\(\int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) [168]

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

Optimal result

Integrand size = 35, antiderivative size = 131 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {a} (4 A+3 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 d}+\frac {a (4 A+3 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {a B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}} \] Output:

1/4*a^(1/2)*(4*A+3*B)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/d+ 
1/4*a*(4*A+3*B)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/2*a 
*B*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} (4 A+3 B) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (4 A+3 B+2 B \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \] Input:

Integrate[Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]) 
,x]
 

Output:

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(Sqrt[2]*(4*A + 3*B)*ArcSin[S 
qrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(4*A + 3*B + 2*B*Cos[c + d 
*x])*Sin[(c + d*x)/2]))/(8*d)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.

\(\displaystyle \int \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

\(\displaystyle \frac {1}{4} (4 A+3 B) \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{4} (4 A+3 B) \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

Input:

Int[Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]
 

Output:

(a*B*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + ((4 
*A + 3*B)*((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
 

Defintions of rubi rules used

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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(449\) vs. \(2(111)=222\).

Time = 13.16 (sec) , antiderivative size = 450, normalized size of antiderivative = 3.44

method result size
parts \(-\frac {A \sqrt {2}\, \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \arctan \left (\frac {\left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )\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}}}}\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}}}\, \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 {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}}}}-\frac {B \sqrt {2}\, \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (3 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \arctan \left (\frac {\left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )\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}}}}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2\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 )}{8 d \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\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}}}}\) \(450\)
default \(\sqrt {2}\, \left (-\frac {A \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \arctan \left (\frac {\left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )\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}}}}\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}}}\, \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 {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}}}}+\frac {B \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \arctan \left (\frac {\left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )\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}}}}\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}}}\, \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 {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}}}}-\frac {B \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (7 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \arctan \left (\frac {\left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )\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}}}}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-10\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 )}{8 d \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\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 )\) \(656\)

Input:

int(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x,method=_RET 
URNVERBOSE)
 

Output:

-1/2*A*2^(1/2)/d*(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)*(a*cos(1/2*d*x+1/2*c)^2) 
^(1/2)/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2 
*c)+1)^2)^(1/2)*(sec(1/2*d*x+1/2*c)*2^(1/2)*arctan(1/((2*cos(1/2*d*x+1/2*c 
)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(cot(1/2*d*x+1/2*c)-csc(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)* 
(-2*sin(1/2*d*x+1/2*c)-2*tan(1/2*d*x+1/2*c)))-1/8*B*2^(1/2)/d*(2*cos(1/2*d 
*x+1/2*c)^2-1)^(1/2)*(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)+1) 
/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(3*sec(1/2*d* 
x+1/2*c)*2^(1/2)*arctan(1/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+ 
1)^2)^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c))*2^(1/2))+tan(1/2*d*x+1 
/2*c)*(-8*cos(1/2*d*x+1/2*c)^3-8*cos(1/2*d*x+1/2*c)^2-2*cos(1/2*d*x+1/2*c) 
-2)*((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.03 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {{\left (2 \, B \cos \left (d x + c\right ) + 4 \, A + 3 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 4 \, A + 3 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algo 
rithm="fricas")
 

Output:

1/4*((2*B*cos(d*x + c) + 4*A + 3*B)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x 
+ c))*sin(d*x + c) + ((4*A + 3*B)*cos(d*x + c) + 4*A + 3*B)*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]

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

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

Output:

Integral(sqrt(a*(cos(c + d*x) + 1))*(A + B*cos(c + d*x))*sqrt(cos(c + d*x) 
), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1851 vs. \(2 (111) = 222\).

Time = 0.33 (sec) , antiderivative size = 1851, normalized size of antiderivative = 14.13 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algo 
rithm="maxima")
 

Output:

1/16*(4*(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(si 
n(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]

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

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algo 
rithm="giac")
 

Output:

integrate((B*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c)) 
, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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