\(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [223]

Optimal result

Integrand size = 33, antiderivative size = 294 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {7 (33 A-17 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(21 A-11 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{2 a^3 d}+\frac {7 (33 A-17 B) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(21 A-11 B) \sin (c+d x)}{2 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {(12 A-7 B) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {3 (21 A-11 B) \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )} \] Output:

7/10*(33*A-17*B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*se 
c(d*x+c)^(1/2)/a^3/d-1/2*(21*A-11*B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2* 
d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/a^3/d+7/30*(33*A-17*B)*sin(d*x+c)/a^3/ 
d/sec(d*x+c)^(3/2)-1/2*(21*A-11*B)*sin(d*x+c)/a^3/d/sec(d*x+c)^(1/2)-1/5*( 
A-B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3-1/15*(12*A-7*B)*sin( 
d*x+c)/a/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2-3/10*(21*A-11*B)*sin(d*x+c) 
/d/sec(d*x+c)^(3/2)/(a^3+a^3*sec(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 10.30 (sec) , antiderivative size = 1032, normalized size of antiderivative = 3.51 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3) 
,x]
 

Output:

(-77*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]^6*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]*Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(5 
*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (119*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/2 + (d*x)/2]^6*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]*Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(15*d*E^(I*d*x)* 
(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) - (42*A*Cos[c/2 + (d*x)/2]^6* 
Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x 
]^(5/2)*(A + B*Sec[c + d*x])*Sin[c])/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c 
+ d*x])^3) + (22*B*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*Ellipt 
icF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*Sin[c 
])/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6 
*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*(((-329*A + 178*B - 133*A*Cos[2*c 
] + 60*B*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) + (8*(-3*A + B)*Cos[2 
*d*x]*Sin[2*c])/(3*d) + (4*A*Cos[3*d*x]*Sin[3*c])/(5*d) - (2*Sec[c/2]*Sec[ 
c/2 + (d*x)/2]^5*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2] 
*Sec[c/2 + (d*x)/2]^3*(-27*A*Sin[(d*x)/2] + 22*B*Sin[(d*x)/2]))/(15*d) ...
 

Rubi [A] (verified)

Time = 1.69 (sec) , antiderivative size = 299, 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, 4508, 27, 3042, 4508, 3042, 4508, 27, 3042, 4274, 3042, 4256, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {5 a (3 A-B)-9 a (A-B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {5 a (3 A-B)-9 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {5 a^2 (21 A-10 B)-7 a^2 (12 A-7 B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {5 a^2 (21 A-10 B)-7 a^2 (12 A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\frac {\int \frac {5 \left (7 a^3 (33 A-17 B)-9 a^3 (21 A-11 B) \sec (c+d x)\right )}{2 \sec ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {5 \int \frac {7 a^3 (33 A-17 B)-9 a^3 (21 A-11 B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 \int \frac {7 a^3 (33 A-17 B)-9 a^3 (21 A-11 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {\frac {5 \left (7 a^3 (33 A-17 B) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)}dx-9 a^3 (21 A-11 B) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx\right )}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 \left (7 a^3 (33 A-17 B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-9 a^3 (21 A-11 B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {\frac {\frac {5 \left (7 a^3 (33 A-17 B) \left (\frac {3}{5} \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-9 a^3 (21 A-11 B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )\right )}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 \left (7 a^3 (33 A-17 B) \left (\frac {3}{5} \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-9 a^3 (21 A-11 B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )\right )}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {\frac {5 \left (7 a^3 (33 A-17 B) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-9 a^3 (21 A-11 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )\right )}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 \left (7 a^3 (33 A-17 B) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-9 a^3 (21 A-11 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )\right )}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {5 \left (7 a^3 (33 A-17 B) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-9 a^3 (21 A-11 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )\right )}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {5 \left (7 a^3 (33 A-17 B) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-9 a^3 (21 A-11 B) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )\right )}{2 a^2}-\frac {9 a^2 (21 A-11 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (12 A-7 B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}\)

Input:

Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3),x]
 

Output:

-1/5*((A - B)*Sin[c + d*x])/(d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^3) 
+ ((-2*a*(12*A - 7*B)*Sin[c + d*x])/(3*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + 
 d*x])^2) + ((-9*a^2*(21*A - 11*B)*Sin[c + d*x])/(d*Sec[c + d*x]^(3/2)*(a 
+ a*Sec[c + d*x])) + (5*(7*a^3*(33*A - 17*B)*((6*Sqrt[Cos[c + d*x]]*Ellipt 
icE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*Sin[c + d*x])/(5*d*Sec[ 
c + d*x]^(3/2))) - 9*a^3*(21*A - 11*B)*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c 
 + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sin[c + d*x])/(3*d*Sqrt[Sec[c 
 + d*x]]))))/(2*a^2))/(3*a^2))/(10*a^2)
 

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 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 4508

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

Maple [A] (verified)

Time = 6.96 (sec) , antiderivative size = 493, normalized size of antiderivative = 1.68

method	result	size
default	\(-\frac {\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}}\, \left (192 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-864 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+160 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-228 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-630 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-1386 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+468 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+330 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+714 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+1590 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-1058 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-744 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+474 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+57 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-47 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 A +3 B \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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}\)	\(493\)

Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x,method=_RETURNV 
ERBOSE)
 

Output:

-1/60/a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(192*A*c 
os(1/2*d*x+1/2*c)^12-864*A*cos(1/2*d*x+1/2*c)^10+160*B*cos(1/2*d*x+1/2*c)^ 
10-228*A*cos(1/2*d*x+1/2*c)^8-630*A*cos(1/2*d*x+1/2*c)^5*(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))-1386*A*cos(1/2*d*x+1/2*c)^5*(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))+468*B*cos 
(1/2*d*x+1/2*c)^8+330*B*cos(1/2*d*x+1/2*c)^5*(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))+71 
4*B*cos(1/2*d*x+1/2*c)^5*(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))+1590*A*cos(1/2*d*x+1/2 
*c)^6-1058*B*cos(1/2*d*x+1/2*c)^6-744*A*cos(1/2*d*x+1/2*c)^4+474*B*cos(1/2 
*d*x+1/2*c)^4+57*A*cos(1/2*d*x+1/2*c)^2-47*B*cos(1/2*d*x+1/2*c)^2-3*A+3*B) 
/cos(1/2*d*x+1/2*c)^5/(-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
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 506, normalized size of antiderivative = 1.72 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorith 
m="fricas")
 

Output:

-1/60*(15*(sqrt(2)*(-21*I*A + 11*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-21*I*A 
+ 11*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-21*I*A + 11*I*B)*cos(d*x + c) + sqr 
t(2)*(-21*I*A + 11*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d 
*x + c)) + 15*(sqrt(2)*(21*I*A - 11*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(21*I* 
A - 11*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(21*I*A - 11*I*B)*cos(d*x + c) + sq 
rt(2)*(21*I*A - 11*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d 
*x + c)) + 21*(sqrt(2)*(-33*I*A + 17*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-33* 
I*A + 17*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-33*I*A + 17*I*B)*cos(d*x + c) + 
 sqrt(2)*(-33*I*A + 17*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4 
, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*(sqrt(2)*(33*I*A - 17*I*B)*cos(d 
*x + c)^3 + 3*sqrt(2)*(33*I*A - 17*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(33*I*A 
 - 17*I*B)*cos(d*x + c) + sqrt(2)*(33*I*A - 17*I*B))*weierstrassZeta(-4, 0 
, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(12*A*cos 
(d*x + c)^5 - 4*(6*A - 5*B)*cos(d*x + c)^4 - 3*(147*A - 79*B)*cos(d*x + c) 
^3 - 2*(357*A - 188*B)*cos(d*x + c)^2 - 15*(21*A - 11*B)*cos(d*x + c))*sin 
(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c) 
^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorith 
m="giac")
 

Output:

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^3*sec(d*x + c)^(5/2)) 
, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(5/2)),x 
)
 

Output:

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(5/2)), 
x)
 

Reduce [F]

\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{6}+3 \sec \left (d x +c \right )^{5}+3 \sec \left (d x +c \right )^{4}+\sec \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{5}+3 \sec \left (d x +c \right )^{4}+3 \sec \left (d x +c \right )^{3}+\sec \left (d x +c \right )^{2}}d x \right ) b}{a^{3}} \] Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x)
 

Output:

(int(sqrt(sec(c + d*x))/(sec(c + d*x)**6 + 3*sec(c + d*x)**5 + 3*sec(c + d 
*x)**4 + sec(c + d*x)**3),x)*a + int(sqrt(sec(c + d*x))/(sec(c + d*x)**5 + 
 3*sec(c + d*x)**4 + 3*sec(c + d*x)**3 + sec(c + d*x)**2),x)*b)/a**3