3.2.28 \(\int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [128]

3.2.28.1 Optimal result
3.2.28.2 Mathematica [C] (warning: unable to verify)
3.2.28.3 Rubi [A] (verified)
3.2.28.4 Maple [B] (verified)
3.2.28.5 Fricas [C] (verification not implemented)
3.2.28.6 Sympy [F(-1)]
3.2.28.7 Maxima [F]
3.2.28.8 Giac [F]
3.2.28.9 Mupad [B] (verification not implemented)

3.2.28.1 Optimal result

Integrand size = 31, antiderivative size = 95 \[ \int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 a (A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a (A+3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

output
-2*a*(A+B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1 
/2*d*x+1/2*c),2^(1/2))/d+2/3*a*(A+3*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/ 
2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*a*A*sin(d*x+c)/d/ 
cos(d*x+c)^(3/2)+2*a*(A+B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
 
3.2.28.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.21 (sec) , antiderivative size = 293, normalized size of antiderivative = 3.08 \[ \int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (6 (A+B) \cos (c+d x) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sec (c) \sin (d x+\arctan (\tan (c)))+\left (-3 (A+B) \cos (c+d x) (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \csc (c) \sec (c)+2 (3 (A+B) \cot (c)+(A \cos (d x)-A \cos (2 c+d x)+3 (A+B) \cos (c+2 d x)) \csc (c)) \sqrt {\sec ^2(c)}-4 (A+3 B) \cos ^2(c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sqrt {\sec ^2(c)} \sec (d x-\arctan (\cot (c))) \sin (c)\right ) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{12 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}} \]

input
Integrate[((a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x 
]
 
output
(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(6*(A + B)*Cos[c + d*x]*Hypergeom 
etricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sec[c]*Sin[d*x 
+ ArcTan[Tan[c]]] + (-3*(A + B)*Cos[c + d*x]*(3*Cos[c - d*x - ArcTan[Tan[c 
]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Csc[c]*Sec[c] + 2*(3*(A + B)*Cot[c] + 
 (A*Cos[d*x] - A*Cos[2*c + d*x] + 3*(A + B)*Cos[c + 2*d*x])*Csc[c])*Sqrt[S 
ec[c]^2] - 4*(A + 3*B)*Cos[c + d*x]^2*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sq 
rt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]] 
]^2]*Sqrt[Sec[c]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c])*Sqrt[Sin[d*x + ArcTa 
n[Tan[c]]]^2]))/(12*d*Cos[c + d*x]^(3/2)*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + Arc 
Tan[Tan[c]]]^2])
 
3.2.28.3 Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 3447, 3042, 3500, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

input
Int[((a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x]
 
output
(2*a*A*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((2*a*(A + 3*B)*EllipticF[ 
(c + d*x)/2, 2])/d + 3*a*(A + B)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Si 
n[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/3
 

3.2.28.3.1 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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3447
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3500
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 
3.2.28.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(398\) vs. \(2(139)=278\).

Time = 7.31 (sec) , antiderivative size = 399, normalized size of antiderivative = 4.20

method result size
default \(-\frac {4 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a \left (\frac {B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{2 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}+\frac {A \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{6 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}\right )}{2}+\frac {\left (\frac {A}{2}+\frac {B}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(399\)
parts \(-\frac {2 \left (a A +B a \right ) \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}+\frac {2 B a \,\operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d}-\frac {2 a A \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}\) \(423\)

input
int((a+cos(d*x+c)*a)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x,method=_RETURNVER 
BOSE)
 
output
-4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(1/2*B*(sin 
(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x 
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) 
)+1/2*A*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2* 
c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2 
*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(1/2*A+1/2*B)/sin(1/2* 
d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d 
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x 
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 
3.2.28.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.06 \[ \int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-i \, \sqrt {2} {\left (A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + A a\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \]

input
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm= 
"fricas")
 
output
1/3*(-I*sqrt(2)*(A + 3*B)*a*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos( 
d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*(A + 3*B)*a*cos(d*x + c)^2*weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(A + B)*a* 
cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + 
 c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(A + B)*a*cos(d*x + c)^2*weierstrassZ 
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2* 
(3*(A + B)*a*cos(d*x + c) + A*a)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d 
*x + c)^2)
 
3.2.28.6 Sympy [F(-1)]

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

input
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)**(5/2),x)
 
output
Timed out
 
3.2.28.7 Maxima [F]

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

input
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm= 
"maxima")
 
output
integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)/cos(d*x + c)^(5/2), x)
 
3.2.28.8 Giac [F]

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

input
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm= 
"giac")
 
output
integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)/cos(d*x + c)^(5/2), x)
 
3.2.28.9 Mupad [B] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.58 \[ \int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,B\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

input
int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x)))/cos(c + d*x)^(5/2),x)
 
output
(2*B*a*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*a*sin(c + d*x)*hypergeom([-1/ 
4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2 
)) + (2*A*a*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d 
*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*B*a*sin(c + d*x)*hypergeo 
m([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2 
)^(1/2))