\(\int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [479]

Optimal result

Integrand size = 45, antiderivative size = 121 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a} (2 B+C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}-\frac {a (2 A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} (2 B+C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 (2 A+C \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d \sqrt {\cos (c+d x)}} \] Input:

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

Output:

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(Sqrt[2]*(2*B + C)*ArcSin[Sqr 
t[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2*(2*A + C*Cos[c + d*x])*Sin[( 
c + d*x)/2]))/(2*d*Sqrt[Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {3042, 3522, 27, 3042, 3460, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3522

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

(2*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + ((a^( 
3/2)*(2*B + C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d 
- (a^2*(2*A - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d* 
x]]))/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 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 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 [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(399\) vs. \(2(107)=214\).

Time = 4.68 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.31

method	result	size
parts	\(\frac {2 A \sqrt {2}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}+\frac {2 B \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right ) \left (1+\sec \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}+\frac {C \sqrt {2}\, \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{2 d \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\)	\(400\)
default	\(\text {Expression too large to display}\)	\(725\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {{\left (C \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (2 \, B + C\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, B + C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )} \] Input:

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

Output:

((C*cos(d*x + c) + 2*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d* 
x + c) + ((2*B + C)*cos(d*x + c)^2 + (2*B + C)*cos(d*x + c))*sqrt(a)*arcta 
n(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)^2 + d*cos(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1035 vs. \(2 (107) = 214\).

Time = 0.30 (sec) , antiderivative size = 1035, normalized size of antiderivative = 8.55 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(4*B*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)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) 
 + 1)) + sin(d*x + c), (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)) + cos(d*x + c)) + (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 + s 
in(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*ar 
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arct 
an2(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*arct 
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*s 
in(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((co...
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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