\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [1234]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 37, antiderivative size = 156 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {2} (A+C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {2 A \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \] Output:

2^(1/2)*(A+C)*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(1/2)/d-2/3*A*sec(d 
*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/3*A*sec(d*x+c)^(3/2)*sin 
(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 6.97 (sec) , antiderivative size = 576, normalized size of antiderivative = 3.69 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )} \left (-\frac {4 C \sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{3/2}}+\frac {(A+C) \csc ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-12 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {7}{2};1,\frac {9}{2};-\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 ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )-12 \operatorname {Hypergeometric2F1}\left (2,\frac {7}{2},\frac {9}{2},-\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 ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (4-7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+3 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )\right )+7 \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 )}} \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3 \left (15-20 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+8 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (\left (3-7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\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 )}}-3 \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 ) \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )\right )}{63 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{7/2}}\right )}{d \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

(2*Cos[c/2 + (d*x)/2]*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*S 
in[c/2 + (d*x)/2]^2]*((-4*C*Sin[c/2 + (d*x)/2]^3)/(3*(1 - 2*Sin[c/2 + (d*x 
)/2]^2)^(3/2)) + ((A + C)*Csc[c/2 + (d*x)/2]^5*(-12*Cos[(c + d*x)/2]^4*Hyp 
ergeometricPFQ[{2, 2, 7/2}, {1, 9/2}, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/ 
2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8 - 12*Hypergeometric2F1[2, 7/2, 9/2, 
 -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^ 
8*(4 - 7*Sin[c/2 + (d*x)/2]^2 + 3*Sin[c/2 + (d*x)/2]^4) + 7*Sqrt[-(Sin[c/2 
 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^ 
3*(15 - 20*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*((3 - 7*Sin[c/2 
+ (d*x)/2]^2)*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))] - 
 3*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*(1 
- 2*Sin[c/2 + (d*x)/2]^2))))/(63*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2))))/(d* 
Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 4709, 3042, 3523, 27, 3042, 3463, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sin[c + d*x])/(3*d*Cos[c + d*x 
]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-3*Sqrt[2]*Sqrt[a]*(A + C)*ArcTan[(S 
qrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) 
])/d + (2*a*A*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]] 
))/(3*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 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 3463
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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
+ 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2))   Int[(a 
 + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 
1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], 
 x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq 
Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m 
 + 1/2, 0])
 

rule 3523
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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])
 

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 [A] (verified)

Time = 2.86 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.18

method result size
default \(-\frac {\sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (3 \cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}\right )+C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (3 \cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}\right )+\sin \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right ) \sqrt {2}\, A \cos \left (d x +c \right )\right )}{3 d a \left (1+\cos \left (d x +c \right )\right )}\) \(184\)
parts \(-\frac {A \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (\sin \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right ) \sqrt {2}\, \cos \left (d x +c \right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \left (3 \cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}\right )\right )}{3 d a \left (1+\cos \left (d x +c \right )\right )}-\frac {C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right )^{2}}{d a}\) \(201\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x,method=_R 
ETURNVERBOSE)
 

Output:

-1/3/d/a*2^(1/2)*((1+cos(d*x+c))*a)^(1/2)*sec(d*x+c)^(5/2)/(1+cos(d*x+c))* 
(A*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(3*cos 
(d*x+c)^3+3*cos(d*x+c)^2)+C*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+c)/(1+ 
cos(d*x+c)))^(1/2)*(3*cos(d*x+c)^3+3*cos(d*x+c)^2)+sin(d*x+c)*(-1+cos(d*x+ 
c))*2^(1/2)*A*cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.87 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{2} + {\left (A + C\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} + \frac {2 \, {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, al 
gorithm="fricas")
 

Output:

-1/3*(3*sqrt(2)*((A + C)*a*cos(d*x + c)^2 + (A + C)*a*cos(d*x + c))*arctan 
(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c) 
))/sqrt(a) + 2*(A*cos(d*x + c) - A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/ 
sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, al 
gorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, al 
gorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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