3.6.41 \(\int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [541]

3.6.41.1 Optimal result
3.6.41.2 Mathematica [C] (warning: unable to verify)
3.6.41.3 Rubi [A] (verified)
3.6.41.4 Maple [B] (verified)
3.6.41.5 Fricas [C] (verification not implemented)
3.6.41.6 Sympy [F(-1)]
3.6.41.7 Maxima [F(-1)]
3.6.41.8 Giac [F]
3.6.41.9 Mupad [F(-1)]

3.6.41.1 Optimal result

Integrand size = 43, antiderivative size = 291 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 a^2 (12 A+9 B+8 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (7 A+6 B+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {4 a^2 (12 A+9 B+8 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {4 a^2 (7 A+6 B+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a^2 (21 A+27 B+19 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 C \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2 (9 B+4 C) \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d} \]

output
4/21*a^2*(7*A+6*B+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/105*a^2*(21*A+27*B+ 
19*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/9*C*sec(d*x+c)^(5/2)*(a+a*sec(d*x+c) 
)^2*sin(d*x+c)/d+2/63*(9*B+4*C)*sec(d*x+c)^(5/2)*(a^2+a^2*sec(d*x+c))*sin( 
d*x+c)/d+4/15*a^2*(12*A+9*B+8*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/15*a^2*(1 
2*A+9*B+8*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)/d+4/21*a^2*(7*A 
+6*B+5*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)/d
 
3.6.41.2 Mathematica [C] (warning: unable to verify)

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

Time = 9.54 (sec) , antiderivative size = 1270, normalized size of antiderivative = 4.36 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, 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(c+d x) \csc (c) \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 ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{15 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {\sqrt {2} B 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(c+d x) \csc (c) \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 ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{5 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {8 \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(c+d x) \csc (c) \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 ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{45 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {7}{2}}(c+d x)}+\frac {4 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{7 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {7}{2}}(c+d x)}+\frac {10 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{21 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {7}{2}}(c+d x)}+\frac {\sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 (12 A+9 B+8 C) \cos (d x) \csc (c)}{15 d}+\frac {C \sec (c) \sec ^4(c+d x) \sin (d x)}{9 d}+\frac {\sec (c) \sec ^3(c+d x) (7 C \sin (c)+9 B \sin (d x)+18 C \sin (d x))}{63 d}+\frac {\sec (c) \sec ^2(c+d x) (45 B \sin (c)+90 C \sin (c)+63 A \sin (d x)+126 B \sin (d x)+112 C \sin (d x))}{315 d}+\frac {\sec (c) \sec (c+d x) (63 A \sin (c)+126 B \sin (c)+112 C \sin (c)+210 A \sin (d x)+180 B \sin (d x)+150 C \sin (d x))}{315 d}+\frac {2 (7 A+6 B+5 C) \tan (c)}{21 d}\right )}{(A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {7}{2}}(c+d x)} \]

input
Integrate[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + 
C*Sec[c + d*x]^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 + d*x]^4*Csc[c]*(-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 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + 
 d*x] + C*Sec[c + d*x]^2))/(15*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A 
*Cos[2*c + 2*d*x])) + (Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d 
*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^4*Csc[c]*(-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 + (d*x)/2]^4*(a + a*Sec[c + d* 
x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(5*d*E^(I*d*x)*(A + 2*C + 2 
*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (8*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 + d*x]^4*C 
sc[c]*(-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 + (d*x)/2 
]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(45*d* 
E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (2*A*Sqrt[C 
os[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + 
 d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*(A + 2*C + 2*B*Cos[ 
c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2)) + (4*B*Sqrt[Cos[c + d*x 
]]*EllipticF[(c + d*x)/2, 2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^...
 
3.6.41.3 Rubi [A] (verified)

Time = 1.46 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4576, 27, 3042, 4506, 27, 3042, 4485, 3042, 4274, 3042, 4255, 3042, 4258, 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.

\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4576

\(\displaystyle \frac {2 \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^2 (3 a (3 A+C)+a (9 B+4 C) \sec (c+d x))dx}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^2 (3 a (3 A+C)+a (9 B+4 C) \sec (c+d x))dx}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {2}{7} \int \frac {3}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a) \left ((21 A+9 B+11 C) a^2+(21 A+27 B+19 C) \sec (c+d x) a^2\right )dx+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3}{7} \int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a) \left ((21 A+9 B+11 C) a^2+(21 A+27 B+19 C) \sec (c+d x) a^2\right )dx+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((21 A+9 B+11 C) a^2+(21 A+27 B+19 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 4485

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \int \sec ^{\frac {3}{2}}(c+d x) \left (7 (12 A+9 B+8 C) a^3+15 (7 A+6 B+5 C) \sec (c+d x) a^3\right )dx+\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (7 (12 A+9 B+8 C) a^3+15 (7 A+6 B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (7 a^3 (12 A+9 B+8 C) \int \sec ^{\frac {3}{2}}(c+d x)dx+15 a^3 (7 A+6 B+5 C) \int \sec ^{\frac {5}{2}}(c+d x)dx\right )+\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (7 a^3 (12 A+9 B+8 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+15 a^3 (7 A+6 B+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )+\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+6 B+5 C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a^3 (12 A+9 B+8 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )\right )+\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+6 B+5 C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a^3 (12 A+9 B+8 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+6 B+5 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a^3 (12 A+9 B+8 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+6 B+5 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a^3 (12 A+9 B+8 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (15 a^3 (7 A+6 B+5 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a^3 (12 A+9 B+8 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2 a^3 (21 A+27 B+19 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2}{5} \left (15 a^3 (7 A+6 B+5 C) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+7 a^3 (12 A+9 B+8 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {2 (9 B+4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{7 d}}{9 a}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d}\)

input
Int[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),x]
 
output
(2*C*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(9*d) + ((2*( 
9*B + 4*C)*Sec[c + d*x]^(5/2)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(7*d) 
 + (3*((2*a^3*(21*A + 27*B + 19*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) 
+ (2*(7*a^3*(12*A + 9*B + 8*C)*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x) 
/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d) + 15 
*a^3*(7*A + 6*B + 5*C)*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sq 
rt[Sec[c + d*x]])/(3*d) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d))))/5)) 
/7)/(9*a)
 

3.6.41.3.1 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 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 4255
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4485
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ 
e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1)   Int[(d*Csc 
[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x 
], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[ 
n, -1]
 

rule 4506
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), 
 x] + Simp[1/(d*(m + n))   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] 
)^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - 
 a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]
 

rule 4576
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs 
c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Cs 
c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* 
B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m 
, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && 
NeQ[m + n + 1, 0]
 
3.6.41.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1155\) vs. \(2(311)=622\).

Time = 8.33 (sec) , antiderivative size = 1156, normalized size of antiderivative = 3.97

method result size
default \(\text {Expression too large to display}\) \(1156\)
parts \(\text {Expression too large to display}\) \(1425\)

input
int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, 
method=_RETURNVERBOSE)
 
output
-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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)^2-1)^(1/2))+2*C*(-1/144*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+s 
in(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*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)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2 
*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(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))-7/15*(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))-Ellipt 
icE(cos(1/2*d*x+1/2*c),2^(1/2))))+8*(1/2*A+1/4*B)*(-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)))+8*(1/4*B+1/2*C)*(-1/56*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)^4-5/42 
*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...
 
3.6.41.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.04 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (7 \, A + 6 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (7 \, A + 6 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (12 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (12 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (42 \, {\left (12 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (7 \, A + 6 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 7 \, {\left (9 \, A + 18 \, B + 16 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 45 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 35 \, C a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{315 \, d \cos \left (d x + c\right )^{4}} \]

input
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="fricas")
 
output
-2/315*(15*I*sqrt(2)*(7*A + 6*B + 5*C)*a^2*cos(d*x + c)^4*weierstrassPInve 
rse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*I*sqrt(2)*(7*A + 6*B + 5*C) 
*a^2*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + 
c)) + 21*I*sqrt(2)*(12*A + 9*B + 8*C)*a^2*cos(d*x + c)^4*weierstrassZeta(- 
4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sq 
rt(2)*(12*A + 9*B + 8*C)*a^2*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierst 
rassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (42*(12*A + 9*B + 8* 
C)*a^2*cos(d*x + c)^4 + 30*(7*A + 6*B + 5*C)*a^2*cos(d*x + c)^3 + 7*(9*A + 
 18*B + 16*C)*a^2*cos(d*x + c)^2 + 45*(B + 2*C)*a^2*cos(d*x + c) + 35*C*a^ 
2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^4)
 
3.6.41.6 Sympy [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+ 
c)**2),x)
 
output
Timed out
 
3.6.41.7 Maxima [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="maxima")
 
output
Timed out
 
3.6.41.8 Giac [F]

\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

input
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) 
^2),x, algorithm="giac")
 
output
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*s 
ec(d*x + c)^(3/2), x)
 
3.6.41.9 Mupad [F(-1)]

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

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