\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [1231]

Optimal result

Integrand size = 43, antiderivative size = 250 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {(20 A-35 B+56 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {5 (A-2 B+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{15 a^2 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {5 (A-2 B+3 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(20 A-35 B+56 C) \sin (c+d x)}{5 a^2 d \sqrt {\cos (c+d x)}}-\frac {(A-2 B+3 C) \sin (c+d x)}{a^2 d \cos ^{\frac {5}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2} \] Output:

-1/5*(20*A-35*B+56*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d-5/3*(A-2 
*B+3*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/d+1/15*(20*A-35*B+56*C) 
*sin(d*x+c)/a^2/d/cos(d*x+c)^(5/2)-5/3*(A-2*B+3*C)*sin(d*x+c)/a^2/d/cos(d* 
x+c)^(3/2)+1/5*(20*A-35*B+56*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)-(A-2*B+3 
*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(5/2)/(1+cos(d*x+c))-1/3*(A-B+C)*sin(d*x+c 
)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^2
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.71 (sec) , antiderivative size = 1810, normalized size of antiderivative = 7.24 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(20*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S 
in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/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[-(Sq 
rt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - Arc 
Tan[Cot[c]]]])/(3*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])^2) - (40*B*Cos[c/2 + (d*x)/2]^4*Csc[c/ 
2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c 
/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]]]])/(3*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 
])^2) + (20*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, 
{5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/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]]]]*S 
qrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d 
*x - ArcTan[Cot[c]]]])/(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])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[ 
Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(10*A - 20*B + 3 
6*C + 10*A*Cos[c] - 15*B*Cos[c] + 20*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/( 
5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2...
 

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.92, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 4600, 3042, 3520, 27, 3042, 3457, 3042, 3227, 3042, 3116, 3042, 3116, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{7/2} (a \sec (c+d x)+a)^2}dx\)

\(\Big \downarrow \) 4600

\(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^2}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3520

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (5 A-5 B+11 C)-a (A-7 B+7 C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (5 A-5 B+11 C)-a (A-7 B+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {a^2 (20 A-35 B+56 C)-15 a^2 (A-2 B+3 C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}dx}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 (20 A-35 B+56 C)-15 a^2 (A-2 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {a^2 (20 A-35 B+56 C) \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x)}dx-15 a^2 (A-2 B+3 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (20 A-35 B+56 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx-15 a^2 (A-2 B+3 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {a^2 (20 A-35 B+56 C) \left (\frac {3}{5} \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-15 a^2 (A-2 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (20 A-35 B+56 C) \left (\frac {3}{5} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-15 a^2 (A-2 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {a^2 (20 A-35 B+56 C) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-15 a^2 (A-2 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (20 A-35 B+56 C) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-15 a^2 (A-2 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {a^2 (20 A-35 B+56 C) \left (\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )-15 a^2 (A-2 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {a^2 (20 A-35 B+56 C) \left (\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )-15 a^2 (A-2 B+3 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{a^2}-\frac {6 (A-2 B+3 C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

Input:

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

Output:

-1/3*((A - B + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x]) 
^2) + ((-6*(A - 2*B + 3*C)*Sin[c + d*x])/(d*Cos[c + d*x]^(5/2)*(1 + Cos[c 
+ d*x])) + (-15*a^2*(A - 2*B + 3*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + 
 (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))) + a^2*(20*A - 35*B + 56*C)*((2 
*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (3*((-2*EllipticE[(c + d*x)/2, 2 
])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/5))/a^2)/(6*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 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1044\) vs. \(2(231)=462\).

Time = 6.05 (sec) , antiderivative size = 1045, normalized size of antiderivative = 4.18

method	result	size
default	\(\text {Expression too large to display}\)	\(1045\)

Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^2,x, 
method=_RETURNVERBOSE)
 

Output:

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*((4*B-8 
*C)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2 
)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2* 
cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 
2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(-2*A+4*B-6*C)*(cos(1/2*d* 
x+1/2*c)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(El 
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(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)/cos(1/2*d*x+1/2*c)/(-2*sin(1 
/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+1/3*(-A+B-C)*(2*(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*EllipticF(cos(1/2*d*x+1 
/2*c),2^(1/2))-3*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)-2*(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*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d 
*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^6+20*sin(1/2* 
d*x+1/2*c)^4-7*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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(sin(1/2*d*x+1/2*c)^2-1)+(4*A-8*B+12*C) 
/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-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*cos(1/2*d*x+1/2*c)-(2* 
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1 
/2*d*x+1/2*c)^2)^(1/2))+4/5*C/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.99 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c) 
)^2,x, algorithm="fricas")
 

Output:

1/30*(2*(3*(20*A - 35*B + 56*C)*cos(d*x + c)^4 + (95*A - 160*B + 261*C)*co 
s(d*x + c)^3 + 2*(15*A - 20*B + 37*C)*cos(d*x + c)^2 + 2*(5*B - 4*C)*cos(d 
*x + c) + 6*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 25*(sqrt(2)*(-I*A + 2*I*B 
 - 3*I*C)*cos(d*x + c)^5 + 2*sqrt(2)*(-I*A + 2*I*B - 3*I*C)*cos(d*x + c)^4 
 + sqrt(2)*(-I*A + 2*I*B - 3*I*C)*cos(d*x + c)^3)*weierstrassPInverse(-4, 
0, cos(d*x + c) + I*sin(d*x + c)) - 25*(sqrt(2)*(I*A - 2*I*B + 3*I*C)*cos( 
d*x + c)^5 + 2*sqrt(2)*(I*A - 2*I*B + 3*I*C)*cos(d*x + c)^4 + sqrt(2)*(I*A 
 - 2*I*B + 3*I*C)*cos(d*x + c)^3)*weierstrassPInverse(-4, 0, cos(d*x + c) 
- I*sin(d*x + c)) - 3*(sqrt(2)*(20*I*A - 35*I*B + 56*I*C)*cos(d*x + c)^5 + 
 2*sqrt(2)*(20*I*A - 35*I*B + 56*I*C)*cos(d*x + c)^4 + sqrt(2)*(20*I*A - 3 
5*I*B + 56*I*C)*cos(d*x + c)^3)*weierstrassZeta(-4, 0, weierstrassPInverse 
(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-20*I*A + 35*I*B - 5 
6*I*C)*cos(d*x + c)^5 + 2*sqrt(2)*(-20*I*A + 35*I*B - 56*I*C)*cos(d*x + c) 
^4 + sqrt(2)*(-20*I*A + 35*I*B - 56*I*C)*cos(d*x + c)^3)*weierstrassZeta(- 
4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*c 
os(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c) 
)^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c) 
)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{4}}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{4}}d x \right ) b}{a^{2}} \] Input:

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

Output:

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