\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\) [1330]

Optimal result

Integrand size = 45, antiderivative size = 242 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 a^{5/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 a^3 (160 A+224 B+245 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (40 A+56 B+35 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a (5 A+7 B) (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \] Output:

2*a^(5/2)*C*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^( 
1/2)*sec(d*x+c)^(1/2)/d+2/105*a^3*(160*A+224*B+245*C)*sec(d*x+c)^(1/2)*sin 
(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/105*a^2*(40*A+56*B+35*C)*(a+a*cos(d*x+c 
))^(1/2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/35*a*(5*A+7*B)*(a+a*cos(d*x+c))^( 
3/2)*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/7*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c) 
^(7/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.80 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.71 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \left (420 \sqrt {2} C \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {7}{2}}(c+d x)+2 (290 A+196 B+70 C+(930 A+987 B+840 C) \cos (c+d x)+2 (115 A+98 B+35 C) \cos (2 (c+d x))+230 A \cos (3 (c+d x))+301 B \cos (3 (c+d x))+280 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{420 d} \] Input:

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

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(7/2)*(420*S 
qrt[2]*C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(7/2) + 2*(290*A + 
196*B + 70*C + (930*A + 987*B + 840*C)*Cos[c + d*x] + 2*(115*A + 98*B + 35 
*C)*Cos[2*(c + d*x)] + 230*A*Cos[3*(c + d*x)] + 301*B*Cos[3*(c + d*x)] + 2 
80*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(420*d)
 

Rubi [A] (verified)

Time = 1.63 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.356, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3459, 3042, 3253, 223}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3522

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{5} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((40 A+56 B+35 C) a^2+35 C \cos (c+d x) a^2\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((40 A+56 B+35 C) a^2+35 C \cos (c+d x) a^2\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((40 A+56 B+35 C) a^2+35 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3454

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((160 A+224 B+245 C) a^3+105 C \cos (c+d x) a^3\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\cos (c+d x) a+a} \left ((160 A+224 B+245 C) a^3+105 C \cos (c+d x) a^3\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((160 A+224 B+245 C) a^3+105 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3459

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {1}{3} \left (105 a^3 C \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^4 (160 A+224 B+245 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {1}{3} \left (105 a^3 C \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^4 (160 A+224 B+245 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3253

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {2 a^4 (160 A+224 B+245 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {210 a^3 C \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (\frac {210 a^{7/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^4 (160 A+224 B+245 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\right )}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^(5/2)*Sin 
[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((2*a^2*(5*A + 7*B)*(a + a*Cos[c + d 
*x])^(3/2)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*a^3*(40*A + 56*B + 
 35*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ( 
(210*a^(7/2)*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d 
+ (2*a^4*(160*A + 224*B + 245*C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[ 
a + a*Cos[c + d*x]]))/3)/5)/(7*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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 3042

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

rule 3253

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

rule 3454

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   Int[Sqrt[a + b*Sin[e + f*x] 
]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1]
 

rule 3522

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

rule 4709

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

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.89

\[\frac {2 a^{2} \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \left (105 \cos \left (d x +c \right )^{5}+105 \cos \left (d x +c \right )^{4}\right )+\left (230 \cos \left (d x +c \right )^{3}+115 \cos \left (d x +c \right )^{2}+60 \cos \left (d x +c \right )+15\right ) \sin \left (d x +c \right ) A \cos \left (d x +c \right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \left (301 \cos \left (d x +c \right )^{2}+98 \cos \left (d x +c \right )+21\right ) B +\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \left (280 \cos \left (d x +c \right )+35\right ) C \right )}{105 d \left (1+\cos \left (d x +c \right )\right )}\]

Input:

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

Output:

2/105/d*a^2*((1+cos(d*x+c))*a)^(1/2)*sec(d*x+c)^(9/2)/(1+cos(d*x+c))*(C*(c 
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*t 
an(d*x+c))*(105*cos(d*x+c)^5+105*cos(d*x+c)^4)+(230*cos(d*x+c)^3+115*cos(d 
*x+c)^2+60*cos(d*x+c)+15)*sin(d*x+c)*A*cos(d*x+c)+sin(d*x+c)*cos(d*x+c)^2* 
(301*cos(d*x+c)^2+98*cos(d*x+c)+21)*B+sin(d*x+c)*cos(d*x+c)^3*(280*cos(d*x 
+c)+35)*C)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.78 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=-\frac {2 \, {\left (105 \, {\left (C a^{2} \cos \left (d x + c\right )^{4} + C a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left ({\left (230 \, A + 301 \, B + 280 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (115 \, A + 98 \, B + 35 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \] Input:

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

Output:

-2/105*(105*(C*a^2*cos(d*x + c)^4 + C*a^2*cos(d*x + c)^3)*sqrt(a)*arctan(s 
qrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - ((230 
*A + 301*B + 280*C)*a^2*cos(d*x + c)^3 + (115*A + 98*B + 35*C)*a^2*cos(d*x 
 + c)^2 + 3*(20*A + 7*B)*a^2*cos(d*x + c) + 15*A*a^2)*sqrt(a*cos(d*x + c) 
+ a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3 
)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2584 vs. \(2 (208) = 416\).

Time = 0.49 (sec) , antiderivative size = 2584, normalized size of antiderivative = 10.68 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/210*(7*(6*(a^2*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) 
+ 25*(a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 
2*c) + a^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos 
(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*sqrt( 
a) + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^ 
(1/4)*((15*a^2*sin(6*d*x + 6*c) + 50*a^2*sin(4*d*x + 4*c) + 58*a^2*sin(2*d 
*x + 2*c) - 20*(3*a^2*sin(6*d*x + 6*c) + 10*a^2*sin(4*d*x + 4*c) + 11*a^2* 
sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2 
0*(3*a^2*cos(6*d*x + 6*c) + 10*a^2*cos(4*d*x + 4*c) + 11*a^2*cos(2*d*x + 2 
*c) + 4*a^2)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(7/2 
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (15*a^2*cos(6*d*x + 6* 
c) + 50*a^2*cos(4*d*x + 4*c) + 58*a^2*cos(2*d*x + 2*c) + 23*a^2 + 20*(3*a^ 
2*cos(6*d*x + 6*c) + 10*a^2*cos(4*d*x + 4*c) + 11*a^2*cos(2*d*x + 2*c) + 4 
*a^2)*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 20*(3*a^2*sin 
(6*d*x + 6*c) + 10*a^2*sin(4*d*x + 4*c) + 11*a^2*sin(2*d*x + 2*c))*sin(7/2 
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(7/2*arctan2(sin(2*d*x + 
 2*c), cos(2*d*x + 2*c) + 1)) + 25*(a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x 
 + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c) 
, cos(2*d*x + 2*c) + 1)))*sqrt(a) + 15*((a^2*cos(2*d*x + 2*c)^4 + a^2*sin( 
2*d*x + 2*c)^4 + 4*a^2*cos(2*d*x + 2*c)^3 + 6*a^2*cos(2*d*x + 2*c)^2 + ...
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) 
+ C*cos(c + d*x)^2),x)
 

Output:

int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) 
+ C*cos(c + d*x)^2), x)
 

Reduce [F]

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

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

Output:

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