\(\int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [1313]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (warning: unable to verify)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [B] (verification not implemented)
Giac [F(-1)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 45, antiderivative size = 140 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 \sqrt {a} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 a (A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \] Output:

2*a^(1/2)*C*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^( 
1/2)*sec(d*x+c)^(1/2)/d+2/3*a*(A+3*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*c 
os(d*x+c))^(1/2)+2/3*A*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^(3/2)*sin(d*x+c)/ 
d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.75 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \left (3 \sqrt {2} C \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+2 (A+(2 A+3 B) \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d} \] Input:

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

Output:

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(3/2)*(3*Sqrt[2] 
*C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(3/2) + 2*(A + (2*A + 3*B 
)*Cos[c + d*x])*Sin[(c + d*x)/2]))/(3*d)
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4709, 3042, 3522, 27, 3042, 3459, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3522

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3459

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {6 a^{3/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^2 (A+3 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c 
 + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((6*a^(3/2)*C*ArcSin[(Sqrt[a]*Sin[c + 
d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a^2*(A + 3*B)*Sin[c + d*x])/(d*Sqr 
t[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(3*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 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 3522
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + 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 - B*d)*( 
a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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])
 

rule 4709
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 
Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(622\) vs. \(2(118)=236\).

Time = 4.11 (sec) , antiderivative size = 623, normalized size of antiderivative = 4.45

method result size
default \(\sqrt {2}\, \left (\frac {2 A \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {2 C \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {2 B \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {4 B \left (5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {8 C \left (5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {C \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \left (\left (-32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+14\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\, \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right ) \left (6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3-3 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\right )\) \(623\)
parts \(\frac {2 A \sqrt {2}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+B \sqrt {2}\, \left (-\frac {2 \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {4 \left (5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\right )+C \sqrt {2}\, \left (\frac {2 \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {8 \left (5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2\right ) \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {\sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \left (\left (-32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+14\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\, \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right ) \left (6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3-3 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\right )\) \(629\)

Input:

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

Output:

2^(1/2)*(2/3*A/d*(4*cos(1/2*d*x+1/2*c)^2-1)*(cos(1/2*d*x+1/2*c)^2*a)^(1/2) 
*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)*tan(1/2*d 
*x+1/2*c)+2/3*C/d*(4*cos(1/2*d*x+1/2*c)^2-1)*(cos(1/2*d*x+1/2*c)^2*a)^(1/2 
)*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)*tan(1/2* 
d*x+1/2*c)-2/3*B/d*(4*cos(1/2*d*x+1/2*c)^2-1)*(cos(1/2*d*x+1/2*c)^2*a)^(1/ 
2)*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)*tan(1/2 
*d*x+1/2*c)+4/3*B/d*(5*cos(1/2*d*x+1/2*c)^2-2)*(cos(1/2*d*x+1/2*c)^2*a)^(1 
/2)*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)*tan(1/ 
2*d*x+1/2*c)-8/3*C/d*(5*cos(1/2*d*x+1/2*c)^2-2)*(cos(1/2*d*x+1/2*c)^2*a)^( 
1/2)*(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)*tan(1 
/2*d*x+1/2*c)-1/3*C/d*(cos(1/2*d*x+1/2*c)^2*a)^(1/2)*(1/(2*cos(1/2*d*x+1/2 
*c)^2-1))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)*((-32*cos(1/2*d*x+1/2*c)^2+14)* 
tan(1/2*d*x+1/2*c)+2^(1/2)*((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c) 
+1)^2)^(1/2)*arctan(2^(1/2)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c 
)+1)^2)^(1/2)*(-csc(1/2*d*x+1/2*c)+cot(1/2*d*x+1/2*c)))*(6*cos(1/2*d*x+1/2 
*c)^2+6*cos(1/2*d*x+1/2*c)-3-3*sec(1/2*d*x+1/2*c))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.89 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {2 \, {\left (3 \, {\left (C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left ({\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \] Input:

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

Output:

-2/3*(3*(C*cos(d*x + c)^2 + C*cos(d*x + c))*sqrt(a)*arctan(sqrt(a*cos(d*x 
+ c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - ((2*A + 3*B)*cos(d* 
x + c) + A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*c 
os(d*x + c)^2 + d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1547 vs. \(2 (118) = 236\).

Time = 0.34 (sec) , antiderivative size = 1547, normalized size of antiderivative = 11.05 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/6*(3*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 
1)^(3/4)*sqrt(a)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) 
+ 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)*(cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(2*d*x + 2 
*c) - (cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c) + 1)))*sqrt(a) + ((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2 
*d*x + 2*c) + 1)*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*arc 
tan2(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)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*co 
s(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*a 
rctan2(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) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^ 
2 + 2*cos(2*d*x + 2*c) + 1)*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*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x +...
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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