3.5.89 \(\int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [489]

3.5.89.1 Optimal result
3.5.89.2 Mathematica [C] (warning: unable to verify)
3.5.89.3 Rubi [A] (verified)
3.5.89.4 Maple [A] (verified)
3.5.89.5 Fricas [C] (verification not implemented)
3.5.89.6 Sympy [F(-1)]
3.5.89.7 Maxima [F]
3.5.89.8 Giac [F]
3.5.89.9 Mupad [B] (verification not implemented)

3.5.89.1 Optimal result

Integrand size = 33, antiderivative size = 194 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {4 a^2 (8 A+9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^2 (5 A+6 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {4 a^2 (5 A+6 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {4 a^2 (8 A+9 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a^2 (11 A+9 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d} \]

output
4/15*a^2*(8*A+9*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti 
cE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/21*a^2*(5*A+6*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+4/45*a^2 
*(8*A+9*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/63*a^2*(11*A+9*B)*cos(d*x+c)^(5 
/2)*sin(d*x+c)/d+2/9*A*cos(d*x+c)^(5/2)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d+ 
4/21*a^2*(5*A+6*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d
 
3.5.89.2 Mathematica [C] (warning: unable to verify)

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

Time = 7.67 (sec) , antiderivative size = 1086, normalized size of antiderivative = 5.60 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\cos ^{\frac {7}{2}}(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \left (-\frac {(8 A+9 B) \cot (c)}{15 d}+\frac {(46 A+51 B) \cos (d x) \sin (c)}{168 d}+\frac {(37 A+36 B) \cos (2 d x) \sin (2 c)}{360 d}+\frac {(2 A+B) \cos (3 d x) \sin (3 c)}{56 d}+\frac {A \cos (4 d x) \sin (4 c)}{144 d}+\frac {(46 A+51 B) \cos (c) \sin (d x)}{168 d}+\frac {(37 A+36 B) \cos (2 c) \sin (2 d x)}{360 d}+\frac {(2 A+B) \cos (3 c) \sin (3 d x)}{56 d}+\frac {A \cos (4 c) \sin (4 d x)}{144 d}\right )}{B+A \cos (c+d x)}-\frac {5 A \cos ^3(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{21 d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)}}-\frac {2 B \cos ^3(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{7 d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)}}-\frac {4 A \cos ^3(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{15 d (B+A \cos (c+d x))}-\frac {3 B \cos ^3(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{10 d (B+A \cos (c+d x))} \]

input
Integrate[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x 
]
 
output
(Cos[c + d*x]^(7/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec 
[c + d*x])*(-1/15*((8*A + 9*B)*Cot[c])/d + ((46*A + 51*B)*Cos[d*x]*Sin[c]) 
/(168*d) + ((37*A + 36*B)*Cos[2*d*x]*Sin[2*c])/(360*d) + ((2*A + B)*Cos[3* 
d*x]*Sin[3*c])/(56*d) + (A*Cos[4*d*x]*Sin[4*c])/(144*d) + ((46*A + 51*B)*C 
os[c]*Sin[d*x])/(168*d) + ((37*A + 36*B)*Cos[2*c]*Sin[2*d*x])/(360*d) + (( 
2*A + B)*Cos[3*c]*Sin[3*d*x])/(56*d) + (A*Cos[4*c]*Sin[4*d*x])/(144*d)))/( 
B + A*Cos[c + d*x]) - (5*A*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4, 1 
/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c 
 + d*x])^2*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x 
 - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[ 
c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*d*(B + A*Cos[c + d*x])*Sqr 
t[1 + Cot[c]^2]) - (2*B*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2} 
, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + 
d*x])^2*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - 
ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]] 
])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*(B + A*Cos[c + d*x])*Sqrt[1 
+ Cot[c]^2]) - (4*A*Cos[c + d*x]^3*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[ 
c + d*x])^2*(A + B*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, 
Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - C 
os[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[...
 
3.5.89.3 Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.98, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 3433, 3042, 3455, 27, 3042, 3447, 3042, 3502, 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.

\(\displaystyle \int \cos ^{\frac {9}{2}}(c+d x) (a \sec (c+d x)+a)^2 (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{9/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3433

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \frac {2}{9} \int \frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a) (a (5 A+9 B)+a (11 A+9 B) \cos (c+d x))dx+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \int \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a) (a (5 A+9 B)+a (11 A+9 B) \cos (c+d x))dx+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (5 A+9 B)+a (11 A+9 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {1}{9} \int \cos ^{\frac {3}{2}}(c+d x) \left ((11 A+9 B) \cos ^2(c+d x) a^2+(5 A+9 B) a^2+\left ((5 A+9 B) a^2+(11 A+9 B) a^2\right ) \cos (c+d x)\right )dx+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left ((11 A+9 B) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+(5 A+9 B) a^2+\left ((5 A+9 B) a^2+(11 A+9 B) a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \int \cos ^{\frac {3}{2}}(c+d x) \left (9 (5 A+6 B) a^2+7 (8 A+9 B) \cos (c+d x) a^2\right )dx+\frac {2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (9 (5 A+6 B) a^2+7 (8 A+9 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (9 a^2 (5 A+6 B) \int \cos ^{\frac {3}{2}}(c+d x)dx+7 a^2 (8 A+9 B) \int \cos ^{\frac {5}{2}}(c+d x)dx\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (9 a^2 (5 A+6 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+7 a^2 (8 A+9 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (7 a^2 (8 A+9 B) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+9 a^2 (5 A+6 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (7 a^2 (8 A+9 B) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+9 a^2 (5 A+6 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \left (9 a^2 (5 A+6 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+7 a^2 (8 A+9 B) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )+\frac {2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{9} \left (\frac {2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {2}{7} \left (7 a^2 (8 A+9 B) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+9 a^2 (5 A+6 B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d}\)

input
Int[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
 
output
(2*A*Cos[c + d*x]^(5/2)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(9*d) + ((2 
*a^2*(11*A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d) + (2*(9*a^2*(5*A 
+ 6*B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c 
+ d*x])/(3*d)) + 7*a^2*(8*A + 9*B)*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + 
(2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d))))/7)/9
 

3.5.89.3.1 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 3433
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + 
c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

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 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 3502
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 
3.5.89.4 Maple [A] (verified)

Time = 22.38 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.13

method result size
default \(-\frac {4 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{2} \left (-560 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (1840 A +360 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-2368 A -1044 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (1568 A +1134 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-387 A -351 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+75 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+90 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-189 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{315 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(413\)

input
int(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 
output
-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-560*A 
*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+(1840*A+360*B)*sin(1/2*d*x+1/2*c 
)^8*cos(1/2*d*x+1/2*c)+(-2368*A-1044*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1 
/2*c)+(1568*A+1134*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-387*A-351* 
B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+75*A*(2*sin(1/2*d*x+1/2*c)^2-1) 
^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))- 
168*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt 
icE(cos(1/2*d*x+1/2*c),2^(1/2))+90*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin 
(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-189*B*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2 
*d*x+1/2*c),2^(1/2)))/(-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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 
3.5.89.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.15 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (5 \, A + 6 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (5 \, A + 6 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 i \, \sqrt {2} {\left (8 \, A + 9 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} {\left (8 \, A + 9 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (35 \, A a^{2} \cos \left (d x + c\right )^{3} + 45 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} + 14 \, {\left (8 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right ) + 30 \, {\left (5 \, A + 6 \, B\right )} a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{315 \, d} \]

input
integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith 
m="fricas")
 
output
-2/315*(15*I*sqrt(2)*(5*A + 6*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + 
c) + I*sin(d*x + c)) - 15*I*sqrt(2)*(5*A + 6*B)*a^2*weierstrassPInverse(-4 
, 0, cos(d*x + c) - I*sin(d*x + c)) - 21*I*sqrt(2)*(8*A + 9*B)*a^2*weierst 
rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) 
 + 21*I*sqrt(2)*(8*A + 9*B)*a^2*weierstrassZeta(-4, 0, weierstrassPInverse 
(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (35*A*a^2*cos(d*x + c)^3 + 45*(2 
*A + B)*a^2*cos(d*x + c)^2 + 14*(8*A + 9*B)*a^2*cos(d*x + c) + 30*(5*A + 6 
*B)*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/d
 
3.5.89.6 Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

input
integrate(cos(d*x+c)**(9/2)*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)
 
output
Timed out
 
3.5.89.7 Maxima [F]

\[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {9}{2}} \,d x } \]

input
integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith 
m="maxima")
 
output
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*cos(d*x + c)^(9/2), 
x)
 
3.5.89.8 Giac [F]

\[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {9}{2}} \,d x } \]

input
integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith 
m="giac")
 
output
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*cos(d*x + c)^(9/2), 
x)
 
3.5.89.9 Mupad [B] (verification not implemented)

Time = 15.46 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.37 \[ \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2\,B\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}-\frac {2\,A\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,A\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a^2\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,B\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

input
int(cos(c + d*x)^(9/2)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^2,x)
 
output
(2*B*a^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/ 
(3*d) - (2*A*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/ 
4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (4*A*a^2*cos(c + d*x)^( 
9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c 
 + d*x)^2)^(1/2)) - (2*A*a^2*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1 
/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2)) - (4*B*a^2 
*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^ 
2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (2*B*a^2*cos(c + d*x)^(9/2)*sin(c + d*x 
)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2) 
)