3.6.18 \(\int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [518]

3.6.18.1 Optimal result
3.6.18.2 Mathematica [A] (verified)
3.6.18.3 Rubi [A] (verified)
3.6.18.4 Maple [A] (verified)
3.6.18.5 Fricas [A] (verification not implemented)
3.6.18.6 Sympy [F(-1)]
3.6.18.7 Maxima [B] (verification not implemented)
3.6.18.8 Giac [F]
3.6.18.9 Mupad [F(-1)]

3.6.18.1 Optimal result

Integrand size = 35, antiderivative size = 247 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^{5/2} (200 A+163 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 d}+\frac {a^3 (104 A+95 B) \sin (c+d x)}{96 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {a^2 (8 A+11 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{24 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {a^3 (200 A+163 B) \sin (c+d x)}{64 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

output
1/4*a*B*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+1/96*a^3*(104 
*A+95*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+1/24*a^2*(8* 
A+11*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/sec(d*x+c)^(3/2)+1/64*a^3*(200 
*A+163*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+1/64*a^(5/2 
)*(200*A+163*B)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+ 
c)^(1/2)*sec(d*x+c)^(1/2)/d
 
3.6.18.2 Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.64 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (3 \sqrt {2} (200 A+163 B) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (632 A+581 B+(272 A+362 B) \cos (c+d x)+4 (8 A+23 B) \cos (2 (c+d x))+12 B \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{384 d} \]

input
Integrate[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Sqrt[Sec[c + d 
*x]],x]
 
output
(a^2*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[S 
ec[c + d*x]]*(3*Sqrt[2]*(200*A + 163*B)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + 
 2*Sqrt[Cos[c + d*x]]*(632*A + 581*B + (272*A + 362*B)*Cos[c + d*x] + 4*(8 
*A + 23*B)*Cos[2*(c + d*x)] + 12*B*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(3 
84*d)
 
3.6.18.3 Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.02, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3440, 3042, 3455, 27, 3042, 3455, 27, 3042, 3460, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3440

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \int \frac {1}{2} \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2} (a (8 A+3 B)+a (8 A+11 B) \cos (c+d x))dx+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \int \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2} (a (8 A+3 B)+a (8 A+11 B) \cos (c+d x))dx+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (8 A+3 B)+a (8 A+11 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 3455

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{3} \int \frac {1}{2} \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a} \left (3 (24 A+17 B) a^2+(104 A+95 B) \cos (c+d x) a^2\right )dx+\frac {a^2 (8 A+11 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a} \left (3 (24 A+17 B) a^2+(104 A+95 B) \cos (c+d x) a^2\right )dx+\frac {a^2 (8 A+11 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (24 A+17 B) a^2+(104 A+95 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {a^2 (8 A+11 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 3460

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a^3 (104 A+95 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (104 A+95 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 3249

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (104 A+95 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (104 A+95 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 3253

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \left (\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (104 A+95 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {a^2 (8 A+11 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}+\frac {1}{6} \left (\frac {a^3 (104 A+95 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}+\frac {3}{4} a^2 (200 A+163 B) \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )\right )\right )+\frac {a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\right )\)

input
Int[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Sqrt[Sec[c + d*x]],x 
]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((a*B*Cos[c + d*x]^(3/2)*(a + a*Cos[ 
c + d*x])^(3/2)*Sin[c + d*x])/(4*d) + ((a^2*(8*A + 11*B)*Cos[c + d*x]^(3/2 
)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((a^3*(104*A + 95*B)*Cos[ 
c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*a^2*(200* 
A + 163*B)*((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)/8)
 

3.6.18.3.1 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 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 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]
 
3.6.18.4 Maple [A] (verified)

Time = 21.16 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.42

method result size
default \(\frac {a^{2} \left (48 B \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+64 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+184 B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+272 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+326 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+600 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+489 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+600 A \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+489 B \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{192 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) \(351\)
parts \(\frac {A \left (8 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+34 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+75 \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+75 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{24 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}+\frac {B \left (48 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+184 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+326 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+489 \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+489 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{192 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) \(403\)

input
int((a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c))/sec(d*x+c)^(1/2),x,method=_RET 
URNVERBOSE)
 
output
1/192*a^2/d*(48*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 
)+64*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+184*B*cos 
(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+272*A*cos(d*x+c)*si 
n(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+326*B*cos(d*x+c)*sin(d*x+c)*(co 
s(d*x+c)/(1+cos(d*x+c)))^(1/2)+600*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin 
(d*x+c)+489*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+600*A*arctan(ta 
n(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+489*B*arctan(tan(d*x+c)*(cos(d 
*x+c)/(1+cos(d*x+c)))^(1/2)))*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/sec( 
d*x+c)^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
 
3.6.18.5 Fricas [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.74 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=-\frac {3 \, {\left ({\left (200 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (200 \, A + 163 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (48 \, B a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (8 \, A + 23 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (136 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (200 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

input
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/sec(d*x+c)^(1/2),x, algo 
rithm="fricas")
 
output
-1/192*(3*((200*A + 163*B)*a^2*cos(d*x + c) + (200*A + 163*B)*a^2)*sqrt(a) 
*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)) 
) - (48*B*a^2*cos(d*x + c)^4 + 8*(8*A + 23*B)*a^2*cos(d*x + c)^3 + 2*(136* 
A + 163*B)*a^2*cos(d*x + c)^2 + 3*(200*A + 163*B)*a^2*cos(d*x + c))*sqrt(a 
*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)
 
3.6.18.6 Sympy [F(-1)]

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

input
integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))/sec(d*x+c)**(1/2),x)
 
output
Timed out
 
3.6.18.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 9390 vs. \(2 (211) = 422\).

Time = 0.96 (sec) , antiderivative size = 9390, normalized size of antiderivative = 38.02 \[ \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\text {Too large to display} \]

input
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/sec(d*x+c)^(1/2),x, algo 
rithm="maxima")
 
output
1/768*(8*(4*(a^2*cos(3/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d 
*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1))*si 
n(3*d*x + 3*c) - (a^2*cos(3*d*x + 3*c) - a^2)*sin(3/2*arctan2(sin(2/3*arct 
an2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c) 
, cos(3*d*x + 3*c))) + 1)))*(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 
 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos 
(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(3/4)*sqrt(a) + 30* 
(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2( 
sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c 
), cos(3*d*x + 3*c))) + 1)^(1/4)*((a^2*sin(2/3*arctan2(sin(3*d*x + 3*c), c 
os(3*d*x + 3*c))) + 5*a^2*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3* 
c))))*cos(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) 
, cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) - (a^2*cos(2/ 
3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 3*a^2*cos(1/3*arctan2(sin 
(3*d*x + 3*c), cos(3*d*x + 3*c))) - 4*a^2)*sin(1/2*arctan2(sin(2/3*arctan2 
(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), c 
os(3*d*x + 3*c))) + 1)))*sqrt(a) + 75*(a^2*arctan2(-(cos(2/3*arctan2(sin(3 
*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos( 
3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) 
+ 1)^(1/4)*(cos(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x...
 
3.6.18.8 Giac [F]

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

input
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/sec(d*x+c)^(1/2),x, algo 
rithm="giac")
 
output
integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(5/2)/sqrt(sec(d*x + c 
)), x)
 
3.6.18.9 Mupad [F(-1)]

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

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