3.13.54 \(\int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1254]

3.13.54.1 Optimal result
3.13.54.2 Mathematica [A] (verified)
3.13.54.3 Rubi [A] (verified)
3.13.54.4 Maple [A] (verified)
3.13.54.5 Fricas [A] (verification not implemented)
3.13.54.6 Sympy [F(-1)]
3.13.54.7 Maxima [B] (verification not implemented)
3.13.54.8 Giac [F]
3.13.54.9 Mupad [F(-1)]

3.13.54.1 Optimal result

Integrand size = 45, antiderivative size = 232 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^2 (136 A+156 B+189 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+156 B+189 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (52 A+72 B+63 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (A+3 B) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{21 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d} \]

output
2/9*A*cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/315*a^2*(52*A 
+72*B+63*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+4/315*a^2 
*(136*A+156*B+189*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+ 
2/315*a^2*(136*A+156*B+189*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c 
))^(1/2)+2/21*a*(A+3*B)*cos(d*x+c)^(5/2)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2) 
/d
 
3.13.54.2 Mathematica [A] (verified)

Time = 4.06 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.53 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \sqrt {\cos (c+d x)} (2689 A+2964 B+3276 C+2 (799 A+759 B+756 C) \cos (c+d x)+4 (137 A+117 B+63 C) \cos (2 (c+d x))+170 A \cos (3 (c+d x))+90 B \cos (3 (c+d x))+35 A \cos (4 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{1260 d} \]

input
Integrate[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x 
] + C*Sec[c + d*x]^2),x]
 
output
(a*Sqrt[Cos[c + d*x]]*(2689*A + 2964*B + 3276*C + 2*(799*A + 759*B + 756*C 
)*Cos[c + d*x] + 4*(137*A + 117*B + 63*C)*Cos[2*(c + d*x)] + 170*A*Cos[3*( 
c + d*x)] + 90*B*Cos[3*(c + d*x)] + 35*A*Cos[4*(c + d*x)])*Sqrt[a*(1 + Sec 
[c + d*x])]*Tan[(c + d*x)/2])/(1260*d)
 
3.13.54.3 Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.13, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 4753, 3042, 4574, 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.

\(\displaystyle \int \cos ^{\frac {9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^{9/2} (a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4753

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )}{\sec ^{\frac {9}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2}{7} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((52 A+72 B+63 C) a^2+(40 A+36 B+63 C) \sec (c+d x) a^2\right )}{2 \sec ^{\frac {5}{2}}(c+d x)}dx+\frac {6 a^2 (A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((52 A+72 B+63 C) a^2+(40 A+36 B+63 C) \sec (c+d x) a^2\right )}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {6 a^2 (A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((52 A+72 B+63 C) a^2+(40 A+36 B+63 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {6 a^2 (A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4503

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \left (\frac {3}{5} a^2 (136 A+156 B+189 C) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^3 (52 A+72 B+63 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^2 (A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \left (\frac {3}{5} a^2 (136 A+156 B+189 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^3 (52 A+72 B+63 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^2 (A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4292

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \left (\frac {3}{5} a^2 (136 A+156 B+189 C) \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (52 A+72 B+63 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^2 (A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{7} \left (\frac {3}{5} a^2 (136 A+156 B+189 C) \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (52 A+72 B+63 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^2 (A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4291

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {6 a^2 (A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 a^3 (52 A+72 B+63 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3}{5} a^2 (136 A+156 B+189 C) \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )}{9 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\right )\)

input
Int[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*(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])^(3/2)*Sin 
[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + ((6*a^2*(A + 3*B)*Sqrt[a + a*Sec[c + 
 d*x]]*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ((2*a^3*(52*A + 72*B + 63* 
C)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (3*a^ 
2*(136*A + 156*B + 189*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*a))
 

3.13.54.3.1 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 
]
 
3.13.54.4 Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.60

method result size
default \(-\frac {2 a \left (\left (35 \cos \left (d x +c \right )^{4}+85 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+136 \cos \left (d x +c \right )+272\right ) A +\left (45 \cos \left (d x +c \right )^{3}+117 \cos \left (d x +c \right )^{2}+156 \cos \left (d x +c \right )+312\right ) B +\left (63 \cos \left (d x +c \right )^{2}+189 \cos \left (d x +c \right )+378\right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{315 d}\) \(139\)

input
int(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 
),x,method=_RETURNVERBOSE)
 
output
-2/315*a/d*((35*cos(d*x+c)^4+85*cos(d*x+c)^3+102*cos(d*x+c)^2+136*cos(d*x+ 
c)+272)*A+(45*cos(d*x+c)^3+117*cos(d*x+c)^2+156*cos(d*x+c)+312)*B+(63*cos( 
d*x+c)^2+189*cos(d*x+c)+378)*C)*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)* 
(cot(d*x+c)-csc(d*x+c))
 
3.13.54.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.57 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (35 \, A a \cos \left (d x + c\right )^{4} + 5 \, {\left (17 \, A + 9 \, B\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (34 \, A + 39 \, B + 21 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (136 \, A + 156 \, B + 189 \, C\right )} a \cos \left (d x + c\right ) + 2 \, {\left (136 \, A + 156 \, B + 189 \, C\right )} a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

input
integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d* 
x+c)^2),x, algorithm="fricas")
 
output
2/315*(35*A*a*cos(d*x + c)^4 + 5*(17*A + 9*B)*a*cos(d*x + c)^3 + 3*(34*A + 
 39*B + 21*C)*a*cos(d*x + c)^2 + (136*A + 156*B + 189*C)*a*cos(d*x + c) + 
2*(136*A + 156*B + 189*C)*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)
 
3.13.54.6 Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

input
integrate(cos(d*x+c)**(9/2)*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec( 
d*x+c)**2),x)
 
output
Timed out
 
3.13.54.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (202) = 404\).

Time = 0.50 (sec) , antiderivative size = 703, normalized size of antiderivative = 3.03 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

input
integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d* 
x+c)^2),x, algorithm="maxima")
 
output
1/5040*(sqrt(2)*(3780*a*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) + 1050*a*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) + 378*a*cos(4/9*arctan2(s 
in(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 135*a*c 
os(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) - 3780*a*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))) - 1050*a*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))) - 378*a*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))) - 135*a*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*sin(9/2*d*x + 9/2*c) + 135*a*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), c 
os(9/2*d*x + 9/2*c))) + 378*a*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/ 
2*d*x + 9/2*c))) + 1050*a*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d* 
x + 9/2*c))) + 3780*a*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 
9/2*c))))*A*sqrt(a) - 6*sqrt(2)*(175*a*cos(7/4*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 5*(35*a*cos(2*d*x + 2*c) + 6*a)*sin(7 
/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 126*a*sin(5/4*arctan2(si 
n(2*d*x + 2*c), cos(2*d*x + 2*c))) - 175*a*sin(3/4*arctan2(sin(2*d*x + 2*c 
), cos(2*d*x + 2*c))) - 1470*a*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
 + 2*c))))*B*sqrt(a) - 504*(10*sqrt(2)*a*cos(5/4*arctan2(sin(2*d*x + 2*...
 
3.13.54.8 Giac [F]

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

input
integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d* 
x+c)^2),x, algorithm="giac")
 
output
sage0*x
 
3.13.54.9 Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{9/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

input
int(cos(c + d*x)^(9/2)*(a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/ 
cos(c + d*x)^2),x)
 
output
int(cos(c + d*x)^(9/2)*(a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/ 
cos(c + d*x)^2), x)