\(\int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [500]

Optimal result

Integrand size = 35, antiderivative size = 196 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a} (6 A+5 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{8 d}+\frac {a B \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {a (6 A+5 B) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {a (6 A+5 B) \sin (c+d x)}{8 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \] Output:

1/8*a^(1/2)*(6*A+5*B)*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))*co 
s(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/3*a*B*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1 
/2)/sec(d*x+c)^(5/2)+1/12*a*(6*A+5*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/ 
sec(d*x+c)^(3/2)+1/8*a*(6*A+5*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d 
*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (3 \sqrt {2} (6 A+5 B) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (18 A+19 B+2 (6 A+5 B) \cos (c+d x)+4 B \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \] Input:

Integrate[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/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]*(6*A + 5*B)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[ 
Cos[c + d*x]]*(18*A + 19*B + 2*(6*A + 5*B)*Cos[c + d*x] + 4*B*Cos[2*(c + d 
*x)])*Sin[(c + d*x)/2]))/(48*d)
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.95, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 3440, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3440

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((a*B*Cos[c + d*x]^(5/2)*Sin[c + d*x 
])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + ((6*A + 5*B)*((a*Cos[c + d*x]^(3/2)*Si 
n[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)
 

Defintions of rubi rules used

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 3440

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Sin[e + f*x])^m*((c + 
d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(739\) vs. \(2(166)=332\).

Time = 16.97 (sec) , antiderivative size = 740, normalized size of antiderivative = 3.78

method	result	size
default	\(\text {Expression too large to display}\)	\(740\)
parts	\(\text {Expression too large to display}\)	\(743\)

Input:

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

Output:

2^(1/2)*(-1/8*A/d*(a*cos(1/2*d*x+1/2*c)^2)^(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)*(3*sec(1/2*d*x+1/2*c)*2^(1/2)*arctan(1/((2*cos(1/2*d* 
x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2* 
d*x+1/2*c))*2^(1/2))+tan(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))+1/8*B/d*(a*cos(1/2*d*x+1/2*c)^2)^(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)*(3*sec(1/2*d*x+1/2*c)*2^(1/2)*arctan(1 
/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(cot(1/2*d*x+ 
1/2*c)-csc(1/2*d*x+1/2*c))*2^(1/2))+tan(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))-1/48*B/d*(a*cos(1/2*d*x+1/2*c)^2)^ 
(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*sec(1/2*d*x+1/2*c) 
*2^(1/2)*arctan(1/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1 
/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c))*2^(1/2))+tan(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)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {3 \, {\left ({\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 6 \, A + 5 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (8 \, B \cos \left (d x + c\right )^{3} + 2 \, {\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

-1/24*(3*((6*A + 5*B)*cos(d*x + c) + 6*A + 5*B)*sqrt(a)*arctan(sqrt(a*cos( 
d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (8*B*cos(d*x + 
c)^3 + 2*(6*A + 5*B)*cos(d*x + c)^2 + 3*(6*A + 5*B)*cos(d*x + c))*sqrt(a*c 
os(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)
 

Sympy [F]

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

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

Output:

Integral(sqrt(a*(cos(c + d*x) + 1))*(A + B*cos(c + d*x))/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. 2981 vs. \(2 (166) = 332\).

Time = 0.75 (sec) , antiderivative size = 2981, normalized size of antiderivative = 15.21 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/96*(6*(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), c 
os(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(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), 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(s 
in(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)*(c 
os(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), cos 
(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))...
 

Giac [F]

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

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

Output:

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

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

Output:

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

Reduce [F]

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

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