\(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [244]

Optimal result

Integrand size = 35, antiderivative size = 249 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(A-49 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(A-13 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A-49 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 (A-4 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(A-13 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:

1/10*(A-49*C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d 
*x+c)^(1/2)/a^3/d+1/6*(A-13*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*(A-49*C)*sec(d*x+c)^(1/2)*sin(d*x 
+c)/a^3/d-1/5*(A+C)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3+2/15* 
(A-4*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2+1/6*(A-13*C)*se 
c(d*x+c)^(3/2)*sin(d*x+c)/d/(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 = 6.82 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.92 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-2 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+98 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+20 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-260 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-2 \sqrt {\sec (c+d x)} \left (6 (A-49 C) \cos (d x) \csc (c)-\frac {1}{4} (11 A-239 C+4 (7 A-73 C) \cos (c+d x)+5 (A-13 C) \cos (2 (c+d x))) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{15 a^3 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^3} \] Input:

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

Output:

(2*Cos[(c + d*x)/2]^6*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((-2*Sqrt[2]*A*S 
qrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x) 
)]*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))]))/E^(I*d*x) + 
(98*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^( 
(2*I)*(c + d*x))]*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))] 
))/E^(I*d*x) + 20*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[ 
c + d*x]] - 260*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c 
+ d*x]] - 2*Sqrt[Sec[c + d*x]]*(6*(A - 49*C)*Cos[d*x]*Csc[c] - ((11*A - 23 
9*C + 4*(7*A - 73*C)*Cos[c + d*x] + 5*(A - 13*C)*Cos[2*(c + d*x)])*Sec[(c 
+ d*x)/2]^4*Tan[(c + d*x)/2])/4)))/(15*a^3*d*(A + 2*C + A*Cos[2*(c + d*x)] 
)*(1 + Sec[c + d*x])^3)
 

Rubi [A] (verified)

Time = 1.57 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4573, 27, 3042, 4507, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

\(\displaystyle -\frac {\int -\frac {\sec ^{\frac {5}{2}}(c+d x) (5 a (A-C)+a (A+11 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (6 (A-4 C) a^2+(A+41 C) \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (6 (A-4 C) a^2+(A+41 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\frac {\int \frac {1}{2} \sqrt {\sec (c+d x)} \left (5 a^3 (A-13 C)-3 a^3 (A-49 C) \sec (c+d x)\right )dx}{a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \sqrt {\sec (c+d x)} \left (5 a^3 (A-13 C)-3 a^3 (A-49 C) \sec (c+d x)\right )dx}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 a^3 (A-13 C)-3 a^3 (A-49 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {\frac {5 a^3 (A-13 C) \int \sqrt {\sec (c+d x)}dx-3 a^3 (A-49 C) \int \sec ^{\frac {3}{2}}(c+d x)dx}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 a^3 (A-13 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a^3 (A-49 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {\frac {5 a^3 (A-13 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 a^3 (A-13 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {\frac {5 a^3 (A-13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 a^3 (A-13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {5 a^3 (A-13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {10 a^3 (A-13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

Input:

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

Output:

-1/5*((A + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^3) 
+ ((4*a*(A - 4*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x 
])^2) + ((5*a^2*(A - 13*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[ 
c + d*x])) + ((10*a^3*(A - 13*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 
 2]*Sqrt[Sec[c + d*x]])/d - 3*a^3*(A - 49*C)*((-2*Sqrt[Cos[c + d*x]]*Ellip 
ticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + 
 d*x])/d))/(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 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. \(678\) vs. \(2(224)=448\).

Time = 55.93 (sec) , antiderivative size = 679, normalized size of antiderivative = 2.73

method	result	size
default	\(\text {Expression too large to display}\)	\(679\)

Input:

int(sec(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,method=_RETUR 
NVERBOSE)
 

Output:

1/60*(-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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(5*A*EllipticF(cos(1/2*d 
*x+1/2*c),2^(1/2))-3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*C*Elliptic 
F(cos(1/2*d*x+1/2*c),2^(1/2))+147*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)^4+4*(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)*(2*sin(1/2*d*x+1/2*c)^2 
-1)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*EllipticE(cos(1/2 
*d*x+1/2*c),2^(1/2))-65*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*C*Elli 
pticE(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) 
-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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2 
*c),2^(1/2))-3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*C*EllipticF(cos( 
1/2*d*x+1/2*c),2^(1/2))+147*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1 
/2*d*x+1/2*c)+12*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A-4 
9*C)*sin(1/2*d*x+1/2*c)^8-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2) 
^(1/2)*(13*A-817*C)*sin(1/2*d*x+1/2*c)^6+12*(-2*sin(1/2*d*x+1/2*c)^4+sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(A-124*C)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2* 
c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A-439*C)*sin(1/2*d*x+1/2*c)^2)/a^3/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 = 0.10 (sec) , antiderivative size = 475, normalized size of antiderivative = 1.91 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algori 
thm="fricas")
 

Output:

-1/60*(5*(sqrt(2)*(I*A - 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A - 13*I*C) 
*cos(d*x + c)^2 + 3*sqrt(2)*(I*A - 13*I*C)*cos(d*x + c) + sqrt(2)*(I*A - 1 
3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqr 
t(2)*(-I*A + 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A + 13*I*C)*cos(d*x + 
c)^2 + 3*sqrt(2)*(-I*A + 13*I*C)*cos(d*x + c) + sqrt(2)*(-I*A + 13*I*C))*w 
eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-I* 
A + 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A + 49*I*C)*cos(d*x + c)^2 + 3* 
sqrt(2)*(-I*A + 49*I*C)*cos(d*x + c) + sqrt(2)*(-I*A + 49*I*C))*weierstras 
sZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 
3*(sqrt(2)*(I*A - 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A - 49*I*C)*cos(d* 
x + c)^2 + 3*sqrt(2)*(I*A - 49*I*C)*cos(d*x + c) + sqrt(2)*(I*A - 49*I*C)) 
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* 
x + c))) + 2*(3*(A - 49*C)*cos(d*x + c)^3 + 4*(A - 94*C)*cos(d*x + c)^2 - 
5*(A + 59*C)*cos(d*x + c) - 60*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. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

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

integrate(sec(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) a}{a^{3}} \] Input:

int(sec(d*x+c)^(5/2)*(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)**4)/(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)**2) 
/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*a)/a**3