\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx\) [102]

Optimal result

Integrand size = 31, antiderivative size = 118 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 (3 A-2 B) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 B \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 a d} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (-3 (A-B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 A \sin \left (\frac {1}{2} (c+d x)\right )-4 B \sin ^3\left (\frac {1}{2} (c+d x)\right )\right )}{3 d \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

(2*Cos[(c + d*x)/2]*(-3*(A - B)*ArcTanh[Sin[(c + d*x)/2]] + 6*A*Sin[(c + d 
*x)/2] - 4*B*Sin[(c + d*x)/2]^3))/(3*d*Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3447, 3042, 3502, 27, 3042, 3230, 3042, 3128, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 a (3 A-2 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \sqrt {2} \sqrt {a} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}+\frac {2 B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\)

Input:

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

Output:

(2*B*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*a*d) + ((-3*Sqrt[2]*Sqrt[a] 
*(A - B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]]) 
])/d + (2*a*(3*A - 2*B)*Sin[c + d*x])/(d*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 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 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3230

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

rule 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.64

method	result	size
default	\(-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 B \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a -6 A \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-3 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{3 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(194\)
parts	\(\frac {A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{3 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(257\)

Input:

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

Output:

-1/3*cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*B*(a*sin 
(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2+3*A*ln(4*(a^(1/2)*(a 
*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a-6*A*(a*sin(1/2*d*x+1 
/2*c)^2)^(1/2)*a^(1/2)-3*B*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a) 
/cos(1/2*d*x+1/2*c))*a)/a^(3/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2 
)^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.26 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {4 \, {\left (B \cos \left (d x + c\right ) + 3 \, A - B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) - \frac {3 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/6*(4*(B*cos(d*x + c) + 3*A - B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) - 
3*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*log(-(cos(d*x + c)^2 - 2*sq 
rt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/ 
(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38386 vs. \(2 (101) = 202\).

Time = 0.83 (sec) , antiderivative size = 38386, normalized size of antiderivative = 325.31 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/60*((20*(cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c)^3 + 8*(cos(d*x + c)^2 + 
sin(d*x + c)^2 + 2*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c)^3 - 20*cos(5/2*d 
*x + 5/2*c)^3*sin(d*x + c) + 2*(15*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d 
*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + 
 sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c)^2 + 15 
*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/ 
2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/ 
2*d*x + 1/2*c) + 1))*sin(d*x + c)^2 + 30*(log(cos(1/2*d*x + 1/2*c)^2 + sin 
(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2* 
c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c) 
+ 4*(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1)*sin(3/2*d*x + 3 
/2*c) - 20*cos(3/2*d*x + 3/2*c)*sin(d*x + c) + 15*log(cos(1/2*d*x + 1/2*c) 
^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 15*log(cos(1/2 
*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*co 
s(5/2*d*x + 5/2*c)^2 + 30*((log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2 
*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2 
*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c)^2 + (log(cos(1 
/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 
 log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2 
*c) + 1))*sin(d*x + c)^2 + 2*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x ...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.36 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2\,A\,\left (2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )-\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )\right )\,\sqrt {\frac {a+a\,\cos \left (c+d\,x\right )}{2\,a}}}{d\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}+\frac {2\,B\,\sin \left (c+d\,x\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{3\,a\,d}-\frac {2\,B\,\left (4\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )-3\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )\right )\,\sqrt {\frac {a+a\,\cos \left (c+d\,x\right )}{2\,a}}}{3\,a^2\,d\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \] Input:

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

Output:

(2*A*(2*ellipticE(c/2 + (d*x)/2, 1) - ellipticF(c/2 + (d*x)/2, 1))*((a + a 
*cos(c + d*x))/(2*a))^(1/2))/(d*(a + a*cos(c + d*x))^(1/2)) + (2*B*sin(c + 
 d*x)*(a + a*cos(c + d*x))^(1/2))/(3*a*d) - (2*B*(4*a^2*ellipticE(c/2 + (d 
*x)/2, 1) - 3*a^2*ellipticF(c/2 + (d*x)/2, 1))*((a + a*cos(c + d*x))/(2*a) 
)^(1/2))/(3*a^2*d*(a + a*cos(c + d*x))^(1/2))
 

Reduce [F]

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

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

Output:

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