\(\int \frac {(a+a \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [190]

Optimal result

Integrand size = 37, antiderivative size = 218 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^{5/2} (304 A+163 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a^3 (432 A+299 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (16 A+17 C) \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{32 d}+\frac {5 a C \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {C \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{4 d} \] Output:

1/64*a^(5/2)*(304*A+163*C)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2 
))/d+1/192*a^3*(432*A+299*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c) 
)^(1/2)+1/32*a^2*(16*A+17*C)*cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)*sin(d 
*x+c)/d+5/24*a*C*cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+1/4* 
C*cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.60 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} (304 A+163 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (528 A+581 C+(96 A+362 C) \cos (c+d x)+92 C \cos (2 (c+d x))+12 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{384 d} \] Input:

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

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*(304*A + 163*C 
)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(528*A + 581*C + 
 (96*A + 362*C)*Cos[c + d*x] + 92*C*Cos[2*(c + d*x)] + 12*C*Cos[3*(c + d*x 
)])*Sin[(c + d*x)/2]))/(384*d)
 

Rubi [A] (verified)

Time = 1.37 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {3042, 3525, 27, 3042, 3455, 27, 3042, 3455, 27, 3042, 3460, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3525

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (a (8 A+C)+5 a C \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx}{4 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (a (8 A+C)+5 a C \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (8 A+C)+5 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {1}{3} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((48 A+11 C) a^2+3 (16 A+17 C) \cos (c+d x) a^2\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \int \frac {(\cos (c+d x) a+a)^{3/2} \left ((48 A+11 C) a^2+3 (16 A+17 C) \cos (c+d x) a^2\right )}{\sqrt {\cos (c+d x)}}dx+\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((48 A+11 C) a^2+3 (16 A+17 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left (5 (48 A+19 C) a^3+(432 A+299 C) \cos (c+d x) a^3\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left (5 (48 A+19 C) a^3+(432 A+299 C) \cos (c+d x) a^3\right )}{\sqrt {\cos (c+d x)}}dx+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (5 (48 A+19 C) a^3+(432 A+299 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (304 A+163 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a^4 (432 A+299 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (304 A+163 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^4 (432 A+299 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3253

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {a^4 (432 A+299 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 a^3 (304 A+163 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {5 a^2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}+\frac {1}{6} \left (\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{7/2} (304 A+163 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a^4 (432 A+299 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )\right )}{8 a}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}{4 d}\)

Input:

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

Output:

(C*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(4*d) + ((5 
*a^2*C*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + 
 ((3*a^3*(16*A + 17*C)*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + 
 d*x])/(2*d) + ((3*a^(7/2)*(304*A + 163*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/S 
qrt[a + a*Cos[c + d*x]]])/d + (a^4*(432*A + 299*C)*Sqrt[Cos[c + d*x]]*Sin[ 
c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/4)/6)/(8*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 3455

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3460

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

rule 3525

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

Maple [A] (verified)

Time = 7.51 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\sqrt {2}\, a^{2} \left (912 A \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+489 C \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\left (96 \cos \left (d x +c \right )+528\right ) \sin \left (d x +c \right ) A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (48 \cos \left (d x +c \right )^{3}+184 \cos \left (d x +c \right )^{2}+326 \cos \left (d x +c \right )+489\right ) \sin \left (d x +c \right ) C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{192 d \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )}\)	\(219\)
parts	\(\frac {A \,a^{2} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+11\right )+19 \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )\right )}{4 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}}+\frac {C \sqrt {2}\, a^{2} \left (489 \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\left (48 \cos \left (d x +c \right )^{3}+184 \cos \left (d x +c \right )^{2}+326 \cos \left (d x +c \right )+489\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{192 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(271\)

Input:

int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/192/d*2^(1/2)*a^2*(912*A*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d* 
x+c))+489*C*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))+(96*cos(d 
*x+c)+528)*sin(d*x+c)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+(48*cos(d*x+c)^3 
+184*cos(d*x+c)^2+326*cos(d*x+c)+489)*sin(d*x+c)*C*(cos(d*x+c)/(1+cos(d*x+ 
c)))^(1/2))*cos(d*x+c)^(1/2)*(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/(cos(d*x+c)/(1 
+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.86 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {{\left (48 \, C a^{2} \cos \left (d x + c\right )^{3} + 184 \, C a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (48 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3 \, {\left (176 \, A + 163 \, C\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (304 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (304 \, A + 163 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, al 
gorithm="fricas")
 

Output:

1/192*((48*C*a^2*cos(d*x + c)^3 + 184*C*a^2*cos(d*x + c)^2 + 2*(48*A + 163 
*C)*a^2*cos(d*x + c) + 3*(176*A + 163*C)*a^2)*sqrt(a*cos(d*x + c) + a)*sqr 
t(cos(d*x + c))*sin(d*x + c) + 3*((304*A + 163*C)*a^2*cos(d*x + c) + (304* 
A + 163*C)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d 
*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))))/(d*cos(d*x + c 
) + d)
 

Sympy [F(-1)]

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

integrate((a+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8557 vs. \(2 (186) = 372\).

Time = 0.74 (sec) , antiderivative size = 8557, normalized size of antiderivative = 39.25 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, al 
gorithm="maxima")
 

Output:

1/768*(48*(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)*((a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin 
(2*d*x + 2*c) + a^2*sin(2*d*x + 2*c) - (a^2*cos(2*d*x + 2*c) - 10*a^2)*sin 
(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d 
*x + 2*c), cos(2*d*x + 2*c) + 1)) + (a^2*sin(2*d*x + 2*c)*sin(1/2*arctan2( 
sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - a^2*cos(2*d*x + 2*c) + 10*a^2 + (a^ 
2*cos(2*d*x + 2*c) - 10*a^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c))))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) 
 + 19*(a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x 
+ 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*si 
n(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), 
cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d* 
x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 
1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2 
(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c) 
, cos(2*d*x + 2*c)))) + 1) - a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 
 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), 
 cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1 
)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*a...
 

Giac [F(-1)]

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

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, al 
gorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a**2*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/cos(c + d*x) 
,x)*a + int(sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x),x)*a + 
int(sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x),x)*c + int(sqrt 
(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**3,x)*c + 2*int(sqrt(co 
s(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*c + 2*int(sqrt(cos(c 
 + d*x) + 1)*sqrt(cos(c + d*x)),x)*a)