\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx\) [506]

Optimal result

Integrand size = 45, antiderivative size = 138 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \] Output:

2*C*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(1/2)/d-2^(1/2)*(A 
-B+C)*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+ 
c))^(1/2))/a^(1/2)/d+2*A*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1 
/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 5.96 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.93 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (5 C \cos ^2(c+d x) \left (\sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )+10 B \cos ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )-\frac {1}{2} (A-B+C) \csc ^3\left (\frac {1}{2} (c+d x)\right ) \left (-5 \cos ^2(c+d x) (2+\cos (c+d x)) \left (1-\cos (c+d x)+\text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos (c+d x) \sqrt {2-2 \sec (c+d x)}\right )+\operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right ) \sin ^2(c+d x)\right )\right )}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a (1+\cos (c+d x))}} \] Input:

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*Sqrt 
[a + a*Cos[c + d*x]]),x]
 

Output:

(2*Cos[(c + d*x)/2]*(5*C*Cos[c + d*x]^2*(Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d 
*x)/2]]*Sqrt[Cos[c + d*x]] - 2*Sin[(c + d*x)/2]) + 10*B*Cos[c + d*x]^2*Sin 
[(c + d*x)/2] - ((A - B + C)*Csc[(c + d*x)/2]^3*(-5*Cos[c + d*x]^2*(2 + Co 
s[c + d*x])*(1 - Cos[c + d*x] + ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/ 
2]^2)]]*Cos[c + d*x]*Sqrt[2 - 2*Sec[c + d*x]]) + Hypergeometric2F1[2, 5/2, 
 7/2, -(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^4*Sin[c + d*x]^ 
2))/2))/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 3522, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3522

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3461

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 218

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

Input:

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*Sqrt[a + a 
*Cos[c + d*x]]),x]
 

Output:

-(((-2*Sqrt[a]*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/ 
d + (Sqrt[2]*Sqrt[a]*(A - B + C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sq 
rt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/d)/a) + (2*A*Sin[c + d*x])/(d 
*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])
 

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 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])
 

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {\left (A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+C \cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+C \cos \left (d x +c \right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{d a \sqrt {\cos \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(202\)
parts	\(\frac {A \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+\sqrt {2}\, \sin \left (d x +c \right )\right )}{d a \sqrt {\cos \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right )}-\frac {B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d a \sqrt {\cos \left (d x +c \right )}}+\frac {C \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(273\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2 
),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 4.13 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.60 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 2 \, {\left (C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right ) - \frac {\sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{2} + {\left (A - B + C\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a}}\right )}{\sqrt {a}}}{a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c) 
)^(1/2),x, algorithm="fricas")
 

Output:

(2*sqrt(a*cos(d*x + c) + a)*A*sqrt(cos(d*x + c))*sin(d*x + c) + 2*(C*cos(d 
*x + c)^2 + C*cos(d*x + 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))) - s 
qrt(2)*((A - B + C)*a*cos(d*x + c)^2 + (A - B + C)*a*cos(d*x + c))*arctan( 
1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/((cos 
(d*x + c)^2 + cos(d*x + c))*sqrt(a)))/sqrt(a))/(a*d*cos(d*x + c)^2 + a*d*c 
os(d*x + c))
 

Sympy [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(a+a*cos(d*x+ 
c))**(1/2),x)
 

Output:

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)/(sqrt(a*(cos(c + d*x) + 
1))*cos(c + d*x)**(3/2)), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.02 (sec) , antiderivative size = 1884, normalized size of antiderivative = 13.65 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c) 
)^(1/2),x, algorithm="maxima")
 

Output:

(sqrt(2)*B*arctan2(((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + sin(d*x 
 + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs(e^( 
I*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos(d*x + c 
) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4)*sin(1 
/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I*c) + 1)^2, ( 
abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + 
 c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sin(d*x + c))/abs(e^(I*d*x + I*c) 
+ 1), ((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + sin(d*x + c)^4 + 2*( 
cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs(e^(I*d*x + I*c) 
+ 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)*sin(d* 
x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4)*sqrt(a)*cos(1/2*ar 
ctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I*c) + 1)^2, (abs(e 
^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 
 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sqrt(a)*cos(d*x + c) - sqrt(a))/(sqrt(a 
)*abs(e^(I*d*x + I*c) + 1)))/sqrt(a) + (2*cos(1/2*arctan2(sin(2*d*x + 2*c) 
, cos(2*d*x + 2*c) + 1))*sin(d*x + c) - 2*(cos(d*x + c) - 1)*sin(1/2*arcta 
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - sqrt(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)*arctan2(((abs(e^(I*d* 
x + I*c) + 1)^4 + cos(d*x + c)^4 + sin(d*x + c)^4 + 2*(cos(d*x + c)^2 - si 
n(d*x + c)^2 - 2*cos(d*x + c) + 1)*abs(e^(I*d*x + I*c) + 1)^2 - 4*cos(d...
 

Giac [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c) 
)^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + a*cos 
(c + d*x))^(1/2)),x)
 

Output:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + a*cos 
(c + d*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}+\cos \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) b \right )}{a} \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2 
),x)
 

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x) + 
1),x)*c + int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**3 
 + cos(c + d*x)**2),x)*a + int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))) 
/(cos(c + d*x)**2 + cos(c + d*x)),x)*b))/a