\(\int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [248]

Optimal result

Integrand size = 35, antiderivative size = 249 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=-\frac {(119 A+9 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(11 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{2 a^3 d}+\frac {(11 A+C) \sin (c+d x)}{2 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3}-\frac {2 A \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac {(119 A+9 C) \sin (c+d x)}{30 d \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )} \] Output:

-1/10*(119*A+9*C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*s 
ec(d*x+c)^(1/2)/a^3/d+1/2*(11*A+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+1/2*(11*A+C)*sin(d*x+c)/a^3/d/sec( 
d*x+c)^(1/2)-1/5*(A+C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3-2/ 
3*A*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^2-1/30*(119*A+9*C)*si 
n(d*x+c)/d/sec(d*x+c)^(1/2)/(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 = 9.25 (sec) , antiderivative size = 608, normalized size of antiderivative = 2.44 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=-\frac {2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (-238 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-18 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-\frac {\left ((1961 A+156 C) \cos \left (\frac {1}{2} (c-d x)\right )+(1609 A+114 C) \cos \left (\frac {1}{2} (3 c+d x)\right )+1165 A \cos \left (\frac {1}{2} (c+3 d x)\right )+90 C \cos \left (\frac {1}{2} (c+3 d x)\right )+620 A \cos \left (\frac {1}{2} (5 c+3 d x)\right )+45 C \cos \left (\frac {1}{2} (5 c+3 d x)\right )+292 A \cos \left (\frac {1}{2} (3 c+5 d x)\right )+27 C \cos \left (\frac {1}{2} (3 c+5 d x)\right )+65 A \cos \left (\frac {1}{2} (7 c+5 d x)\right )+5 A \cos \left (\frac {1}{2} (5 c+7 d x)\right )-5 A \cos \left (\frac {1}{2} (9 c+7 d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right )}{8 \sqrt {\sec (c+d x)}}-660 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-60 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{15 a^3 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^3} \] Input:

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

Output:

(-2*Cos[(c + d*x)/2]^6*((-238*Sqrt[2]*A*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) - (18*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))*Hypergeometri 
c2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) - (((1961*A + 156*C) 
*Cos[(c - d*x)/2] + (1609*A + 114*C)*Cos[(3*c + d*x)/2] + 1165*A*Cos[(c + 
3*d*x)/2] + 90*C*Cos[(c + 3*d*x)/2] + 620*A*Cos[(5*c + 3*d*x)/2] + 45*C*Co 
s[(5*c + 3*d*x)/2] + 292*A*Cos[(3*c + 5*d*x)/2] + 27*C*Cos[(3*c + 5*d*x)/2 
] + 65*A*Cos[(7*c + 5*d*x)/2] + 5*A*Cos[(5*c + 7*d*x)/2] - 5*A*Cos[(9*c + 
7*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2]^5)/(8*Sqrt[Sec[c + d*x]]) - 
660*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 60 
*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])*Sec[c 
+ d*x]*(A + C*Sec[c + d*x]^2))/(15*a^3*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 
 + Sec[c + d*x])^3)
 

Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.05, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-1/5*((A + C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3) 
+ ((-20*a*A*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2) 
+ (-((a^2*(119*A + 9*C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + 
 d*x]))) + (3*((-2*a^3*(119*A + 9*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x 
)/2, 2]*Sqrt[Sec[c + d*x]])/d + 15*a^3*(11*A + C)*((2*Sqrt[Cos[c + d*x]]*E 
llipticF[(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]
 

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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(464\) vs. \(2(224)=448\).

Time = 4.31 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.87

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 (160 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+468 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+330 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 )+714 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 )+108 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+30 C \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 )+54 C \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 )-1058 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-198 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+474 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+114 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-47 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-27 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A +3 C \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}\)	\(465\)

Input:

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

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)*(160*A*c 
os(1/2*d*x+1/2*c)^10+468*A*cos(1/2*d*x+1/2*c)^8+330*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))+714*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))+108*C*cos(1/2*d*x+1/2*c)^8+30*C*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))+54*C*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*c 
os(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1058*A* 
cos(1/2*d*x+1/2*c)^6-198*C*cos(1/2*d*x+1/2*c)^6+474*A*cos(1/2*d*x+1/2*c)^4 
+114*C*cos(1/2*d*x+1/2*c)^4-47*A*cos(1/2*d*x+1/2*c)^2-27*C*cos(1/2*d*x+1/2 
*c)^2+3*A+3*C)/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 = 0.11 (sec) , antiderivative size = 487, normalized size of antiderivative = 1.96 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/60*(15*(sqrt(2)*(11*I*A + I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(11*I*A + I*C 
)*cos(d*x + c)^2 + 3*sqrt(2)*(11*I*A + I*C)*cos(d*x + c) + sqrt(2)*(11*I*A 
 + I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*(s 
qrt(2)*(-11*I*A - I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-11*I*A - I*C)*cos(d*x 
+ c)^2 + 3*sqrt(2)*(-11*I*A - I*C)*cos(d*x + c) + sqrt(2)*(-11*I*A - I*C)) 
*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(1 
19*I*A + 9*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(119*I*A + 9*I*C)*cos(d*x + c)^ 
2 + 3*sqrt(2)*(119*I*A + 9*I*C)*cos(d*x + c) + sqrt(2)*(119*I*A + 9*I*C))* 
weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
 + c))) + 3*(sqrt(2)*(-119*I*A - 9*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-119*I 
*A - 9*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-119*I*A - 9*I*C)*cos(d*x + c) + s 
qrt(2)*(-119*I*A - 9*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
0, cos(d*x + c) - I*sin(d*x + c))) - 2*(20*A*cos(d*x + c)^4 + 3*(79*A + 9* 
C)*cos(d*x + c)^3 + 4*(94*A + 9*C)*cos(d*x + c)^2 + 15*(11*A + C)*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. \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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