\(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [1126]

Optimal result

Integrand size = 35, antiderivative size = 242 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {(9 A+119 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+11 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{2 a^3 d}+\frac {(A+11 C) \sin (c+d x)}{2 a^3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(9 A+119 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac {2 C \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {(9 A+119 C) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

1/10*(9*A+119*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/2*(A+11*C)* 
InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d+1/2*(A+11*C)*sin(d*x+c)/a^3/d 
/cos(d*x+c)^(3/2)-1/10*(9*A+119*C)*sin(d*x+c)/a^3/d/cos(d*x+c)^(1/2)-1/5*( 
A+C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3-2/3*C*sin(d*x+c)/a/d 
/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2-1/30*(9*A+119*C)*sin(d*x+c)/d/cos(d*x 
+c)^(3/2)/(a^3+a^3*cos(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 = 8.90 (sec) , antiderivative size = 1269, normalized size of antiderivative = 5.24 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(-4*A*Cos[c/2 + (d*x)/2]^6*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]*(A + C*Sec[c + d*x]^2)*S 
ec[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]]]])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec 
[c + d*x])^3) - (44*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4 
, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + C*S 
ec[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 + S 
in[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[ 
c]^2]*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^ 
2)*((-4*(60*C + 9*A*Cos[c] + 59*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(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 
) - (16*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(3*A*Sin[(d*x)/2] + 8*C*Sin[(d*x)/2] 
))/(15*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(9*A*Sin[(d*x)/2] + 59*C*Sin[(d 
*x)/2]))/(5*d) + (32*C*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3*d) + (32*Sec[c]* 
Sec[c + d*x]*(C*Sin[c] - 9*C*Sin[d*x]))/(3*d) - (16*(3*A + 8*C)*Sec[c/2 + 
(d*x)/2]^2*Tan[c/2])/(15*d) - (4*(A + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5 
*d)))/(Sqrt[Cos[c + d*x]]*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d* 
x])^3) - (18*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(A +...
 

Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4602, 3042, 3521, 27, 3042, 3457, 3042, 3457, 27, 3042, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-1/5*((A + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3) 
+ ((-20*a*C*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) 
+ (-((a^2*(9*A + 119*C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + 
 d*x]))) + (3*(15*a^3*(A + 11*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2 
*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))) - a^3*(9*A + 119*C)*((-2*Elliptic 
E[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[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 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 3521

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[a*(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))   I 
nt[(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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 4602

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (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 + A*Cos 
[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, 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. \(875\) vs. \(2(221)=442\).

Time = 5.12 (sec) , antiderivative size = 876, normalized size of antiderivative = 3.62

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

Input:

int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(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 = 542, normalized size of antiderivative = 2.24 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/60*(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) - 20*C)*sqrt(cos(d* 
x + c))*sin(d*x + c) + 15*(sqrt(2)*(I*A + 11*I*C)*cos(d*x + c)^5 + 3*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 
 + sqrt(2)*(I*A + 11*I*C)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d 
*x + c) + I*sin(d*x + c)) + 15*(sqrt(2)*(-I*A - 11*I*C)*cos(d*x + c)^5 + 3 
*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 + sqrt(2)*(-I*A - 11*I*C)*cos(d*x + c)^2)*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)^5 + 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 + sqrt(2)*(-9*I*A - 119*I*C)*cos(d*x + c)^2)*wei 
erstrassZeta(-4, 0, 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)^5 + 3*sqrt(2)*(9*I*A + 11 
9*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(9*I*A + 119*I*C)*cos(d*x + c)^3 + sqrt( 
2)*(9*I*A + 119*I*C)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPIn 
verse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^3*d*cos(d*x + c)^5 + 3*a^ 
3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{4}}d x \right ) c}{a^{3}} \] Input:

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

Output:

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