\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx\) [1243]

Optimal result

Integrand size = 43, antiderivative size = 234 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx=\frac {(B+8 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(3 A+4 B+17 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{42 a^4 d}+\frac {(15 A-B-83 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac {(B+8 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 a^4 d (1+\cos (c+d x))}-\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(5 A+2 B-9 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3} \] Output:

1/10*(B+8*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d+1/42*(3*A+4*B+17* 
C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^4/d+1/210*(15*A-B-83*C)*cos(d* 
x+c)^(1/2)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2-1/10*(B+8*C)*cos(d*x+c)^(1/2) 
*sin(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A-B+C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d 
/(a+a*cos(d*x+c))^4+1/35*(5*A+2*B-9*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+ 
a*cos(d*x+c))^3
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 12.74 (sec) , antiderivative size = 1626, normalized size of antiderivative = 6.95 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + 
 a*Sec[c + d*x])^4),x]
 

Output:

(-8*A*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S 
in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + 
C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[ 
c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 
+ Sin[d*x - ArcTan[Cot[c]]]])/(7*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c 
 + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^4) - (32*B*Cos[c/2 + (d 
*x)/2]^8*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Co 
t[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)* 
Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[ 
1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan 
[Cot[c]]]])/(21*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 
 + Cot[c]^2]*(a + a*Sec[c + d*x])^4) - (136*C*Cos[c/2 + (d*x)/2]^8*Csc[c/2 
]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/ 
2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan 
[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Si 
n[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21* 
d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a 
+ a*Sec[c + d*x])^4) + (Cos[c/2 + (d*x)/2]^8*(A + B*Sec[c + d*x] + C*Sec[c 
 + d*x]^2)*((-16*(B + 8*C)*Csc[c])/(5*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^ 
3*(15*A*Sin[(d*x)/2] - B*Sin[(d*x)/2] - 83*C*Sin[(d*x)/2]))/(105*d) + (...
 

Rubi [A] (verified)

Time = 1.66 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.09, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4600, 3042, 3520, 27, 3042, 3457, 25, 3042, 3457, 27, 3042, 3457, 3042, 3227, 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.

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{5/2} (a \sec (c+d x)+a)^4}dx\)

\(\Big \downarrow \) 4600

\(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\)

\(\Big \downarrow \) 3520

\(\displaystyle \frac {\int -\frac {a (A-B-13 C)-a (9 A+5 B-5 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {a (A-B-13 C)-a (9 A+5 B-5 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^3}dx}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {a (A-B-13 C)-a (9 A+5 B-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {\frac {\int -\frac {7 (B+8 C) a^2+3 (5 A+2 B-9 C) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {-\frac {\int \frac {7 (B+8 C) a^2+3 (5 A+2 B-9 C) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {7 (B+8 C) a^2+3 (5 A+2 B-9 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {-\frac {\frac {\int \frac {(15 A+41 B+253 C) a^3+(15 A-B-83 C) \cos (c+d x) a^3}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{3 a^2}+\frac {(15 A-B-83 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\frac {\int \frac {(15 A+41 B+253 C) a^3+(15 A-B-83 C) \cos (c+d x) a^3}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{6 a^2}+\frac {(15 A-B-83 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\int \frac {(15 A+41 B+253 C) a^3+(15 A-B-83 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}+\frac {(15 A-B-83 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {-\frac {\frac {\frac {\int \frac {5 (3 A+4 B+17 C) a^4+21 (B+8 C) \cos (c+d x) a^4}{\sqrt {\cos (c+d x)}}dx}{a^2}-\frac {42 a^3 (B+8 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{6 a^2}+\frac {(15 A-B-83 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {\int \frac {5 (3 A+4 B+17 C) a^4+21 (B+8 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {42 a^3 (B+8 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{6 a^2}+\frac {(15 A-B-83 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3227

\(\displaystyle -\frac {-\frac {\frac {\frac {5 a^4 (3 A+4 B+17 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+21 a^4 (B+8 C) \int \sqrt {\cos (c+d x)}dx}{a^2}-\frac {42 a^3 (B+8 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{6 a^2}+\frac {(15 A-B-83 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {5 a^4 (3 A+4 B+17 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+21 a^4 (B+8 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {42 a^3 (B+8 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{6 a^2}+\frac {(15 A-B-83 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {-\frac {\frac {\frac {5 a^4 (3 A+4 B+17 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {42 a^4 (B+8 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}-\frac {42 a^3 (B+8 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{6 a^2}+\frac {(15 A-B-83 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {-\frac {\frac {\frac {\frac {10 a^4 (3 A+4 B+17 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {42 a^4 (B+8 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}-\frac {42 a^3 (B+8 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{6 a^2}+\frac {(15 A-B-83 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {2 a (5 A+2 B-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4}\)

Input:

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec 
[c + d*x])^4),x]
 

Output:

-1/7*((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) 
^4) - ((-2*a*(5*A + 2*B - 9*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(5*d*(a + 
a*Cos[c + d*x])^3) - (((15*A - B - 83*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/ 
(3*d*(1 + Cos[c + d*x])^2) + (((42*a^4*(B + 8*C)*EllipticE[(c + d*x)/2, 2] 
)/d + (10*a^4*(3*A + 4*B + 17*C)*EllipticF[(c + d*x)/2, 2])/d)/a^2 - (42*a 
^3*(B + 8*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])))/(6 
*a^2))/(5*a^2))/(14*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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3457

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3520

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* 
Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x 
] + Simp[1/(b*(b*c - a*d)*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c 
+ d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a 
*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c 
*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c 
^2 - d^2, 0] && LtQ[m, -2^(-1)]
 

rule 4600

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) 
*(x_)]^2), x_Symbol] :> Simp[d^(m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[ 
e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(594\) vs. \(2(217)=434\).

Time = 6.39 (sec) , antiderivative size = 595, normalized size of antiderivative = 2.54

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 (60 A \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-168 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+80 B \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-84 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} \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 )-1344 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+340 C \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-672 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} \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 )+60 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+248 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+1684 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-30 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-54 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-282 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-90 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-8 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-34 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+75 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-33 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-15 A +15 B -15 C \right )}{840 a^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} \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}\)	\(595\)

Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^4,x, 
method=_RETURNVERBOSE)
 

Output:

-1/840*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(60*A*(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))*cos(1/2*d*x+1/2*c)^7-168*B*cos(1/2*d*x+1/2*c)^10+80* 
B*(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))*cos(1/2*d*x+1/2*c)^7-84*B*cos(1/2*d*x+1/2*c)^ 
7*(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))-1344*C*cos(1/2*d*x+1/2*c)^10+340*C*(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))*cos(1/2*d*x+1/2*c)^7-672*C*cos(1/2*d*x+1/2*c)^7*(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))+60*A*cos(1/2*d*x+1/2*c)^8+248*B*cos(1/2*d*x+1/2*c)^8+168 
4*C*cos(1/2*d*x+1/2*c)^8-30*A*cos(1/2*d*x+1/2*c)^6-54*B*cos(1/2*d*x+1/2*c) 
^6-282*C*cos(1/2*d*x+1/2*c)^6-90*A*cos(1/2*d*x+1/2*c)^4-8*B*cos(1/2*d*x+1/ 
2*c)^4-34*C*cos(1/2*d*x+1/2*c)^4+75*A*cos(1/2*d*x+1/2*c)^2-33*B*cos(1/2*d* 
x+1/2*c)^2-9*C*cos(1/2*d*x+1/2*c)^2-15*A+15*B-15*C)/a^4/cos(1/2*d*x+1/2*c) 
^7/(-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 = 615, normalized size of antiderivative = 2.63 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) 
)^4,x, algorithm="fricas")
 

Output:

-1/420*(2*(21*(B + 8*C)*cos(d*x + c)^3 - (15*A - 64*B - 587*C)*cos(d*x + c 
)^2 - (60*A - 53*B - 724*C)*cos(d*x + c) - 15*A - 20*B + 335*C)*sqrt(cos(d 
*x + c))*sin(d*x + c) + 5*(sqrt(2)*(3*I*A + 4*I*B + 17*I*C)*cos(d*x + c)^4 
 + 4*sqrt(2)*(3*I*A + 4*I*B + 17*I*C)*cos(d*x + c)^3 + 6*sqrt(2)*(3*I*A + 
4*I*B + 17*I*C)*cos(d*x + c)^2 + 4*sqrt(2)*(3*I*A + 4*I*B + 17*I*C)*cos(d* 
x + c) + sqrt(2)*(3*I*A + 4*I*B + 17*I*C))*weierstrassPInverse(-4, 0, cos( 
d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-3*I*A - 4*I*B - 17*I*C)*cos(d*x 
+ c)^4 + 4*sqrt(2)*(-3*I*A - 4*I*B - 17*I*C)*cos(d*x + c)^3 + 6*sqrt(2)*(- 
3*I*A - 4*I*B - 17*I*C)*cos(d*x + c)^2 + 4*sqrt(2)*(-3*I*A - 4*I*B - 17*I* 
C)*cos(d*x + c) + sqrt(2)*(-3*I*A - 4*I*B - 17*I*C))*weierstrassPInverse(- 
4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*(sqrt(2)*(-I*B - 8*I*C)*cos(d*x 
+ c)^4 + 4*sqrt(2)*(-I*B - 8*I*C)*cos(d*x + c)^3 + 6*sqrt(2)*(-I*B - 8*I*C 
)*cos(d*x + c)^2 + 4*sqrt(2)*(-I*B - 8*I*C)*cos(d*x + c) + sqrt(2)*(-I*B - 
 8*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + 
I*sin(d*x + c))) + 21*(sqrt(2)*(I*B + 8*I*C)*cos(d*x + c)^4 + 4*sqrt(2)*(I 
*B + 8*I*C)*cos(d*x + c)^3 + 6*sqrt(2)*(I*B + 8*I*C)*cos(d*x + c)^2 + 4*sq 
rt(2)*(I*B + 8*I*C)*cos(d*x + c) + sqrt(2)*(I*B + 8*I*C))*weierstrassZeta( 
-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^4*d* 
cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d 
*cos(d*x + c) + a^4*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*sec(d*x+ 
c))**4,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) 
)^4,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) 
)^4,x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^4* 
cos(d*x + c)^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^4} \,d x \] Input:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos 
(c + d*x))^4),x)
 

Output:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos 
(c + d*x))^4), x)
 

Reduce [F]

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

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^4,x)
 

Output:

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