\(\int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [243]

Optimal result

Integrand size = 35, antiderivative size = 282 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(9 A+119 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(A+11 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{2 a^3 d}-\frac {(9 A+119 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}+\frac {(A+11 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A+C) \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 C \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {(9 A+119 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:

1/10*(9*A+119*C)*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*(A+11*C)*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-1/10*(9*A+119*C)*sec(d*x+c)^(1/2)*s 
in(d*x+c)/a^3/d+1/2*(A+11*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^3/d-1/5*(A+C)*s 
ec(d*x+c)^(9/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-2/3*C*sec(d*x+c)^(7/2)*sin 
(d*x+c)/a/d/(a+a*sec(d*x+c))^2-1/30*(9*A+119*C)*sec(d*x+c)^(5/2)*sin(d*x+c 
)/d/(a^3+a^3*sec(d*x+c))
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.80 (sec) , antiderivative size = 984, normalized size of antiderivative = 3.49 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

Integrate[(Sec[c + d*x]^(7/2)*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]) 
^3,x]
 

Output:

(-6*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]*(A + C*Sec[c + d*x]^2))/(5* 
d*E^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) - (238* 
Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I 
)*(c + d*x))]*Cos[c/2 + (d*x)/2]^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]*(A + C*Sec[c + d*x]^2))/(15*d*E 
^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (4*A*Cos 
[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Se 
c[c/2]*Sec[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + A*C 
os[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (44*C*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]^(3 
/2)*(A + C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + 
a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*Sec[c + d*x]^(3/2)*(A + C*Sec[c 
 + d*x]^2)*((-4*(9*A + 119*C)*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) + (4*Sec[c 
/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) + (8*Sec 
[c/2]*Sec[c/2 + (d*x)/2]^3*(3*A*Sin[(d*x)/2] + 13*C*Sin[(d*x)/2]))/(15*d) 
+ (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(3*A*Sin[(d*x)/2] + 29*C*Sin[(d*x)/2])...
 

Rubi [A] (verified)

Time = 1.64 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 4573, 27, 3042, 4507, 25, 3042, 4507, 27, 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.

\(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\)

\(\Big \downarrow \) 4573

\(\displaystyle -\frac {\int -\frac {\sec ^{\frac {7}{2}}(c+d x) (a (3 A-7 C)+a (3 A+13 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^{\frac {7}{2}}(c+d x) (a (3 A-7 C)+a (3 A+13 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int -\frac {\sec ^{\frac {5}{2}}(c+d x) \left (50 a^2 C-3 a^2 (3 A+23 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (50 a^2 C-3 a^2 (3 A+23 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (50 a^2 C-3 a^2 (3 A+23 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {-\frac {\frac {\int \frac {3}{2} \sec ^{\frac {3}{2}}(c+d x) \left (a^3 (9 A+119 C)-15 a^3 (A+11 C) \sec (c+d x)\right )dx}{a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {3 \int \sec ^{\frac {3}{2}}(c+d x) \left (a^3 (9 A+119 C)-15 a^3 (A+11 C) \sec (c+d x)\right )dx}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a^3 (9 A+119 C)-15 a^3 (A+11 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \int \sec ^{\frac {3}{2}}(c+d x)dx-15 a^3 (A+11 C) \int \sec ^{\frac {5}{2}}(c+d x)dx\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-15 a^3 (A+11 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )-15 a^3 (A+11 C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-15 a^3 (A+11 C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )-15 a^3 (A+11 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-15 a^3 (A+11 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-15 a^3 (A+11 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {-\frac {\frac {a^2 (9 A+119 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {3 \left (a^3 (9 A+119 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-15 a^3 (A+11 C) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )\right )}{2 a^2}}{3 a^2}-\frac {20 a C \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

Input:

Int[(Sec[c + d*x]^(7/2)*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]
 

Output:

-1/5*((A + C)*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^3) 
+ ((-20*a*C*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) 
- ((a^2*(9*A + 119*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d 
*x])) + (3*(a^3*(9*A + 119*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*(A + 11*C)*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c 
 + d*x]])/(3*d) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d))))/(2*a^2))/(3 
*a^2))/(10*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 4507

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

rule 4573

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(875\) vs. \(2(253)=506\).

Time = 57.04 (sec) , antiderivative size = 876, normalized size of antiderivative = 3.11

method	result	size
default	\(\text {Expression too large to display}\)	\(876\)

Input:

int(sec(d*x+c)^(7/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,method=_RETUR 
NVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/60*(12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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)*(5*A*EllipticF(cos(1/2*d 
*x+1/2*c),2^(1/2))-9*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+55*C*Elliptic 
F(cos(1/2*d*x+1/2*c),2^(1/2))-119*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))) 
*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-30*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/ 
2)*(sin(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)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*A*EllipticE(cos(1/ 
2*d*x+1/2*c),2^(1/2))+55*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-119*C*Ell 
ipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c 
)+24*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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)*(5*A*EllipticF(cos(1/2*d*x+1 
/2*c),2^(1/2))-9*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+55*C*EllipticF(co 
s(1/2*d*x+1/2*c),2^(1/2))-119*C*EllipticE(cos(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)-6*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(s 
in(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)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*A*EllipticE(cos(1/2*d*x 
+1/2*c),2^(1/2))+55*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-119*C*Elliptic 
E(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-24*(-2*sin(1/2*d*x+1/2*c 
)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(9*A+119*C)*sin(1/2*d*x+1/2*c)^10+24*(-2*s 
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(29*A+389*C)*sin(1/2*d*...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 532, normalized size of antiderivative = 1.89 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(7/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algori 
thm="fricas")
 

Output:

-1/60*(15*(sqrt(2)*(I*A + 11*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(I*A + 11*I*C 
)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 11*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A 
+ 11*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d* 
x + c)) + 15*(sqrt(2)*(-I*A - 11*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-I*A - 1 
1*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - 11*I*C)*cos(d*x + c)^2 + sqrt(2) 
*(-I*A - 11*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I 
*sin(d*x + c)) + 3*(sqrt(2)*(-9*I*A - 119*I*C)*cos(d*x + c)^4 + 3*sqrt(2)* 
(-9*I*A - 119*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-9*I*A - 119*I*C)*cos(d*x + 
 c)^2 + sqrt(2)*(-9*I*A - 119*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, we 
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(9*I 
*A + 119*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(9*I*A + 119*I*C)*cos(d*x + c)^3 
+ 3*sqrt(2)*(9*I*A + 119*I*C)*cos(d*x + c)^2 + sqrt(2)*(9*I*A + 119*I*C)*c 
os(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c 
) - I*sin(d*x + c))) + 2*(3*(9*A + 119*C)*cos(d*x + c)^4 + 6*(11*A + 151*C 
)*cos(d*x + c)^3 + 5*(9*A + 139*C)*cos(d*x + c)^2 + 120*C*cos(d*x + c) - 2 
0*C)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos( 
d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**(7/2)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)^(7/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate(sec(d*x+c)^(7/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algori 
thm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^(7/2)/(a*sec(d*x + c) + a)^3 
, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \] Input:

int(((A + C/cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2))/(a + a/cos(c + d*x))^3 
,x)
 

Output:

int(((A + C/cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2))/(a + a/cos(c + d*x))^3 
, x)
 

Reduce [F]

\[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) a}{a^{3}} \] Input:

int(sec(d*x+c)^(7/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)
 

Output:

(int((sqrt(sec(c + d*x))*sec(c + d*x)**5)/(sec(c + d*x)**3 + 3*sec(c + d*x 
)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x))*sec(c + d*x)**3) 
/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*a)/a**3