3.12.96 \(\int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx\) [1196]

3.12.96.1 Optimal result
3.12.96.2 Mathematica [C] (verified)
3.12.96.3 Rubi [A] (verified)
3.12.96.4 Maple [A] (verified)
3.12.96.5 Fricas [C] (verification not implemented)
3.12.96.6 Sympy [F(-1)]
3.12.96.7 Maxima [F]
3.12.96.8 Giac [F]
3.12.96.9 Mupad [F(-1)]

3.12.96.1 Optimal result

Integrand size = 35, antiderivative size = 236 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 (5 A+14 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^2 d}-\frac {5 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {7}{2}}(c+d x)}-\frac {(A+3 C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 (5 A+14 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (A+3 C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}} \]

output
-1/3*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2/sec(d*x+c)^(7/2)-(A+3*C)*sin(d* 
x+c)/a^2/d/(1+cos(d*x+c))/sec(d*x+c)^(5/2)+4/15*(5*A+14*C)*sin(d*x+c)/a^2/ 
d/sec(d*x+c)^(3/2)-5/3*(A+3*C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)+4/5*(5*A+ 
14*C)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d-5/3*(A+3*C)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d
 
3.12.96.2 Mathematica [C] (verified)

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

Time = 10.26 (sec) , antiderivative size = 813, normalized size of antiderivative = 3.44 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {4 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right )}{3 d (a+a \cos (c+d x))^2}-\frac {56 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right )}{15 d (a+a \cos (c+d x))^2}-\frac {10 A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \sqrt {\sec (c+d x)} \sin (c)}{3 d (a+a \cos (c+d x))^2}-\frac {10 C \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \sqrt {\sec (c+d x)} \sin (c)}{d (a+a \cos (c+d x))^2}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\sec (c+d x)} \left (-\frac {(60 A+151 C+20 A \cos (2 c)+73 C \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{10 d}-\frac {8 C \cos (2 d x) \sin (2 c)}{3 d}+\frac {2 C \cos (3 d x) \sin (3 c)}{5 d}-\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{3 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (7 A \sin \left (\frac {d x}{2}\right )+13 C \sin \left (\frac {d x}{2}\right )\right )}{3 d}+\frac {2 (20 A+73 C) \cos (c) \sin (d x)}{5 d}-\frac {8 C \cos (2 c) \sin (2 d x)}{3 d}+\frac {2 C \cos (3 c) \sin (3 d x)}{5 d}+\frac {4 (7 A+13 C) \tan \left (\frac {c}{2}\right )}{3 d}-\frac {2 (A+C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{3 d}\right )}{(a+a \cos (c+d x))^2} \]

input
Integrate[(A + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2 
)),x]
 
output
(-4*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^( 
(2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + 
 d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4 
, -E^((2*I)*(c + d*x))])*Sec[c/2])/(3*d*E^(I*d*x)*(a + a*Cos[c + d*x])^2) 
- (56*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E 
^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c 
 + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7 
/4, -E^((2*I)*(c + d*x))])*Sec[c/2])/(15*d*E^(I*d*x)*(a + a*Cos[c + d*x])^ 
2) - (10*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + 
 d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x])^ 
2) - (10*C*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + 
 d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(d*(a + a*Cos[c + d*x])^2) 
 + (Cos[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*(-1/10*((60*A + 151*C + 20*A*C 
os[2*c] + 73*C*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2])/d - (8*C*Cos[2*d*x]*S 
in[2*c])/(3*d) + (2*C*Cos[3*d*x]*Sin[3*c])/(5*d) - (2*Sec[c/2]*Sec[c/2 + ( 
d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) + (4*Sec[c/2]*Sec[c/2 + 
 (d*x)/2]*(7*A*Sin[(d*x)/2] + 13*C*Sin[(d*x)/2]))/(3*d) + (2*(20*A + 73*C) 
*Cos[c]*Sin[d*x])/(5*d) - (8*C*Cos[2*c]*Sin[2*d*x])/(3*d) + (2*C*Cos[3*c]* 
Sin[3*d*x])/(5*d) + (4*(7*A + 13*C)*Tan[c/2])/(3*d) - (2*(A + C)*Sec[c/2 + 
 (d*x)/2]^2*Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2
 
3.12.96.3 Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.91, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4709, 3042, 3521, 27, 3042, 3456, 3042, 3227, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int -\frac {\cos ^{\frac {5}{2}}(c+d x) (a (A+7 C)-a (5 A+11 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\cos ^{\frac {5}{2}}(c+d x) (a (A+7 C)-a (5 A+11 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (A+7 C)-a (5 A+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 3456

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \cos ^{\frac {3}{2}}(c+d x) \left (15 a^2 (A+3 C)-4 a^2 (5 A+14 C) \cos (c+d x)\right )dx}{a^2}+\frac {6 (A+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (15 a^2 (A+3 C)-4 a^2 (5 A+14 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {6 (A+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 3227

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {15 a^2 (A+3 C) \int \cos ^{\frac {3}{2}}(c+d x)dx-4 a^2 (5 A+14 C) \int \cos ^{\frac {5}{2}}(c+d x)dx}{a^2}+\frac {6 (A+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {15 a^2 (A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-4 a^2 (5 A+14 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx}{a^2}+\frac {6 (A+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {15 a^2 (A+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-4 a^2 (5 A+14 C) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (A+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {15 a^2 (A+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-4 a^2 (5 A+14 C) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (A+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {15 a^2 (A+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-4 a^2 (5 A+14 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (A+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {15 a^2 (A+3 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-4 a^2 (5 A+14 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (A+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\)

input
Int[(A + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)),x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/3*((A + C)*Cos[c + d*x]^(7/2)*Si 
n[c + d*x])/(d*(a + a*Cos[c + d*x])^2) - ((6*(A + 3*C)*Cos[c + d*x]^(5/2)* 
Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + (15*a^2*(A + 3*C)*((2*EllipticF[(c 
+ d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)) - 4*a^2*( 
5*A + 14*C)*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*S 
in[c + d*x])/(5*d)))/a^2)/(6*a^2))
 

3.12.96.3.1 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 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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 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 3521
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[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
 + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1))   I 
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) 
- b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* 
c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] && LtQ[m, -2^(-1)]
 

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]
 
3.12.96.4 Maple [A] (verified)

Time = 3.77 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.91

method result size
default \(\frac {\sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-96 C \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+352 C \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A +50 A \left (\cos ^{3}\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 (\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 )+120 A \left (\cos ^{3}\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 (\cos ^{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 )-120 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+150 C \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 ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+336 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C \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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-190 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A -266 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+75 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+135 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 A -5 C \right )}{30 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{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 )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(451\)

input
int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2),x,method=_RETUR 
NVERBOSE)
 
output
1/30*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-96*C*cos(1 
/2*d*x+1/2*c)^10+352*C*cos(1/2*d*x+1/2*c)^8+120*cos(1/2*d*x+1/2*c)^6*A+50* 
A*cos(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+120*A*cos(1/2*d*x+1/2*c) 
^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)*Elliptic 
E(cos(1/2*d*x+1/2*c),2^(1/2))-120*C*cos(1/2*d*x+1/2*c)^6+150*C*(sin(1/2*d* 
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+ 
1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+336*cos(1/2*d*x+1/2*c)^3*C*(sin(1/2*d 
*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x 
+1/2*c),2^(1/2))-190*cos(1/2*d*x+1/2*c)^4*A-266*C*cos(1/2*d*x+1/2*c)^4+75* 
A*cos(1/2*d*x+1/2*c)^2+135*C*cos(1/2*d*x+1/2*c)^2-5*A-5*C)/a^2/cos(1/2*d*x 
+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x 
+1/2*c)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 
3.12.96.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.64 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 \, {\left (\sqrt {2} {\left (-5 i \, A - 14 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-5 i \, A - 14 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A - 14 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 12 \, {\left (\sqrt {2} {\left (5 i \, A + 14 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (5 i \, A + 14 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + 14 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, C \cos \left (d x + c\right )^{4} - 8 \, C \cos \left (d x + c\right )^{3} - 2 \, {\left (15 \, A + 47 \, C\right )} \cos \left (d x + c\right )^{2} - 25 \, {\left (A + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

input
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algori 
thm="fricas")
 
output
-1/30*(25*(sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-I*A - 3*I*C 
)*cos(d*x + c) + sqrt(2)*(-I*A - 3*I*C))*weierstrassPInverse(-4, 0, cos(d* 
x + c) + I*sin(d*x + c)) + 25*(sqrt(2)*(I*A + 3*I*C)*cos(d*x + c)^2 + 2*sq 
rt(2)*(I*A + 3*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 3*I*C))*weierstrassPInve 
rse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 12*(sqrt(2)*(-5*I*A - 14*I*C)* 
cos(d*x + c)^2 + 2*sqrt(2)*(-5*I*A - 14*I*C)*cos(d*x + c) + sqrt(2)*(-5*I* 
A - 14*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c 
) + I*sin(d*x + c))) + 12*(sqrt(2)*(5*I*A + 14*I*C)*cos(d*x + c)^2 + 2*sqr 
t(2)*(5*I*A + 14*I*C)*cos(d*x + c) + sqrt(2)*(5*I*A + 14*I*C))*weierstrass 
Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2 
*(6*C*cos(d*x + c)^4 - 8*C*cos(d*x + c)^3 - 2*(15*A + 47*C)*cos(d*x + c)^2 
 - 25*(A + 3*C)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos( 
d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 
3.12.96.6 Sympy [F(-1)]

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

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

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

input
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algori 
thm="maxima")
 
output
integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2 
)), x)
 
3.12.96.8 Giac [F]

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

input
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algori 
thm="giac")
 
output
integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2 
)), x)
 
3.12.96.9 Mupad [F(-1)]

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

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