\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx\) [1244]

Optimal result

Integrand size = 43, antiderivative size = 276 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx=\frac {(A+8 B-57 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(4 A+17 B-108 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{42 a^4 d}-\frac {(A+8 B-57 C) \sin (c+d x)}{10 a^4 d \sqrt {\cos (c+d x)}}+\frac {(13 A+29 B-141 C) \sin (c+d x)}{210 a^4 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))^2}+\frac {(4 A+17 B-108 C) \sin (c+d x)}{42 a^4 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \] Output:

1/10*(A+8*B-57*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d+1/42*(4*A+17 
*B-108*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^4/d-1/10*(A+8*B-57*C)*s 
in(d*x+c)/a^4/d/cos(d*x+c)^(1/2)+1/210*(13*A+29*B-141*C)*sin(d*x+c)/a^4/d/ 
cos(d*x+c)^(1/2)/(1+cos(d*x+c))^2+1/42*(4*A+17*B-108*C)*sin(d*x+c)/a^4/d/c 
os(d*x+c)^(1/2)/(1+cos(d*x+c))-1/7*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/( 
a+a*cos(d*x+c))^4+1/35*(3*A+4*B-11*C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(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 = 12.35 (sec) , antiderivative size = 1962, normalized size of antiderivative = 7.11 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + 
 a*Sec[c + d*x])^4),x]
 

Output:

(-32*A*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]*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*B*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[-(Sq 
rt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - Arc 
Tan[Cot[c]]]])/(21*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqr 
t[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^4) + (288*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 - Arc 
Tan[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) + (Cos[c/2 + (d*x)/2]^8*(A + B*Sec[c + d*x] + C*Sec 
[c + d*x]^2)*((8*(20*C - A*Cos[c] - 8*B*Cos[c] + 37*C*Cos[c])*Csc[c/2]*Sec 
[c/2]*Sec[c])/(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] ...
 

Rubi [A] (verified)

Time = 1.90 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.04, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.442, Rules used = {3042, 4600, 3042, 3520, 27, 3042, 3457, 3042, 3457, 27, 3042, 3457, 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.

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{7/2} (a \sec (c+d x)+a)^4}dx\)

\(\Big \downarrow \) 4600

\(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^4}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\)

\(\Big \downarrow \) 3520

\(\displaystyle \frac {\int \frac {a (A-B+15 C)+7 a (A+B-C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (A-B+15 C)+7 a (A+B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {(2 A-9 B+86 C) a^2+5 (3 A+4 B-11 C) \cos (c+d x) a^2}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {(2 A-9 B+86 C) a^2+5 (3 A+4 B-11 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\frac {\int -\frac {a^3 (A+83 B-657 C)-3 a^3 (13 A+29 B-141 C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}+\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\int \frac {a^3 (A+83 B-657 C)-3 a^3 (13 A+29 B-141 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\int \frac {a^3 (A+83 B-657 C)-3 a^3 (13 A+29 B-141 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\frac {\int \frac {21 a^4 (A+8 B-57 C)-5 a^4 (4 A+17 B-108 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {10 a^3 (4 A+17 B-108 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\frac {\int \frac {21 a^4 (A+8 B-57 C)-5 a^4 (4 A+17 B-108 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}-\frac {10 a^3 (4 A+17 B-108 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\frac {21 a^4 (A+8 B-57 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx-5 a^4 (4 A+17 B-108 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}-\frac {10 a^3 (4 A+17 B-108 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\frac {21 a^4 (A+8 B-57 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx-5 a^4 (4 A+17 B-108 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {10 a^3 (4 A+17 B-108 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\frac {21 a^4 (A+8 B-57 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )-5 a^4 (4 A+17 B-108 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {10 a^3 (4 A+17 B-108 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\frac {21 a^4 (A+8 B-57 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-5 a^4 (4 A+17 B-108 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {10 a^3 (4 A+17 B-108 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\frac {21 a^4 (A+8 B-57 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-5 a^4 (4 A+17 B-108 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {10 a^3 (4 A+17 B-108 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {(13 A+29 B-141 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}-\frac {\frac {21 a^4 (A+8 B-57 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {10 a^4 (4 A+17 B-108 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}-\frac {10 a^3 (4 A+17 B-108 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{6 a^2}}{5 a^2}+\frac {2 a (3 A+4 B-11 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}}{14 a^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4}\)

Input:

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Sec 
[c + d*x])^4),x]
 

Output:

-1/7*((A - B + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x]) 
^4) + ((2*a*(3*A + 4*B - 11*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]*(a + 
a*Cos[c + d*x])^3) + (((13*A + 29*B - 141*C)*Sin[c + d*x])/(3*d*Sqrt[Cos[c 
 + d*x]]*(1 + Cos[c + d*x])^2) - ((-10*a^3*(4*A + 17*B - 108*C)*Sin[c + d* 
x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])) + ((-10*a^4*(4*A + 17*B - 
108*C)*EllipticF[(c + d*x)/2, 2])/d + 21*a^4*(A + 8*B - 57*C)*((-2*Ellipti 
cE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/a^2)/(6* 
a^2))/(5*a^2))/(14*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. \(1016\) vs. \(2(255)=510\).

Time = 6.12 (sec) , antiderivative size = 1017, normalized size of antiderivative = 3.68

method	result	size
default	\(\text {Expression too large to display}\)	\(1017\)

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))^4,x, 
method=_RETURNVERBOSE)
 

Output:

1/840*(4*(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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(20*A*EllipticF(cos(1/2* 
d*x+1/2*c),2^(1/2))-21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+85*B*Ellipt 
icF(cos(1/2*d*x+1/2*c),2^(1/2))-168*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2) 
)-540*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1197*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)^6-12*(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)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(20*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-21 
*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+85*B*EllipticF(cos(1/2*d*x+1/2*c) 
,2^(1/2))-168*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-540*C*EllipticF(cos( 
1/2*d*x+1/2*c),2^(1/2))+1197*C*EllipticE(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)+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*si 
n(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)^2-1)^ 
(1/2)*(20*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-21*A*EllipticE(cos(1/2*d 
*x+1/2*c),2^(1/2))+85*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-168*B*Ellipt 
icE(cos(1/2*d*x+1/2*c),2^(1/2))-540*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) 
)+1197*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)-4*(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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(20*A*EllipticF(c 
os(1/2*d*x+1/2*c),2^(1/2))-21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+8...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 700, normalized size of antiderivative = 2.54 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c) 
)^4,x, algorithm="fricas")
 

Output:

-1/420*(2*(21*(A + 8*B - 57*C)*cos(d*x + c)^4 + (64*A + 587*B - 4248*C)*co 
s(d*x + c)^3 + (53*A + 724*B - 5421*C)*cos(d*x + c)^2 - 5*(4*A - 67*B + 56 
4*C)*cos(d*x + c) - 420*C)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(4 
*I*A + 17*I*B - 108*I*C)*cos(d*x + c)^5 + 4*sqrt(2)*(4*I*A + 17*I*B - 108* 
I*C)*cos(d*x + c)^4 + 6*sqrt(2)*(4*I*A + 17*I*B - 108*I*C)*cos(d*x + c)^3 
+ 4*sqrt(2)*(4*I*A + 17*I*B - 108*I*C)*cos(d*x + c)^2 + sqrt(2)*(4*I*A + 1 
7*I*B - 108*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I 
*sin(d*x + c)) + 5*(sqrt(2)*(-4*I*A - 17*I*B + 108*I*C)*cos(d*x + c)^5 + 4 
*sqrt(2)*(-4*I*A - 17*I*B + 108*I*C)*cos(d*x + c)^4 + 6*sqrt(2)*(-4*I*A - 
17*I*B + 108*I*C)*cos(d*x + c)^3 + 4*sqrt(2)*(-4*I*A - 17*I*B + 108*I*C)*c 
os(d*x + c)^2 + sqrt(2)*(-4*I*A - 17*I*B + 108*I*C)*cos(d*x + c))*weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*(sqrt(2)*(-I*A - 8* 
I*B + 57*I*C)*cos(d*x + c)^5 + 4*sqrt(2)*(-I*A - 8*I*B + 57*I*C)*cos(d*x + 
 c)^4 + 6*sqrt(2)*(-I*A - 8*I*B + 57*I*C)*cos(d*x + c)^3 + 4*sqrt(2)*(-I*A 
 - 8*I*B + 57*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A - 8*I*B + 57*I*C)*cos(d* 
x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I 
*sin(d*x + c))) + 21*(sqrt(2)*(I*A + 8*I*B - 57*I*C)*cos(d*x + c)^5 + 4*sq 
rt(2)*(I*A + 8*I*B - 57*I*C)*cos(d*x + c)^4 + 6*sqrt(2)*(I*A + 8*I*B - 57* 
I*C)*cos(d*x + c)^3 + 4*sqrt(2)*(I*A + 8*I*B - 57*I*C)*cos(d*x + c)^2 + sq 
rt(2)*(I*A + 8*I*B - 57*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weier...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx=\text {Timed out} \] 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))**4,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c) 
)^4,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

Mupad [F(-1)]

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

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(7/2)*(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)^(7/2)*(a + a/cos 
(c + d*x))^4), x)
 

Reduce [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{4}}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{4}}d x \right ) b}{a^{4}} \] 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))^4,x)
 

Output:

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