\(\int \frac {\sqrt {a+a \cos (c+d x)} (A+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [177]

Optimal result

Integrand size = 37, antiderivative size = 168 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a A \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a (24 A+35 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a (24 A+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \] Output:

2/35*a*A*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)+2/105*a*(24* 
A+35*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+4/105*a*(24*A 
+35*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)+2/7*A*(a+a*cos 
(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(7/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (54 A+35 C+3 (36 A+35 C) \cos (c+d x)+(24 A+35 C) \cos (2 (c+d x))+24 A \cos (3 (c+d x))+35 C \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{105 d \cos ^{\frac {7}{2}}(c+d x)} \] Input:

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

Output:

(Sqrt[a*(1 + Cos[c + d*x])]*(54*A + 35*C + 3*(36*A + 35*C)*Cos[c + d*x] + 
(24*A + 35*C)*Cos[2*(c + d*x)] + 24*A*Cos[3*(c + d*x)] + 35*C*Cos[3*(c + d 
*x)])*Tan[(c + d*x)/2])/(105*d*Cos[c + d*x]^(7/2))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {3042, 3523, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

\(\displaystyle \frac {2 \int \frac {\sqrt {\cos (c+d x) a+a} (a A+a (4 A+7 C) \cos (c+d x))}{2 \cos ^{\frac {7}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x) a+a} (a A+a (4 A+7 C) \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3459

\(\displaystyle \frac {\frac {1}{5} a (24 A+35 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} a (24 A+35 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {\frac {1}{5} a (24 A+35 C) \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} a (24 A+35 C) \left (\frac {2}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3250

\(\displaystyle \frac {\frac {2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {1}{5} a (24 A+35 C) \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

Input:

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

Output:

(2*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((2 
*a^2*A*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + ( 
a*(24*A + 35*C)*((2*a*Sin[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*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 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 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 3523

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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])
 

Maple [A] (verified)

Time = 13.79 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01

method	result	size
parts	\(\frac {2 A \sqrt {2}\, \left (128 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-160 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+76 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35 d \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{\frac {7}{2}}}+\frac {2 C \sqrt {2}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{\frac {3}{2}}}\)	\(169\)
default	\(\sqrt {2}\, \left (\frac {2 A \left (128 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-160 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+76 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35 d \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{\frac {7}{2}}}+\frac {2 C \left (128 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-160 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+76 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35 d \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{\frac {7}{2}}}-\frac {8 C \left (416 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-520 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+247 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-38\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{105 d \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{\frac {7}{2}}}+\frac {8 C \left (92 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-115 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+52 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-8\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{21 d \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{\frac {7}{2}}}\right )\)	\(358\)

Input:

int((a+a*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x,method=_R 
ETURNVERBOSE)
 

Output:

2/35*A*2^(1/2)/d/(-1+2*cos(1/2*d*x+1/2*c)^2)^(7/2)*(128*cos(1/2*d*x+1/2*c) 
^6-160*cos(1/2*d*x+1/2*c)^4+76*cos(1/2*d*x+1/2*c)^2-9)*(a*cos(1/2*d*x+1/2* 
c)^2)^(1/2)*tan(1/2*d*x+1/2*c)+2/3*C*2^(1/2)/d/(-1+2*cos(1/2*d*x+1/2*c)^2) 
^(3/2)*(2*cos(1/2*d*x+1/2*c)+1)*(-1+2*cos(1/2*d*x+1/2*c))*(a*cos(1/2*d*x+1 
/2*c)^2)^(1/2)*tan(1/2*d*x+1/2*c)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (2 \, {\left (24 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (24 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 18 \, A \cos \left (d x + c\right ) + 15 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, al 
gorithm="fricas")
 

Output:

2/105*(2*(24*A + 35*C)*cos(d*x + c)^3 + (24*A + 35*C)*cos(d*x + c)^2 + 18* 
A*cos(d*x + c) + 15*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x 
 + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)
 

Sympy [F(-1)]

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

integrate((a+a*cos(d*x+c))**(1/2)*(A+C*cos(d*x+c)**2)/cos(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. 475 vs. \(2 (144) = 288\).

Time = 0.18 (sec) , antiderivative size = 475, normalized size of antiderivative = 2.83 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, al 
gorithm="maxima")
 

Output:

2/105*(35*C*(3*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sqrt(2) 
*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sqrt(2)*sqrt(a)*sin(d*x + c 
)^5/(cos(d*x + c) + 1)^5)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((si 
n(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) 
 + 1)^(5/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d 
*x + c) + 1)^4 + 1)) + 3*A*(35*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) 
+ 1) - 70*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 84*sqrt(2) 
*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 58*sqrt(2)*sqrt(a)*sin(d*x 
+ c)^7/(cos(d*x + c) + 1)^7 + 9*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + 
c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos 
(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(4* 
sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^ 
4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 
 1)^8 + 1)))/d
 

Giac [F(-1)]

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

integrate((a+a*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, al 
gorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 3.41 (sec) , antiderivative size = 479, normalized size of antiderivative = 2.85 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

((a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(((96*A + 1 
40*C)*1i)/(105*d) - (C*exp(c*3i + d*x*3i)*4i)/(3*d) + (C*exp(c*4i + d*x*4i 
)*4i)/(3*d) - (exp(c*7i + d*x*7i)*(96*A + 140*C)*1i)/(105*d) + (exp(c*2i + 
 d*x*2i)*(336*A + 280*C)*1i)/(105*d) - (exp(c*5i + d*x*5i)*(336*A + 280*C) 
*1i)/(105*d)))/((exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + ex 
p(c*1i + d*x*1i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 3 
*exp(c*2i + d*x*2i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) 
+ 3*exp(c*3i + d*x*3i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/ 
2) + 3*exp(c*4i + d*x*4i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^ 
(1/2) + 3*exp(c*5i + d*x*5i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/ 
2)^(1/2) + exp(c*6i + d*x*6i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i) 
/2)^(1/2) + exp(c*7i + d*x*7i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i 
)/2)^(1/2))
 

Reduce [F]

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

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

Output:

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