3.6.4 \(\int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx\) [504]

3.6.4.1 Optimal result

Integrand size = 45, antiderivative size = 195 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {(8 A-4 B+7 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {(4 B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}} \]

output

1/4*(8*A-4*B+7*C)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d/a^(1 
/2)-(A-B+C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*co 
s(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+1/2*C*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+ 
a*cos(d*x+c))^(1/2)+1/4*(4*B-C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x 
+c))^(1/2)
 

3.6.4.2 Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\left ((4 B-C) \arcsin \left (\sqrt {1-\cos (c+d x)}\right )-8 (A-B+C) \arcsin \left (\sqrt {\cos (c+d x)}\right )+4 \sqrt {2} A \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right )-4 \sqrt {2} B \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right )+4 \sqrt {2} C \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right )+2 C \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+4 B \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}-C \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}\right ) \sin (c+d x)}{4 d \sqrt {1-\cos (c+d x)} \sqrt {a (1+\cos (c+d x))}} \]

input

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

output

(((4*B - C)*ArcSin[Sqrt[1 - Cos[c + d*x]]] - 8*(A - B + C)*ArcSin[Sqrt[Cos 
[c + d*x]]] + 4*Sqrt[2]*A*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^ 
2]] - 4*Sqrt[2]*B*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]] + 4* 
Sqrt[2]*C*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]] + 2*C*Sqrt[1 
 - Cos[c + d*x]]*Cos[c + d*x]^(3/2) + 4*B*Sqrt[-((-1 + Cos[c + d*x])*Cos[c 
 + d*x])] - C*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])])*Sin[c + d*x])/(4* 
d*Sqrt[1 - Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])])
 

3.6.4.3 Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 3524, 27, 3042, 3462, 27, 3042, 3461, 3042, 3253, 223, 3261, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3524

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3462

\(\displaystyle \frac {\frac {\int \frac {(4 B-C) a^2+(8 A-4 B+7 C) \cos (c+d x) a^2}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {a (4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {(4 B-C) a^2+(8 A-4 B+7 C) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {(4 B-C) a^2+(8 A-4 B+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3461

\(\displaystyle \frac {\frac {a (8 A-4 B+7 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-8 a^2 (A-B+C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a (8 A-4 B+7 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-8 a^2 (A-B+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3253

\(\displaystyle \frac {\frac {-8 a^2 (A-B+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a (8 A-4 B+7 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {\frac {2 a^{3/2} (8 A-4 B+7 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-8 a^2 (A-B+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {\frac {\frac {16 a^3 (A-B+C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^{3/2} (8 A-4 B+7 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {2 a^{3/2} (8 A-4 B+7 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} (A-B+C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\)

input

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

output

(C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (((2* 
a^(3/2)*(8*A - 4*B + 7*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + 
 d*x]]])/d - (8*Sqrt[2]*a^(3/2)*(A - B + C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/ 
(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/d)/(2*a) + (a*(4*B 
 - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(4*a)
 

3.6.4.3.1 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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[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 3461

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim 
p[(A*b - a*B)/b   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) 
, x], x] + Simp[B/b   Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]] 
, 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]
 

rule 3462

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Sin[e + f*x])^m*(c + d*S 
in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m 
 + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
 

rule 3524

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)*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)) + b*B*d*(m + 
n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n} 
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !Lt 
Q[m, -2^(-1)] && NeQ[m + n + 2, 0]
 

3.6.4.4 Maple [A] (verified)

Time = 27.86 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.56

method	result	size
default	\(\frac {\left (2 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+4 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-4 B \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+4 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+4 C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-C \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+8 A \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-4 B \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+7 C \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{4 d \left (1+\cos \left (d x +c \right )\right ) a \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(305\)
parts	\(\frac {A \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a}+\frac {B \left (\sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-2 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{2 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a}+\frac {C \left (2 \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+7 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+8 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {2}}{8 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a}\)	\(429\)

input

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

output

1/4/d*(2*C*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+4*A*2^( 
1/2)*arcsin(cot(d*x+c)-csc(d*x+c))-4*B*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c 
))+4*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+4*C*2^(1/2)*arcsin(cot 
(d*x+c)-csc(d*x+c))-C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+8*A*arc 
tan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-4*B*arctan(tan(d*x+c)*(c 
os(d*x+c)/(1+cos(d*x+c)))^(1/2))+7*C*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos( 
d*x+c)))^(1/2)))*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/ 
a/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
 

3.6.4.5 Fricas [A] (verification not implemented)

Time = 15.75 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {{\left (2 \, C \cos \left (d x + c\right ) + 4 \, B - C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left ({\left (8 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 8 \, A - 4 \, B + 7 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {4 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) + {\left (A - B + C\right )} a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

input

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

output

1/4*((2*C*cos(d*x + c) + 4*B - C)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + 
c))*sin(d*x + c) - ((8*A - 4*B + 7*C)*cos(d*x + c) + 8*A - 4*B + 7*C)*sqrt 
(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + 
c))) + 4*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*arctan(sqrt( 
2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqr 
t(a))/(a*d*cos(d*x + c) + a*d)
 

3.6.4.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.6.4.7 Maxima [F]

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

input

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

output

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

3.6.4.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.6.4.9 Mupad [F(-1)]

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

input

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

output

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