\(\int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [172]

Optimal result

Integrand size = 37, antiderivative size = 169 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {a} (8 A+5 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {a (8 A+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {C \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d} \] Output:

1/8*a^(1/2)*(8*A+5*C)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/d+ 
1/8*a*(8*A+5*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/12* 
a*C*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/3*C*cos(d*x+c)^ 
(3/2)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.66 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} (8 A+5 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (24 A+19 C+10 C \cos (c+d x)+4 C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \] Input:

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

Output:

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*(8*A + 5*C)*ArcSin 
[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(24*A + 19*C + 10*C*Cos[ 
c + d*x] + 4*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {3042, 3525, 27, 3042, 3460, 3042, 3249, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3525

\(\displaystyle \frac {\int \frac {1}{2} \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a} (3 a (2 A+C)+a C \cos (c+d x))dx}{3 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a} (3 a (2 A+C)+a C \cos (c+d x))dx}{6 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

\(\displaystyle \frac {\frac {3}{4} a (8 A+5 C) \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {a^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}+\frac {3}{4} a (8 A+5 C) \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )}{6 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

Input:

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

Output:

(C*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((a^2 
*C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*a* 
(8*A + 5*C)*((Sqrt[a]*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x 
]]])/d + (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])) 
)/4)/(6*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 3249

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 
2*n + 1)))   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] && GtQ[n, 0] && IntegerQ[2*n]
 

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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(475\) vs. \(2(143)=286\).

Time = 7.40 (sec) , antiderivative size = 476, normalized size of antiderivative = 2.82

method	result	size
parts	\(\frac {A \sqrt {2}\, \sqrt {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sqrt {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 {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{2 d \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}+\frac {C \sqrt {2}\, \sqrt {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (15 \sqrt {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 {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (64 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+64 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+26 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+26\right ) \sqrt {\frac {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\)	\(476\)
default	\(\text {Expression too large to display}\)	\(916\)

Input:

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

Output:

1/2*A*2^(1/2)/d*(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)*(a*cos(1/2*d*x+1/2*c)^2) 
^(1/2)/(cos(1/2*d*x+1/2*c)+1)/((-1+2*cos(1/2*d*x+1/2*c)^2)/(cos(1/2*d*x+1/ 
2*c)+1)^2)^(1/2)*(2^(1/2)*arctan(2^(1/2)*(csc(1/2*d*x+1/2*c)-cot(1/2*d*x+1 
/2*c))/((-1+2*cos(1/2*d*x+1/2*c)^2)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))*sec(1 
/2*d*x+1/2*c)+((-1+2*cos(1/2*d*x+1/2*c)^2)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2) 
*(2*sin(1/2*d*x+1/2*c)+2*tan(1/2*d*x+1/2*c)))+1/48*C*2^(1/2)/d*(-1+2*cos(1 
/2*d*x+1/2*c)^2)^(1/2)*(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)+ 
1)/((-1+2*cos(1/2*d*x+1/2*c)^2)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(15*2^(1/2 
)*arctan(2^(1/2)*(csc(1/2*d*x+1/2*c)-cot(1/2*d*x+1/2*c))/((-1+2*cos(1/2*d* 
x+1/2*c)^2)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))*sec(1/2*d*x+1/2*c)+(64*cos(1/ 
2*d*x+1/2*c)^5+64*cos(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)^3-24*cos(1/2* 
d*x+1/2*c)^2+26*cos(1/2*d*x+1/2*c)+26)*((-1+2*cos(1/2*d*x+1/2*c)^2)/(cos(1 
/2*d*x+1/2*c)+1)^2)^(1/2)*tan(1/2*d*x+1/2*c))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.87 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {{\left (8 \, C \cos \left (d x + c\right )^{2} + 10 \, C \cos \left (d x + c\right ) + 24 \, A + 15 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (8 \, A + 5 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 5 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

1/24*((8*C*cos(d*x + c)^2 + 10*C*cos(d*x + c) + 24*A + 15*C)*sqrt(a*cos(d* 
x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((8*A + 5*C)*cos(d*x + c) 
+ 8*A + 5*C)*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. \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2713 vs. \(2 (143) = 286\).

Time = 0.47 (sec) , antiderivative size = 2713, normalized size of antiderivative = 16.05 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/96*(24*(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(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d 
*x + c) - (cos(d*x + c) - 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c) + 1)))*sqrt(a) + sqrt(a)*(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) + 1))*sin(d*x + c) - cos(d*x + c)*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)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d* 
x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x 
+ 2*c), cos(2*d*x + 2*c) + 1))) + 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) + 1))*sin(d*x + c) - cos(d*x + c)*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)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin 
(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 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)*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)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c) + 1)) + 1) + arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c...
 

Giac [F]

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

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

Output:

integrate((C*cos(d*x + c)^2 + A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c 
)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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