\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [1238]

Optimal result

Integrand size = 43, antiderivative size = 268 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {(9 A-49 B+119 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A-13 B+33 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(3 A-13 B+33 C) \sin (c+d x)}{6 a^3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(9 A-49 B+119 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac {(B-2 C) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {(9 A-49 B+119 C) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

1/10*(9*A-49*B+119*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(3*A 
-13*B+33*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d+1/6*(3*A-13*B+33* 
C)*sin(d*x+c)/a^3/d/cos(d*x+c)^(3/2)-1/10*(9*A-49*B+119*C)*sin(d*x+c)/a^3/ 
d/cos(d*x+c)^(1/2)-1/5*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+ 
c))^3+1/3*(B-2*C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2-1/30* 
(9*A-49*B+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 = 9.34 (sec) , antiderivative size = 1894, normalized size of antiderivative = 7.07 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] 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])^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 + 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]]]])/(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])^3) + (52*B*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 + 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 + Co 
t[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])^3) - (44*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Hyperg 
eometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c 
 + d*x]*(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]]]])/(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])^3) + (Cos[c/2 + (d*x)/2]^6*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(( 
-4*(-20*B + 60*C + 9*A*Cos[c] - 29*B*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] - B...
 

Rubi [A] (verified)

Time = 1.56 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4600, 3042, 3520, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4600

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {3 (3 A-8 B+23 C) a^2+25 (B-2 C) \cos (c+d x) a^2}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {3 (3 A-8 B+23 C) a^2+25 (B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\frac {\int \frac {3 \left (5 a^3 (3 A-13 B+33 C)-a^3 (9 A-49 B+119 C) \cos (c+d x)\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (3 A-13 B+33 C)-a^3 (9 A-49 B+119 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (3 A-13 B+33 C)-a^3 (9 A-49 B+119 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx-a^3 (9 A-49 B+119 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-a^3 (9 A-49 B+119 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (9 A-49 B+119 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (9 A-49 B+119 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (9 A-49 B+119 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (9 A-49 B+119 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

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])^3),x]
 

Output:

-1/5*((A - B + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x]) 
^3) + ((10*a*(B - 2*C)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c 
+ d*x])^2) + (-((a^2*(9*A - 49*B + 119*C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3 
/2)*(a + a*Cos[c + d*x]))) + (3*(5*a^3*(3*A - 13*B + 33*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 - 49*B + 119*C)*((-2*EllipticE[(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 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. \(1012\) vs. \(2(247)=494\).

Time = 5.36 (sec) , antiderivative size = 1013, normalized size of antiderivative = 3.78

method	result	size
default	\(\text {Expression too large to display}\)	\(1013\)

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))^3,x, 
method=_RETURNVERBOSE)
 

Output:

-1/4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^3*((-4*B+ 
12*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)*(EllipticF(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*(-2*B+4*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*Elli 
pticF(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)+(8*B-24*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))+(A-B+C)*(1/5*(-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)^5+4/5*(-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)^3+18/5*(-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)-8/5*(sin(1/2*d* 
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.25 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] 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) 
)^3,x, algorithm="fricas")
 

Output:

-1/60*(2*(3*(9*A - 49*B + 119*C)*cos(d*x + c)^4 + 2*(33*A - 188*B + 453*C) 
*cos(d*x + c)^3 + 5*(9*A - 59*B + 139*C)*cos(d*x + c)^2 - 60*(B - 2*C)*cos 
(d*x + c) - 20*C)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(3*I*A - 13 
*I*B + 33*I*C)*cos(d*x + c)^5 + 3*sqrt(2)*(3*I*A - 13*I*B + 33*I*C)*cos(d* 
x + c)^4 + 3*sqrt(2)*(3*I*A - 13*I*B + 33*I*C)*cos(d*x + c)^3 + sqrt(2)*(3 
*I*A - 13*I*B + 33*I*C)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x 
 + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-3*I*A + 13*I*B - 33*I*C)*cos(d*x + 
c)^5 + 3*sqrt(2)*(-3*I*A + 13*I*B - 33*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-3 
*I*A + 13*I*B - 33*I*C)*cos(d*x + c)^3 + sqrt(2)*(-3*I*A + 13*I*B - 33*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 + 49*I*B - 119*I*C)*cos(d*x + c)^5 + 3*sqrt(2)*(-9* 
I*A + 49*I*B - 119*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-9*I*A + 49*I*B - 119* 
I*C)*cos(d*x + c)^3 + sqrt(2)*(-9*I*A + 49*I*B - 119*I*C)*cos(d*x + c)^2)* 
weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
 + c))) + 3*(sqrt(2)*(9*I*A - 49*I*B + 119*I*C)*cos(d*x + c)^5 + 3*sqrt(2) 
*(9*I*A - 49*I*B + 119*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(9*I*A - 49*I*B + 1 
19*I*C)*cos(d*x + c)^3 + sqrt(2)*(9*I*A - 49*I*B + 119*I*C)*cos(d*x + c)^2 
)*weierstrassZeta(-4, 0, weierstrassPInverse(-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. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:

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

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

Output:

Timed out
 

Giac [F]

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

Output:

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

Mupad [F(-1)]

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

Reduce [F]

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