\(\int \frac {(a+a \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac {13}{2}}(c+d x)} \, dx\) [196]

Optimal result

Integrand size = 37, antiderivative size = 266 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\frac {2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (568 A+759 C) \sin (c+d x)}{693 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (32 A+33 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{231 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {10 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{99 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{11 d \cos ^{\frac {11}{2}}(c+d x)} \] Output:

2/693*a^3*(232*A+297*C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/ 
2)+2/693*a^3*(568*A+759*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^ 
(1/2)+4/693*a^3*(568*A+759*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c 
))^(1/2)+2/231*a^2*(32*A+33*C)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x 
+c)^(7/2)+10/99*a*A*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(9/2)+2 
/11*A*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d/cos(d*x+c)^(11/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.56 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (3628 A+2673 C+2 (5014 A+4983 C) \cos (c+d x)+52 (71 A+66 C) \cos (2 (c+d x))+3692 A \cos (3 (c+d x))+4587 C \cos (3 (c+d x))+568 A \cos (4 (c+d x))+759 C \cos (4 (c+d x))+568 A \cos (5 (c+d x))+759 C \cos (5 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{2772 d \cos ^{\frac {11}{2}}(c+d x)} \] Input:

Integrate[((a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x] 
^(13/2),x]
 

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(3628*A + 2673*C + 2*(5014*A + 4983*C)*Cos 
[c + d*x] + 52*(71*A + 66*C)*Cos[2*(c + d*x)] + 3692*A*Cos[3*(c + d*x)] + 
4587*C*Cos[3*(c + d*x)] + 568*A*Cos[4*(c + d*x)] + 759*C*Cos[4*(c + d*x)] 
+ 568*A*Cos[5*(c + d*x)] + 759*C*Cos[5*(c + d*x)])*Tan[(c + d*x)/2])/(2772 
*d*Cos[c + d*x]^(11/2))
 

Rubi [A] (verified)

Time = 1.61 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 3523, 27, 3042, 3454, 27, 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.

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{13/2}}dx\)

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {2}{9} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (3 (32 A+33 C) a^2+(56 A+99 C) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {9}{2}}(c+d x)}dx+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{9} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (3 (32 A+33 C) a^2+(56 A+99 C) \cos (c+d x) a^2\right )}{\cos ^{\frac {9}{2}}(c+d x)}dx+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{9} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 (32 A+33 C) a^2+(56 A+99 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{9} \left (\frac {2}{7} \int \frac {\sqrt {\cos (c+d x) a+a} \left (5 (232 A+297 C) a^3+(776 A+1089 C) \cos (c+d x) a^3\right )}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{9} \left (\frac {1}{7} \int \frac {\sqrt {\cos (c+d x) a+a} \left (5 (232 A+297 C) a^3+(776 A+1089 C) \cos (c+d x) a^3\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{9} \left (\frac {1}{7} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (5 (232 A+297 C) a^3+(776 A+1089 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {\frac {1}{9} \left (\frac {1}{7} \left (3 a^3 (568 A+759 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{9} \left (\frac {1}{7} \left (3 a^3 (568 A+759 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {\frac {1}{9} \left (\frac {1}{7} \left (3 a^3 (568 A+759 C) \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{9} \left (\frac {1}{7} \left (3 a^3 (568 A+759 C) \left (\frac {2}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

\(\Big \downarrow \) 3250

\(\displaystyle \frac {\frac {10 a^2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {1}{9} \left (\frac {6 a^3 (32 A+33 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 a^4 (232 A+297 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+3 a^3 (568 A+759 C) \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\right )\right )}{11 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{11 d \cos ^{\frac {11}{2}}(c+d x)}\)

Input:

Int[((a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(13/2 
),x]
 

Output:

(2*A*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(11*d*Cos[c + d*x]^(11/2)) + 
 ((10*a^2*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/ 
2)) + ((6*a^3*(32*A + 33*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d*Co 
s[c + d*x]^(7/2)) + ((2*a^4*(232*A + 297*C)*Sin[c + d*x])/(d*Cos[c + d*x]^ 
(5/2)*Sqrt[a + a*Cos[c + d*x]]) + 3*a^3*(568*A + 759*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]])))/7)/9)/(11*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 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 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 = 3.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.52

method	result	size
default	\(\frac {2 a^{2} \sin \left (d x +c \right ) \left (\left (1136 \cos \left (d x +c \right )^{5}+568 \cos \left (d x +c \right )^{4}+426 \cos \left (d x +c \right )^{3}+355 \cos \left (d x +c \right )^{2}+224 \cos \left (d x +c \right )+63\right ) A +\cos \left (d x +c \right )^{2} \left (1518 \cos \left (d x +c \right )^{3}+759 \cos \left (d x +c \right )^{2}+396 \cos \left (d x +c \right )+99\right ) C \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{693 d \cos \left (d x +c \right )^{\frac {11}{2}} \left (1+\cos \left (d x +c \right )\right )}\)	\(138\)
parts	\(\frac {2 A \,a^{2} \sin \left (d x +c \right ) \left (1136 \cos \left (d x +c \right )^{5}+568 \cos \left (d x +c \right )^{4}+426 \cos \left (d x +c \right )^{3}+355 \cos \left (d x +c \right )^{2}+224 \cos \left (d x +c \right )+63\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{693 d \cos \left (d x +c \right )^{\frac {11}{2}} \left (1+\cos \left (d x +c \right )\right )}+\frac {2 C \,a^{2} \sin \left (d x +c \right ) \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{21 d \cos \left (d x +c \right )^{\frac {7}{2}} \left (1+\cos \left (d x +c \right )\right )}\)	\(172\)

Input:

int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x,method=_ 
RETURNVERBOSE)
 

Output:

2/693/d*a^2*sin(d*x+c)*((1136*cos(d*x+c)^5+568*cos(d*x+c)^4+426*cos(d*x+c) 
^3+355*cos(d*x+c)^2+224*cos(d*x+c)+63)*A+cos(d*x+c)^2*(1518*cos(d*x+c)^3+7 
59*cos(d*x+c)^2+396*cos(d*x+c)+99)*C)*(a*(1+cos(d*x+c)))^(1/2)/cos(d*x+c)^ 
(11/2)/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.56 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\frac {2 \, {\left (2 \, {\left (568 \, A + 759 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (568 \, A + 759 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 6 \, {\left (71 \, A + 66 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (355 \, A + 99 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 224 \, A a^{2} \cos \left (d x + c\right ) + 63 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x, a 
lgorithm="fricas")
 

Output:

2/693*(2*(568*A + 759*C)*a^2*cos(d*x + c)^5 + (568*A + 759*C)*a^2*cos(d*x 
+ c)^4 + 6*(71*A + 66*C)*a^2*cos(d*x + c)^3 + (355*A + 99*C)*a^2*cos(d*x + 
 c)^2 + 224*A*a^2*cos(d*x + c) + 63*A*a^2)*sqrt(a*cos(d*x + c) + a)*sqrt(c 
os(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^7 + d*cos(d*x + c)^6)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(13/2),x 
)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 579 vs. \(2 (230) = 460\).

Time = 0.21 (sec) , antiderivative size = 579, normalized size of antiderivative = 2.18 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x, a 
lgorithm="maxima")
 

Output:

8/693*(33*(21*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 56*sqrt(2) 
*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sqrt(2)*a^(5/2)*sin(d*x 
+ c)^5/(cos(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + 
 c) + 1)^7 + 8*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*C*(sin 
(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)/(cos(d*x + c) + 1) 
+ 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(2*sin(d*x + c)^2/ 
(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + (693*sq 
rt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 2310*sqrt(2)*a^(5/2)*sin(d 
*x + c)^3/(cos(d*x + c) + 1)^3 + 4620*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos( 
d*x + c) + 1)^5 - 5478*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 
 + 3575*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 1300*sqrt(2) 
*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 200*sqrt(2)*a^(5/2)*sin(d 
*x + c)^13/(cos(d*x + c) + 1)^13)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 
 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2)*(-sin(d*x + c)/(cos(d* 
x + c) + 1) + 1)^(13/2)*(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. \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x, a 
lgorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 4.19 (sec) , antiderivative size = 773, normalized size of antiderivative = 2.91 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^(13/2 
),x)
 

Output:

((a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*(568* 
A + 759*C)*4i)/(693*d) - (a^2*exp(c*5i + d*x*5i)*(3*A + 5*C)*16i)/(3*d) + 
(a^2*exp(c*6i + d*x*6i)*(3*A + 5*C)*16i)/(3*d) + (a^2*exp(c*4i + d*x*4i)*( 
32*A + 33*C)*8i)/(7*d) - (a^2*exp(c*7i + d*x*7i)*(32*A + 33*C)*8i)/(7*d) + 
 (a^2*exp(c*2i + d*x*2i)*(71*A + 87*C)*16i)/(63*d) - (a^2*exp(c*9i + d*x*9 
i)*(71*A + 87*C)*16i)/(63*d) - (a^2*exp(c*11i + d*x*11i)*(568*A + 759*C)*4 
i)/(693*d) - (C*a^2*exp(c*3i + d*x*3i)*20i)/(3*d) + (C*a^2*exp(c*8i + d*x* 
8i)*20i)/(3*d)))/((exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 
exp(c*1i + d*x*1i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 
 5*exp(c*2i + d*x*2i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2 
) + 5*exp(c*3i + d*x*3i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^( 
1/2) + 10*exp(c*4i + d*x*4i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/ 
2)^(1/2) + 10*exp(c*5i + d*x*5i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x* 
1i)/2)^(1/2) + 10*exp(c*6i + d*x*6i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + 
d*x*1i)/2)^(1/2) + 10*exp(c*7i + d*x*7i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1 
i + d*x*1i)/2)^(1/2) + 5*exp(c*8i + d*x*8i)*(exp(- c*1i - d*x*1i)/2 + exp( 
c*1i + d*x*1i)/2)^(1/2) + 5*exp(c*9i + d*x*9i)*(exp(- c*1i - d*x*1i)/2 + e 
xp(c*1i + d*x*1i)/2)^(1/2) + exp(c*10i + d*x*10i)*(exp(- c*1i - d*x*1i)/2 
+ exp(c*1i + d*x*1i)/2)^(1/2) + exp(c*11i + d*x*11i)*(exp(- c*1i - d*x*1i) 
/2 + exp(c*1i + d*x*1i)/2)^(1/2))
 

Reduce [F]

\[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\sqrt {a}\, a^{2} \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{7}}d x \right ) a +2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{6}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}}d x \right ) c +2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) c \right ) \] Input:

int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x)
 

Output:

sqrt(a)*a**2*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/cos(c + d*x) 
**7,x)*a + 2*int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/cos(c + d*x)* 
*6,x)*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)**5,x)* 
c + 2*int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/cos(c + d*x)**4,x)*c 
 + int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/cos(c + d*x)**3,x)*c)