\(\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [727]

Optimal result

Integrand size = 33, antiderivative size = 454 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (10 a^4 C-21 b^4 (9 A+7 C)-3 a^2 b^2 (161 A+93 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3 C+21 b^3 (9 A+7 C)+15 a^2 b (21 A+11 C)-6 a b^2 (28 A+19 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}+\frac {4 a \left (84 A b^2-5 a^2 C+57 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac {4 a C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 C (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d} \] Output:

2/315*(a-b)*(a+b)^(1/2)*(10*a^4*C-21*b^4*(9*A+7*C)-3*a^2*b^2*(161*A+93*C)) 
*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/ 
2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^3/d+2 
/315*(a-b)*(a+b)^(1/2)*(10*a^3*C+21*b^3*(9*A+7*C)+15*a^2*b*(21*A+11*C)-6*a 
*b^2*(28*A+19*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2), 
((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a- 
b))^(1/2)/b^2/d+4/315*a*(84*A*b^2-5*C*a^2+57*C*b^2)*(a+b*sec(d*x+c))^(1/2) 
*tan(d*x+c)/b/d-2/315*(10*C*a^2-7*b^2*(9*A+7*C))*(a+b*sec(d*x+c))^(3/2)*ta 
n(d*x+c)/b/d-4/63*a*C*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d+2/9*C*(a+b*sec 
(d*x+c))^(7/2)*tan(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 19.94 (sec) , antiderivative size = 710, normalized size of antiderivative = 1.56 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \sqrt {2} \sqrt {\frac {\cos (c+d x)}{(1+\cos (c+d x))^2}} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \left ((a+b) \left (\left (10 a^4 C-21 b^4 (9 A+7 C)-3 a^2 b^2 (161 A+93 C)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+b \left (-10 a^3 C+21 b^3 (9 A+7 C)+15 a^2 b (21 A+11 C)+6 a b^2 (28 A+19 C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )\right ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}} \sec (c+d x)+\left (10 a^4 C-21 b^4 (9 A+7 C)-3 a^2 b^2 (161 A+93 C)\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{315 b^2 d \sqrt {\frac {1}{1+\cos (c+d x)}} (b+a \cos (c+d x))^3 (A+2 C+A \cos (2 c+2 d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \sec ^{\frac {9}{2}}(c+d x)}+\frac {\cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (483 a^2 A b^2+189 A b^4-10 a^4 C+279 a^2 b^2 C+147 b^4 C\right ) \sin (c+d x)}{315 b^2}+\frac {4}{315} \sec ^2(c+d x) \left (63 A b^2 \sin (c+d x)+75 a^2 C \sin (c+d x)+49 b^2 C \sin (c+d x)\right )+\frac {4 \sec (c+d x) \left (231 a A b^2 \sin (c+d x)+5 a^3 C \sin (c+d x)+163 a b^2 C \sin (c+d x)\right )}{315 b}+\frac {76}{63} a b C \sec ^2(c+d x) \tan (c+d x)+\frac {4}{9} b^2 C \sec ^3(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x))} \] Input:

Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x 
]
 

Output:

(4*Sqrt[2]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])^2]*Sqrt[Cos[c + d*x]*Sec[( 
c + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(a + b*Sec[c + d*x] 
)^(5/2)*(A + C*Sec[c + d*x]^2)*((a + b)*((10*a^4*C - 21*b^4*(9*A + 7*C) - 
3*a^2*b^2*(161*A + 93*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + 
 b)] + b*(-10*a^3*C + 21*b^3*(9*A + 7*C) + 15*a^2*b*(21*A + 11*C) + 6*a*b^ 
2*(28*A + 19*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*(Co 
s[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d 
*x)/2]^2)/(a + b)]*Sec[c + d*x] + (10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b 
^2*(161*A + 93*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Ta 
n[(c + d*x)/2]))/(315*b^2*d*Sqrt[(1 + Cos[c + d*x])^(-1)]*(b + a*Cos[c + d 
*x])^3*(A + 2*C + A*Cos[2*c + 2*d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Sec[c + d 
*x]^(9/2)) + (Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x 
]^2)*((4*(483*a^2*A*b^2 + 189*A*b^4 - 10*a^4*C + 279*a^2*b^2*C + 147*b^4*C 
)*Sin[c + d*x])/(315*b^2) + (4*Sec[c + d*x]^2*(63*A*b^2*Sin[c + d*x] + 75* 
a^2*C*Sin[c + d*x] + 49*b^2*C*Sin[c + d*x]))/315 + (4*Sec[c + d*x]*(231*a* 
A*b^2*Sin[c + d*x] + 5*a^3*C*Sin[c + d*x] + 163*a*b^2*C*Sin[c + d*x]))/(31 
5*b) + (76*a*b*C*Sec[c + d*x]^2*Tan[c + d*x])/63 + (4*b^2*C*Sec[c + d*x]^3 
*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + A*Cos[2*c + 2*d*x] 
))
 

Rubi [A] (verified)

Time = 2.02 (sec) , antiderivative size = 464, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 4571, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4493, 3042, 4319, 4492}

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.

\(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4571

\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} (b (9 A+7 C)-2 a C \sec (c+d x))dx}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (b (9 A+7 C)-2 a C \sec (c+d x))dx}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (b (9 A+7 C)-2 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {2}{7} \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 a b (21 A+13 C)-\left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \sec (c+d x)\right )dx-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 a b (21 A+13 C)-\left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \sec (c+d x)\right )dx-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 a b (21 A+13 C)+\left (7 b^2 (9 A+7 C)-10 a^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{7} \left (\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (5 (21 A+11 C) a^2+7 b^2 (9 A+7 C)\right )+2 a \left (-5 C a^2+84 A b^2+57 b^2 C\right ) \sec (c+d x)\right )dx-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (5 (21 A+11 C) a^2+7 b^2 (9 A+7 C)\right )+2 a \left (-5 C a^2+84 A b^2+57 b^2 C\right ) \sec (c+d x)\right )dx-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (5 (21 A+11 C) a^2+7 b^2 (9 A+7 C)\right )+2 a \left (-5 C a^2+84 A b^2+57 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (a b \left (5 (63 A+31 C) a^2+3 b^2 (119 A+87 C)\right )-\left (10 C a^4-3 b^2 (161 A+93 C) a^2-21 b^4 (9 A+7 C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {4 a \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (a b \left (5 (63 A+31 C) a^2+3 b^2 (119 A+87 C)\right )-\left (10 C a^4-3 b^2 (161 A+93 C) a^2-21 b^4 (9 A+7 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {4 a \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b \left (5 (63 A+31 C) a^2+3 b^2 (119 A+87 C)\right )+\left (-10 C a^4+3 b^2 (161 A+93 C) a^2+21 b^4 (9 A+7 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {4 a \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (10 a^3 C+15 a^2 b (21 A+11 C)-6 a b^2 (28 A+19 C)+21 b^3 (9 A+7 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\left (10 a^4 C-3 a^2 b^2 (161 A+93 C)-21 b^4 (9 A+7 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {4 a \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (10 a^3 C+15 a^2 b (21 A+11 C)-6 a b^2 (28 A+19 C)+21 b^3 (9 A+7 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (10 a^4 C-3 a^2 b^2 (161 A+93 C)-21 b^4 (9 A+7 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {4 a \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (10 a^3 C+15 a^2 b (21 A+11 C)-6 a b^2 (28 A+19 C)+21 b^3 (9 A+7 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\left (10 a^4 C-3 a^2 b^2 (161 A+93 C)-21 b^4 (9 A+7 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {4 a \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {4 a \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (10 a^4 C-3 a^2 b^2 (161 A+93 C)-21 b^4 (9 A+7 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3 C+15 a^2 b (21 A+11 C)-6 a b^2 (28 A+19 C)+21 b^3 (9 A+7 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )\right )-\frac {2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

Input:

Int[Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]
 

Output:

(2*C*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d) + ((-4*a*C*(a + b*Se 
c[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((-2*(10*a^2*C - 7*b^2*(9*A + 7*C) 
)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (3*(((2*(a - b)*Sqrt[a 
+ b]*(10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Cot[c + d* 
x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b) 
]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - 
b))])/(b^2*d) + (2*(a - b)*Sqrt[a + b]*(10*a^3*C + 21*b^3*(9*A + 7*C) + 15 
*a^2*b*(21*A + 11*C) - 6*a*b^2*(28*A + 19*C))*Cot[c + d*x]*EllipticF[ArcSi 
n[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec 
[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/3 + ( 
4*a*(84*A*b^2 - 5*a^2*C + 57*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x]) 
/(3*d)))/5)/7)/(9*b)
 

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 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4490

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

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4493

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B)   Int[Csc[e 
 + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B   Int[Csc[e + f*x]*((1 + 
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B} 
, x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
 

rule 4571

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[ 
(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x] 
*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   In 
t[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2357\) vs. \(2(416)=832\).

Time = 116.82 (sec) , antiderivative size = 2358, normalized size of antiderivative = 5.19

method	result	size
default	\(\text {Expression too large to display}\)	\(2358\)
parts	\(\text {Expression too large to display}\)	\(2363\)

Input:

int(sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x,method=_RETURNV 
ERBOSE)
 

Output:

2/315/d/b^2*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*cos(d*x+ 
c)+b)*((279*cos(d*x+c)^2+80*cos(d*x+c)+80)*C*a^3*b^2*tan(d*x+c)+21*(9*cos( 
d*x+c)^2+14*cos(d*x+c)+14)*a*A*b^4*tan(d*x+c)+21*sin(d*x+c)*(34+11*cos(d*x 
+c))*a^2*A*b^3+5*sin(d*x+c)*(cos(d*x+c)-1)*a^4*b*C+7*(21*cos(d*x+c)^4+7*co 
s(d*x+c)^3+7*cos(d*x+c)^2+5*cos(d*x+c)+5)*C*b^5*tan(d*x+c)*sec(d*x+c)^3+(1 
63*cos(d*x+c)^3+442*cos(d*x+c)^2+170*cos(d*x+c)+170)*C*a^2*b^3*tan(d*x+c)* 
sec(d*x+c)+(147*cos(d*x+c)^4+212*cos(d*x+c)^3+212*cos(d*x+c)^2+130*cos(d*x 
+c)+130)*C*a*b^4*tan(d*x+c)*sec(d*x+c)^2+10*(-cos(d*x+c)^2-2*cos(d*x+c)-1) 
*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c) 
+1))^(1/2)*a^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+189*( 
cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)* 
(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^5*EllipticE(-csc(d*x+c)+cot(d*x+c 
),((a-b)/(a+b))^(1/2))+147*(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^5*Ell 
ipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+189*(-cos(d*x+c)^2-2*co 
s(d*x+c)-1)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/ 
(cos(d*x+c)+1))^(1/2)*b^5*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^( 
1/2))+147*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/ 
2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^5*EllipticF(-csc(d*x+ 
c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+483*A*a^3*b^2*cos(d*x+c)*sin(d*x+c)+...
 

Fricas [F]

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

integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorith 
m="fricas")
 

Output:

integral((C*b^2*sec(d*x + c)^5 + 2*C*a*b*sec(d*x + c)^4 + 2*A*a*b*sec(d*x 
+ c)^2 + A*a^2*sec(d*x + c) + (C*a^2 + A*b^2)*sec(d*x + c)^3)*sqrt(b*sec(d 
*x + c) + a), x)
 

Sympy [F]

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

integrate(sec(d*x+c)*(a+b*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)
 

Output:

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**(5/2)*sec(c + d*x), 
 x)
 

Maxima [F(-1)]

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

integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorith 
m="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c) + a)^(5/2)*sec(d*x + c), 
x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{5}d x \right ) b^{2} c +2 \left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{4}d x \right ) a b c +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{3}d x \right ) a^{2} c +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{3}d x \right ) a \,b^{2}+2 \left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{2}d x \right ) a^{2} b +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )d x \right ) a^{3} \] Input:

int(sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)
 

Output:

int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**5,x)*b**2*c + 2*int(sqrt(sec(c 
+ d*x)*b + a)*sec(c + d*x)**4,x)*a*b*c + int(sqrt(sec(c + d*x)*b + a)*sec( 
c + d*x)**3,x)*a**2*c + int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**3,x)*a* 
b**2 + 2*int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**2,x)*a**2*b + int(sqrt 
(sec(c + d*x)*b + a)*sec(c + d*x),x)*a**3