\(\int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [1317]

Optimal result

Integrand size = 45, antiderivative size = 247 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a} (48 A+40 B+35 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 d}+\frac {a (8 B+C) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a (48 A+40 B+35 C) \sin (c+d x)}{96 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {a (48 A+40 B+35 C) \sin (c+d x)}{64 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \] Output:

1/64*a^(1/2)*(48*A+40*B+35*C)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^( 
1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/24*a*(8*B+C)*sin(d*x+c)/d/(a+a 
*cos(d*x+c))^(1/2)/sec(d*x+c)^(5/2)+1/4*C*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c 
)/d/sec(d*x+c)^(5/2)+1/96*a*(48*A+40*B+35*C)*sin(d*x+c)/d/(a+a*cos(d*x+c)) 
^(1/2)/sec(d*x+c)^(3/2)+1/64*a*(48*A+40*B+35*C)*sin(d*x+c)/d/(a+a*cos(d*x+ 
c))^(1/2)/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (3 \sqrt {2} (48 A+40 B+35 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (144 A+152 B+133 C+2 (48 A+40 B+53 C) \cos (c+d x)+4 (8 B+7 C) \cos (2 (c+d x))+12 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{384 d} \] Input:

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

Output:

(Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c 
 + d*x]]*(3*Sqrt[2]*(48*A + 40*B + 35*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] 
+ 2*Sqrt[Cos[c + d*x]]*(144*A + 152*B + 133*C + 2*(48*A + 40*B + 53*C)*Cos 
[c + d*x] + 4*(8*B + 7*C)*Cos[2*(c + d*x)] + 12*C*Cos[3*(c + d*x)])*Sin[(c 
 + d*x)/2]))/(384*d)
 

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 4709, 3042, 3524, 27, 3042, 3460, 3042, 3249, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3524

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a} (a (8 A+5 C)+a (8 B+C) \cos (c+d x))dx}{4 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a} (a (8 A+5 C)+a (8 B+C) \cos (c+d x))dx}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (8 A+5 C)+a (8 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3460

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}dx+\frac {a^2 (8 B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (8 B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3249

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3249

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3253

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (8 B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {1}{6} a (48 A+40 B+35 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )}{8 a}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Cos[c + d*x]^(5/2)*Sqrt[a + a*Co 
s[c + d*x]]*Sin[c + d*x])/(4*d) + ((a^2*(8*B + C)*Cos[c + d*x]^(5/2)*Sin[c 
 + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (a*(48*A + 40*B + 35*C)*((a*Cos[ 
c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*((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))/6)/(8*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 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]
 

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]
 

rule 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1487\) vs. \(2(211)=422\).

Time = 19.39 (sec) , antiderivative size = 1488, normalized size of antiderivative = 6.02

method	result	size
parts	\(\text {Expression too large to display}\)	\(1488\)
default	\(\text {Expression too large to display}\)	\(1526\)

Input:

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

Output:

1/8*A*2^(1/2)/d*(cos(1/2*d*x+1/2*c)^2*a)^(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)/(1/(2*cos(1/2*d*x+ 
1/2*c)^2-1))^(1/2)*(-3*2^(1/2)*arctan(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)*(-csc(1/2*d*x+1/2*c)+cot(1/2*d*x+1/2*c)))* 
sec(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 
)*tan(1/2*d*x+1/2*c))+B*2^(1/2)*(-1/8/d*(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/(co 
s(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)/(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)*(-3*2^(1/2)*arctan(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)*(-csc(1/2*d*x+1/2* 
c)+cot(1/2*d*x+1/2*c)))*sec(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)*tan(1/2*d*x+1/2*c))-1/48/d*(cos(1/2*d*x+1/2*c)^2 
*a)^(1/2)/(cos(1/2*d*x+1/2*c)+1)/(1/(2*cos(1/2*d*x+1/2*c)^2-1))^(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)*(33*2^(1/2)*arctan 
(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)*(-csc 
(1/2*d*x+1/2*c)+cot(1/2*d*x+1/2*c)))*sec(1/2*d*x+1/2*c)+(-64*cos(1/2*d*x+1 
/2*c)^5-64*cos(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)^3-24*cos(1/2*d*x+1/2 
*c)^2-38*cos(1/2*d*x+1/2*c)-38)*((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1 
/2*c)+1)^2)^(1/2)*tan(1/2*d*x+1/2*c)))+17/16*C*2^(1/2)/d*(cos(1/2*d*x+1...
 

Fricas [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {3 \, {\left ({\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right ) + 48 \, A + 40 \, B + 35 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (48 \, C \cos \left (d x + c\right )^{4} + 8 \, {\left (8 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

-1/192*(3*((48*A + 40*B + 35*C)*cos(d*x + c) + 48*A + 40*B + 35*C)*sqrt(a) 
*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)) 
) - (48*C*cos(d*x + c)^4 + 8*(8*B + 7*C)*cos(d*x + c)^3 + 2*(48*A + 40*B + 
 35*C)*cos(d*x + c)^2 + 3*(48*A + 40*B + 35*C)*cos(d*x + c))*sqrt(a*cos(d* 
x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)
 

Sympy [F]

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

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 9619 vs. \(2 (211) = 422\).

Time = 0.80 (sec) , antiderivative size = 9619, normalized size of antiderivative = 38.94 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/768*(48*(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)))*sin(2*d 
*x + 2*c) - (cos(2*d*x + 2*c) - 2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2 
*d*x + 2*c))) + sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2* 
d*x + 2*c) + 1)) + ((cos(2*d*x + 2*c) - 2)*cos(1/2*arctan2(sin(2*d*x + 2*c 
), cos(2*d*x + 2*c))) + sin(2*d*x + 2*c)*sin(1/2*arctan2(sin(2*d*x + 2*c), 
 cos(2*d*x + 2*c))) - cos(2*d*x + 2*c) + 2)*sin(1/2*arctan2(sin(2*d*x + 2* 
c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 3*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(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 
1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*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(1/2*arctan2(sin(2*d*x + 2*c) 
, cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) 
+ 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 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)))*sin(1/2*arctan2(sin( 
2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Reduce [F]

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

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

Output:

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