\(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\) [503]

Optimal result

Integrand size = 35, antiderivative size = 181 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {4 a^2 (52 A+63 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (52 A+63 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (8 A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \] Output:

4/105*a^2*(52*A+63*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2) 
+2/105*a^2*(52*A+63*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2 
)+2/35*a^2*(8*A+7*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+ 
2/7*a*A*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^(7/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.56 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} (82 A+63 B+3 (78 A+77 B) \cos (c+d x)+(52 A+63 B) \cos (2 (c+d x))+52 A \cos (3 (c+d x))+63 B \cos (3 (c+d x))) \sec ^{\frac {7}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{105 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/ 
2),x]
 

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*(82*A + 63*B + 3*(78*A + 77*B)*Cos[c + d*x] 
+ (52*A + 63*B)*Cos[2*(c + d*x)] + 52*A*Cos[3*(c + d*x)] + 63*B*Cos[3*(c + 
 d*x)])*Sec[c + d*x]^(7/2)*Tan[(c + d*x)/2])/(105*d)
 

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 3440, 3042, 3454, 27, 3042, 3459, 3042, 3251, 3042, 3250}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3440

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{7} \int \frac {\sqrt {\cos (c+d x) a+a} (a (8 A+7 B)+a (4 A+7 B) \cos (c+d x))}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \int \frac {\sqrt {\cos (c+d x) a+a} (a (8 A+7 B)+a (4 A+7 B) \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (8 A+7 B)+a (4 A+7 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3459

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3251

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \left (\frac {2}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3250

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {1}{5} a (52 A+63 B) \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\)

Input:

Int[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sin 
[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((2*a^2*(8*A + 7*B)*Sin[c + d*x])/(5 
*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (a*(52*A + 63*B)*((2*a*S 
in[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a*Sin[ 
c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])))/5)/7)
 

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 3250

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

rule 3251

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

rule 3440

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Sin[e + f*x])^m*((c + 
d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

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]
 

Maple [A] (verified)

Time = 14.00 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.64

method	result	size
default	\(\frac {2 \sin \left (d x +c \right ) \left (\left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) A +\cos \left (d x +c \right ) \left (126 \cos \left (d x +c \right )^{2}+63 \cos \left (d x +c \right )+21\right ) B \right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{\frac {9}{2}} a \sqrt {2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\)	\(116\)
parts	\(\frac {2 A \sin \left (d x +c \right ) \left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sec \left (d x +c \right )^{\frac {9}{2}} \cos \left (d x +c \right ) a \sqrt {2}}{105 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \sin \left (d x +c \right ) \left (4+3 \cos \left (2 d x +2 c \right )+3 \cos \left (d x +c \right )\right ) \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {9}{2}} \cos \left (d x +c \right )^{2} a}{5 d \left (\cos \left (d x +c \right )+1\right )}\)	\(159\)

Input:

int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x,method=_RET 
URNVERBOSE)
 

Output:

2/105/d*sin(d*x+c)*((104*cos(d*x+c)^3+52*cos(d*x+c)^2+39*cos(d*x+c)+15)*A+ 
cos(d*x+c)*(126*cos(d*x+c)^2+63*cos(d*x+c)+21)*B)*(a*cos(1/2*d*x+1/2*c)^2) 
^(1/2)*cos(d*x+c)*sec(d*x+c)^(9/2)/(cos(d*x+c)+1)*a*2^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.59 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {2 \, {\left (2 \, {\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (13 \, A + 7 \, B\right )} a \cos \left (d x + c\right ) + 15 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algo 
rithm="fricas")
 

Output:

2/105*(2*(52*A + 63*B)*a*cos(d*x + c)^3 + (52*A + 63*B)*a*cos(d*x + c)^2 + 
 3*(13*A + 7*B)*a*cos(d*x + c) + 15*A*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x 
+ c)/((d*cos(d*x + c)^4 + d*cos(d*x + c)^3)*sqrt(cos(d*x + c)))
 

Sympy [F(-1)]

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

integrate((a+a*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c))*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. 527 vs. \(2 (157) = 314\).

Time = 0.23 (sec) , antiderivative size = 527, normalized size of antiderivative = 2.91 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx =\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algo 
rithm="maxima")
 

Output:

4/105*((105*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 245*sqrt(2)* 
a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 273*sqrt(2)*a^(3/2)*sin(d*x 
+ c)^5/(cos(d*x + c) + 1)^5 - 171*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x 
+ c) + 1)^7 + 38*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*A*(s 
in(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1 
) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(3*sin(d*x + c)^ 
2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + 
 c)^6/(cos(d*x + c) + 1)^6 + 1)) + 21*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos 
(d*x + c) + 1) - 15*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 
17*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 9*sqrt(2)*a^(3/2) 
*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 2*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(c 
os(d*x + c) + 1)^9)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d* 
x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1 
)^(9/2)*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x 
 + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)))/d
 

Giac [F(-1)]

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

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algo 
rithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 32.11 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.43 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {\sqrt {\frac {1}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (-\frac {8\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A+3\,B\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{3\,d}+\frac {16\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (13\,A+12\,B\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{15\,d}+\frac {8\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (52\,A+63\,B\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{105\,d}\right )}{6\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )+2\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )+2\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )} \] Input:

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(3/2) 
,x)
 

Output:

((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((16*a*exp((c*7 
i)/2 + (d*x*7i)/2)*sin((3*c)/2 + (3*d*x)/2)*(13*A + 12*B)*(a + a*cos(c + d 
*x))^(1/2))/(15*d) - (8*a*exp((c*7i)/2 + (d*x*7i)/2)*sin(c/2 + (d*x)/2)*(2 
*A + 3*B)*(a + a*cos(c + d*x))^(1/2))/(3*d) + (8*a*exp((c*7i)/2 + (d*x*7i) 
/2)*sin((7*c)/2 + (7*d*x)/2)*(52*A + 63*B)*(a + a*cos(c + d*x))^(1/2))/(10 
5*d)))/(6*exp((c*7i)/2 + (d*x*7i)/2)*cos(c/2 + (d*x)/2) + 6*exp((c*7i)/2 + 
 (d*x*7i)/2)*cos((3*c)/2 + (3*d*x)/2) + 2*exp((c*7i)/2 + (d*x*7i)/2)*cos(( 
5*c)/2 + (5*d*x)/2) + 2*exp((c*7i)/2 + (d*x*7i)/2)*cos((7*c)/2 + (7*d*x)/2 
))
 

Reduce [F]

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

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

Output:

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