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{9} \left (\frac {6 a^3 (32 A+44 B+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 a^4 (1160 A+1364 B+1485 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3}{5} a^3 (2840 A+3212 B+3795 C) \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )}{11 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)}\right )\)

input

Int[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + 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)) + ((2*a^2*(5*A + 11*B)*(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 + 44*B 
 + 33*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + 
 ((2*a^4*(1160*A + 1364*B + 1485*C)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)* 
Sqrt[a + a*Sec[c + d*x]]) + (3*a^3*(2840*A + 3212*B + 3795*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*Sec[c + d*x]])))/5)/7)/9)/(11*a))
 

3.13.63.3.1 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 4574

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(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*((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*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] 
, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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 
]
 

3.13.63.4 Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.60

\[-\frac {2 a^{2} \left (\left (315 \cos \left (d x +c \right )^{5}+1120 \cos \left (d x +c \right )^{4}+1775 \cos \left (d x +c \right )^{3}+2130 \cos \left (d x +c \right )^{2}+2840 \cos \left (d x +c \right )+5680\right ) A +\left (385 \cos \left (d x +c \right )^{4}+1430 \cos \left (d x +c \right )^{3}+2409 \cos \left (d x +c \right )^{2}+3212 \cos \left (d x +c \right )+6424\right ) B +\left (495 \cos \left (d x +c \right )^{3}+1980 \cos \left (d x +c \right )^{2}+3795 \cos \left (d x +c \right )+7590\right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{3465 d}\]

input

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

output

-2/3465*a^2/d*((315*cos(d*x+c)^5+1120*cos(d*x+c)^4+1775*cos(d*x+c)^3+2130* 
cos(d*x+c)^2+2840*cos(d*x+c)+5680)*A+(385*cos(d*x+c)^4+1430*cos(d*x+c)^3+2 
409*cos(d*x+c)^2+3212*cos(d*x+c)+6424)*B+(495*cos(d*x+c)^3+1980*cos(d*x+c) 
^2+3795*cos(d*x+c)+7590)*C)*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)*(cot 
(d*x+c)-csc(d*x+c))
 

3.13.63.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.58 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (315 \, A a^{2} \cos \left (d x + c\right )^{5} + 35 \, {\left (32 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (355 \, A + 286 \, B + 99 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (710 \, A + 803 \, B + 660 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (2840 \, A + 3212 \, B + 3795 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (2840 \, A + 3212 \, B + 3795 \, C\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

input

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

output

2/3465*(315*A*a^2*cos(d*x + c)^5 + 35*(32*A + 11*B)*a^2*cos(d*x + c)^4 + 5 
*(355*A + 286*B + 99*C)*a^2*cos(d*x + c)^3 + 3*(710*A + 803*B + 660*C)*a^2 
*cos(d*x + c)^2 + (2840*A + 3212*B + 3795*C)*a^2*cos(d*x + c) + 2*(2840*A 
+ 3212*B + 3795*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)
 

3.13.63.6 Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.13.63.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 925 vs. \(2 (248) = 496\).

Time = 0.52 (sec) , antiderivative size = 925, normalized size of antiderivative = 3.26 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

input

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

output

1/110880*(5*sqrt(2)*(31878*a^2*cos(10/11*arctan2(sin(11/2*d*x + 11/2*c), c 
os(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 8778*a^2*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) + 
 3465*a^2*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) + 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*arct 
an2(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*a 
rctan2(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(2/1 
1*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 126*a^2*sin(1 
1/2*d*x + 11/2*c) + 385*a^2*sin(9/11*arctan2(sin(11/2*d*x + 11/2*c), cos(1 
1/2*d*x + 11/2*c))) + 1287*a^2*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), co 
s(11/2*d*x + 11/2*c))) + 3465*a^2*sin(5/11*arctan2(sin(11/2*d*x + 11/2*c), 
 cos(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 + 1 
1/2*c), cos(11/2*d*x + 11/2*c))))*A*sqrt(a) + 44*sqrt(2)*(225*a^2*sin(7...
 

3.13.63.8 Giac [F]

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

input

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

output

sage0*x
 

3.13.63.9 Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{11/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

input

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

output

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