\(\int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [399]

Optimal result

Integrand size = 35, antiderivative size = 284 \[ \int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 (a-b) \sqrt {a+b} (A b+3 a B) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 d}+\frac {2 (a-b) \sqrt {a+b} (A-3 B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a d}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:

2/3*(a-b)*(a+b)^(1/2)*(A*b+3*B*a)*cot(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1 
/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/( 
a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d+2/3*(a-b)*(a+b)^(1/2)*(A- 
3*B)*cot(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1 
/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c)) 
/(a-b))^(1/2)/a/d+2/3*A*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(3/ 
2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 14.60 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.43 \[ \int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right )^{5/2} \left (\frac {\cos (c+d x)}{1+\cos (c+d x)}\right )^{3/2} \sqrt {1+\cos (c+d x)} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (-2 (a+b) (A b+3 a B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+2 a (a+b) (A+3 B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-(A b+3 a B) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {2 \sec (c+d x) (A b \sin (c+d x)+3 a B \sin (c+d x))}{3 a}+\frac {2}{3} A \sec (c+d x) \tan (c+d x)\right )}{d} \] Input:

Integrate[(Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/ 
2),x]
 

Output:

(4*(Cos[(c + d*x)/2]^2)^(5/2)*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(3/2)*Sqrt 
[1 + Cos[c + d*x]]*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*(-2*(a + b)*(A*b 
+ 3*a*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/( 
(a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/ 
(a + b)] + 2*a*(a + b)*(A + 3*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqr 
t[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[ 
(c + d*x)/2]], (-a + b)/(a + b)] - (A*b + 3*a*B)*Cos[c + d*x]*(a + b*Cos[c 
 + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*a*d*Cos[c + d*x]^(5/2)*S 
qrt[a + b*Cos[c + d*x]]) + (Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*(( 
2*Sec[c + d*x]*(A*b*Sin[c + d*x] + 3*a*B*Sin[c + d*x]))/(3*a) + (2*A*Sec[c 
 + d*x]*Tan[c + d*x])/3))/d
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 3478, 27, 3042, 3477, 3042, 3295, 3473}

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.

\(\displaystyle \int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3478

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {A b+3 a B+(a A+3 b B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

\(\displaystyle \frac {1}{3} \left ((3 a B+A b) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} (A-3 B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3473

\(\displaystyle \frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} (3 a B+A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 (a-b) \sqrt {a+b} (A-3 B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

Input:

Int[(Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x]
 

Output:

((2*(a - b)*Sqrt[a + b]*(A*b + 3*a*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a 
 + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]* 
Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)]) 
/(a^2*d) + (2*(a - b)*Sqrt[a + b]*(A - 3*B)*Cot[c + d*x]*EllipticF[ArcSin[ 
Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - 
 b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a 
- b)])/(a*d))/3 + (2*A*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + 
 d*x]^(3/2))
 

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*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] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, B]
 

rule 3478

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(B*a - A*b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f* 
x])^n/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2))   Int[(a 
+ b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[c*(a*A - b*B)*( 
m + 1) + d*n*(A*b - a*B) + (d*(a*A - b*B)*(m + 1) - c*(A*b - a*B)*(m + 2))* 
Sin[e + f*x] - d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x]^2, x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
 NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1021\) vs. \(2(252)=504\).

Time = 30.07 (sec) , antiderivative size = 1022, normalized size of antiderivative = 3.60

method	result	size
parts	\(\text {Expression too large to display}\)	\(1022\)
default	\(\text {Expression too large to display}\)	\(1040\)

Input:

int((a+cos(d*x+c)*b)^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x,method=_RET 
URNVERBOSE)
 

Output:

2*B/d*((cos(d*x+c)^2+2*cos(d*x+c)+1)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b 
))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*EllipticE(cot(d*x+c)-csc(d*x+ 
c),(-(a-b)/(a+b))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*((a+cos(d*x+c)*b)/( 
cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b*EllipticE(c 
ot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)* 
((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^ 
(1/2)*a*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(-cos(d*x+c) 
^2-2*cos(d*x+c)-1)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+ 
c)/(cos(d*x+c)+1))^(1/2)*b*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^ 
(1/2))+b*cos(d*x+c)*sin(d*x+c)+a*sin(d*x+c))*(a+cos(d*x+c)*b)^(1/2)/cos(d* 
x+c)^(1/2)/(b*cos(d*x+c)^2+a*cos(d*x+c)+cos(d*x+c)*b+a)+2/3*A/d*((cos(d*x+ 
c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a*b 
*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)^3+2*cos 
(d*x+c)^2+cos(d*x+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/ 
(cos(d*x+c)+1)/(a+b))^(1/2)*b^2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a 
+b))^(1/2))*(cos(d*x+c)^3+2*cos(d*x+c)^2+cos(d*x+c))+(cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^2*EllipticF(c 
ot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-cos(d*x+c)^3-2*cos(d*x+c)^2-c 
os(d*x+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c) 
+1)/(a+b))^(1/2)*a*b*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/...
 

Fricas [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algo 
rithm="fricas")
 

Output:

integral((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)/cos(d*x + c)^(5/2), 
 x)
 

Sympy [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sqrt {a + b \cos {\left (c + d x \right )}}}{\cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate((a+b*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(5/2),x)
 

Output:

Integral((A + B*cos(c + d*x))*sqrt(a + b*cos(c + d*x))/cos(c + d*x)**(5/2) 
, x)
 

Maxima [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algo 
rithm="maxima")
 

Output:

integrate((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)/cos(d*x + c)^(5/2) 
, x)
 

Giac [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algo 
rithm="giac")
 

Output:

integrate((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)/cos(d*x + c)^(5/2) 
, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(1/2))/cos(c + d*x)^(5/2),x 
)
 

Output:

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(1/2))/cos(c + d*x)^(5/2), 
x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) b \] Input:

int((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x)
 

Output:

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