\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\) [516]

Optimal result

Integrand size = 35, antiderivative size = 198 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a^{5/2} (20 A+19 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}-\frac {a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \] Output:

1/4*a^(5/2)*(20*A+19*B)*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-1/4*a^3*(4*A-9*B)*sin(d*x+c)/d/(a+a*co 
s(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-1/2*a^2*(4*A-B)*(a+a*cos(d*x+c))^(1/2)*si 
n(d*x+c)/d/sec(d*x+c)^(1/2)+2*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(1/2)* 
sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.64 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sqrt {2} (20 A+19 B) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 (8 A+B+(4 A+11 B) \cos (c+d x)+B \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(3/ 
2),x]
 

Output:

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(Sqrt[ 
2]*(20*A + 19*B)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2*( 
8*A + B + (4*A + 11*B)*Cos[c + d*x] + B*Cos[2*(c + d*x)])*Sin[(c + d*x)/2] 
))/(8*d)
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3440, 3042, 3454, 27, 3042, 3455, 27, 3042, 3460, 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.

\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3440

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^2 (12 A+5 B)-a^2 (4 A-9 B) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^2 (12 A+5 B)-a^2 (4 A-9 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^2 (12 A+5 B)-a^2 (4 A-9 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3460

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {1}{2} a^2 (20 A+19 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-\frac {a^3 (4 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (\frac {1}{2} a^2 (20 A+19 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^3 (4 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3253

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \left (-\frac {a^2 (20 A+19 B) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {a^3 (4 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {a^{5/2} (20 A+19 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^3 (4 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{d \sqrt {\cos (c+d x)}}\right )\)

Input:

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*(a^2*(4*A - B)*Sqrt[Cos[c + d* 
x]]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/d + (2*a*A*(a + a*Cos[c + d*x]) 
^(3/2)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + ((a^(5/2)*(20*A + 19*B)*ArcS 
in[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^3*(4*A - 9*B)* 
Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/4)
 

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 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 3454

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{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] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3455

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*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] && GtQ[m, 1/2] &&  !LtQ[n, -1 
] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

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 [A] (verified)

Time = 21.98 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {\sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (20 A \cos \left (d x +c \right )^{2} \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+19 B \cos \left (d x +c \right )^{2} \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\cos \left (d x +c \right ) \left (4 \cos \left (d x +c \right )+8\right ) \sin \left (d x +c \right ) A \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+11\right ) B \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) a^{2}}{4 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(223\)
parts	\(\frac {A \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (5 \cos \left (d x +c \right )^{2} \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\cos \left (d x +c \right ) \left (\cos \left (d x +c \right )+2\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) a^{2}}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}+\frac {B \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+11\right )+\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \left (19 \cos \left (d x +c \right )^{2}+19 \cos \left (d x +c \right )\right )\right ) a^{2}}{4 d \left (\cos \left (d x +c \right )+1\right )}\)	\(268\)

Input:

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/4/d*(a*(cos(d*x+c)+1))^(1/2)*sec(d*x+c)^(3/2)/(cos(d*x+c)+1)/(cos(d*x+c) 
/(cos(d*x+c)+1))^(1/2)*(20*A*cos(d*x+c)^2*arctan(tan(d*x+c)*(cos(d*x+c)/(c 
os(d*x+c)+1))^(1/2))+19*B*cos(d*x+c)^2*arctan(tan(d*x+c)*(cos(d*x+c)/(cos( 
d*x+c)+1))^(1/2))+cos(d*x+c)*(4*cos(d*x+c)+8)*sin(d*x+c)*A*(cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)+cos(d*x+c)^2*sin(d*x+c)*(2*cos(d*x+c)+11)*B*(cos(d*x+c) 
/(cos(d*x+c)+1))^(1/2))*a^2
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.74 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {{\left ({\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (20 \, A + 19 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algo 
rithm="fricas")
 

Output:

-1/4*(((20*A + 19*B)*a^2*cos(d*x + c) + (20*A + 19*B)*a^2)*sqrt(a)*arctan( 
sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (2*B 
*a^2*cos(d*x + c)^2 + (4*A + 11*B)*a^2*cos(d*x + c) + 8*A*a^2)*sqrt(a*cos( 
d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(3/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algo 
rithm="maxima")
 

Output:

Timed out
 

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algo 
rithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(5/2) 
,x)
                                                                                    
                                                                                    
 

Output:

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(5/2) 
, x)
 

Reduce [F]

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

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x)
 

Output:

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