\(\int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\) [204]

Optimal result

Integrand size = 35, antiderivative size = 154 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(5 A+3 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(A+7 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \] Output:

1/32*(5*A+3*B)*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))*2^(1/2)/a^(5/2)/d+1/4*(A-B)*cos(d*x+c)^(1/2)*sin(d*x+c 
)/d/(a+a*cos(d*x+c))^(5/2)+1/16*(A+7*B)*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a 
+a*cos(d*x+c))^(3/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(491\) vs. \(2(154)=308\).

Time = 6.42 (sec) , antiderivative size = 491, normalized size of antiderivative = 3.19 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {3 B \arcsin \left (\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{4 d (a (1+\cos (c+d x)))^{5/2}}+\frac {5 B \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {1-\sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}{4 d \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )} (a (1+\cos (c+d x)))^{5/2}}-\frac {B \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {1-\sec ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}{2 d \cos ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a (1+\cos (c+d x)))^{5/2}}+\frac {A \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )} \left (3-\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-5 \text {arctanh}\left (\sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right )}{4 d (a (1+\cos (c+d x)))^{5/2}} \] Input:

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

Output:

(3*B*ArcSin[Sin[c/2 + (d*x)/2]/Sqrt[Cos[(c + d*x)/2]^2]]*Cos[c/2 + (d*x)/2 
]^5)/(4*d*(a*(1 + Cos[c + d*x]))^(5/2)) + (5*B*Cos[c/2 + (d*x)/2]^5*Sin[c/ 
2 + (d*x)/2]*Sqrt[1 - Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]^2])/(4*d*Sqrt[ 
Cos[(c + d*x)/2]^2]*(a*(1 + Cos[c + d*x]))^(5/2)) - (B*Cos[c/2 + (d*x)/2]^ 
5*Sin[c/2 + (d*x)/2]^3*Sqrt[1 - Sec[(c + d*x)/2]^2*Sin[c/2 + (d*x)/2]^2])/ 
(2*d*(Cos[(c + d*x)/2]^2)^(3/2)*(a*(1 + Cos[c + d*x]))^(5/2)) + (A*Cos[c/2 
 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*Sqrt[1 - 2*Sin[c/2 + ( 
d*x)/2]^2]*(3 - Sin[c/2 + (d*x)/2]^2 - 5*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2] 
^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*Cos[(c + d*x)/2]^4*Csc[c/2 + (d*x)/2]^2 
*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(4*d*(a*(1 + 
 Cos[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3456, 27, 3042, 3457, 27, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\int \frac {a (A-B)+2 a (A+3 B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {a^2 (5 A+3 B)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}+\frac {a (A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{4} (5 A+3 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx+\frac {a (A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} (5 A+3 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {a (A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {\frac {a (A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {a (5 A+3 B) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {(5 A+3 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}+\frac {a (A+7 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\)

Input:

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

Output:

((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) 
 + (((5*A + 3*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]] 
*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*Sqrt[a]*d) + (a*(A + 7*B)*Sqrt[Cos 
[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)))/(8*a^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 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 3042

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

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 3456

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{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] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In 
tegerQ[2*n] || EqQ[c, 0])
 

rule 3457

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], 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[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

Maple [A] (verified)

Time = 8.88 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.53

method	result	size
default	\(\frac {\left (\sin \left (d x +c \right ) \left (\cos \left (d x +c \right )+5\right ) \sqrt {2}\, A \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \left (7 \cos \left (d x +c \right )+3\right ) \sqrt {2}\, B \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (-5 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-5\right ) A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+\left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sqrt {\cos \left (d x +c \right )}}{32 d \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{3}}\)	\(236\)
parts	\(\frac {A \left (\sqrt {2}\, \sin \left (d x +c \right ) \left (\cos \left (d x +c \right )+5\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (-5 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-5\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{32 d \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{3}}+\frac {B \left (\sin \left (d x +c \right ) \left (7 \cos \left (d x +c \right )+3\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{32 d \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{3}}\)	\(316\)

Input:

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

Output:

1/32/d*(sin(d*x+c)*(cos(d*x+c)+5)*2^(1/2)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1 
/2)+sin(d*x+c)*(7*cos(d*x+c)+3)*2^(1/2)*B*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2 
)+(-5*cos(d*x+c)^2-10*cos(d*x+c)-5)*A*arcsin(cot(d*x+c)-csc(d*x+c))+(-3*co 
s(d*x+c)^2-6*cos(d*x+c)-3)*B*arcsin(cot(d*x+c)-csc(d*x+c)))*2^(1/2)*(a*(co 
s(d*x+c)+1))^(1/2)*cos(d*x+c)^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x 
+c)+1)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/a^3
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.40 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 5 \, A + 3 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) + 2 \, {\left ({\left (A + 7 \, B\right )} \cos \left (d x + c\right ) + 5 \, A + 3 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

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

Output:

1/32*(sqrt(2)*((5*A + 3*B)*cos(d*x + c)^3 + 3*(5*A + 3*B)*cos(d*x + c)^2 + 
 3*(5*A + 3*B)*cos(d*x + c) + 5*A + 3*B)*sqrt(a)*arctan(1/2*sqrt(2)*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))) + 2*((A + 7*B)*cos(d*x + c) + 5*A + 3*B)*sqrt(a*cos( 
d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a 
^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
 

Sympy [F]

\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:

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

Output:

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