\(\int \sqrt {a+a \cos (c+d x)} (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\) [82]

Optimal result

Integrand size = 35, antiderivative size = 196 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\sqrt {a} (35 A+48 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a (35 A+48 C) \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a (35 A+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:

1/64*a^(1/2)*(35*A+48*C)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2) 
)/d+1/64*a*(35*A+48*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/96*a*(35*A+48 
*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/24*a*A*sec(d*x+c)^2*t 
an(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/4*A*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c) 
^3*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.77 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (6 \sqrt {2} (35 A+48 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)+(332 A+192 C+(539 A+432 C) \cos (c+d x)+4 (35 A+48 C) \cos (2 (c+d x))+105 A \cos (3 (c+d x))+144 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{768 d} \] Input:

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

Output:

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^4*(6*Sqrt[2]*(35 
*A + 48*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^4 + (332*A + 192 
*C + (539*A + 432*C)*Cos[c + d*x] + 4*(35*A + 48*C)*Cos[2*(c + d*x)] + 105 
*A*Cos[3*(c + d*x)] + 144*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(768*d)
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3523, 27, 3042, 3459, 3042, 3251, 3042, 3251, 3042, 3252, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

\(\displaystyle \frac {\int \frac {1}{2} \sqrt {\cos (c+d x) a+a} (a A+a (5 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} (a A+a (5 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a A+a (5 A+8 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {\frac {1}{6} a (35 A+48 C) \int \sqrt {\cos (c+d x) a+a} \sec ^3(c+d x)dx+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} a (35 A+48 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {\frac {1}{6} a (35 A+48 C) \left (\frac {3}{4} \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} a (35 A+48 C) \left (\frac {3}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {\frac {1}{6} a (35 A+48 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} a (35 A+48 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {1}{6} a (35 A+48 C) \left (\frac {3}{4} \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {1}{6} a (35 A+48 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\)

Input:

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

Output:

(A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((a^2*A*S 
ec[c + d*x]^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (a*(35*A + 48 
*C)*((a*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*((S 
qrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a*Ta 
n[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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

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 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1446\) vs. \(2(172)=344\).

Time = 0.60 (sec) , antiderivative size = 1447, normalized size of antiderivative = 7.38

method	result	size
parts	\(\text {Expression too large to display}\)	\(1447\)
default	\(\text {Expression too large to display}\)	\(1651\)

Input:

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

Output:

1/24*A*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(1680*a*(ln(-4/(2 
*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2) 
*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))+ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))* 
(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/ 
2)+2*a)))*sin(1/2*d*x+1/2*c)^8-1680*(a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2 
*a)^(1/2)+2*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2 
*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a+2*ln(-4/(2*cos( 
1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin 
(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a)*sin(1/2*d*x+1/2*c)^6+280*(11*a^(1/2)*2 
^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+9*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2) 
)*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^( 
1/2)+2*a))*a+9*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x 
+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a)*sin(1/2*d* 
x+1/2*c)^4+(-2044*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-840*ln(-4 
/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1 
/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a-840*ln(4/(2*cos(1/2*d*x+1/2*c)+ 
2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c) 
^2*a)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^2+558*a^(1/2)*2^(1/2)*(sin(1/2*d*x 
+1/2*c)^2*a)^(1/2)+105*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos 
(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a+...
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.06 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left ({\left (35 \, A + 48 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (35 \, A + 48 \, C\right )} \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (35 \, A + 48 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (35 \, A + 48 \, C\right )} \cos \left (d x + c\right )^{2} + 56 \, A \cos \left (d x + c\right ) + 48 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \] Input:

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

Output:

1/768*(3*((35*A + 48*C)*cos(d*x + c)^5 + (35*A + 48*C)*cos(d*x + c)^4)*sqr 
t(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + 
a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d* 
x + c)^2)) + 4*(3*(35*A + 48*C)*cos(d*x + c)^3 + 2*(35*A + 48*C)*cos(d*x + 
 c)^2 + 56*A*cos(d*x + c) + 48*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/( 
d*cos(d*x + c)^5 + d*cos(d*x + c)^4)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.49 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=-\frac {\sqrt {2} {\left (3 \, \sqrt {2} {\left (35 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 48 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) + \frac {4 \, {\left (840 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1152 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1540 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2112 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1022 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1248 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 279 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 240 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}\right )} \sqrt {a}}{768 \, d} \] Input:

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

Output:

-1/768*sqrt(2)*(3*sqrt(2)*(35*A*sgn(cos(1/2*d*x + 1/2*c)) + 48*C*sgn(cos(1 
/2*d*x + 1/2*c)))*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt( 
2) + 4*sin(1/2*d*x + 1/2*c))) + 4*(840*A*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2 
*d*x + 1/2*c)^7 + 1152*C*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 
- 1540*A*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 2112*C*sgn(cos 
(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 1022*A*sgn(cos(1/2*d*x + 1/2*c 
))*sin(1/2*d*x + 1/2*c)^3 + 1248*C*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 
 1/2*c)^3 - 279*A*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 240*C*s 
gn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))/(2*sin(1/2*d*x + 1/2*c)^2 - 
 1)^4)*sqrt(a)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*(int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**5,x)*c + 
 int(sqrt(cos(c + d*x) + 1)*sec(c + d*x)**5,x)*a)