\(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [527]

Optimal result

Integrand size = 43, antiderivative size = 277 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(75 A-163 B+283 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(465 A-985 B+1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(45 A-85 B+157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(195 A-475 B+787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d} \] Output:

-1/32*(75*A-163*B+283*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d* 
x+c))^(1/2))*2^(1/2)/a^(5/2)/d-1/4*(A-B+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a* 
sec(d*x+c))^(5/2)-1/16*(5*A-13*B+21*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*se 
c(d*x+c))^(3/2)+1/120*(465*A-985*B+1729*C)*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c 
))^(1/2)+1/80*(45*A-85*B+157*C)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(a+a*sec(d*x 
+c))^(1/2)-1/240*(195*A-475*B+787*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^3 
/d
 

Mathematica [A] (warning: unable to verify)

Time = 13.17 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-15 \sqrt {2} (75 A-163 B+283 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{(1+\cos (c+d x))^2}} \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}+\frac {1}{16} (4125 A-7685 B+15053 C+10 (765 A-1381 B+2605 C) \cos (c+d x)+108 (45 A-85 B+157 C) \cos (2 (c+d x))+2550 A \cos (3 (c+d x))-5030 B \cos (3 (c+d x))+9110 C \cos (3 (c+d x))+735 A \cos (4 (c+d x))-1495 B \cos (4 (c+d x))+2671 C \cos (4 (c+d x))) \sec ^3(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{30 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

(Cos[(c + d*x)/2]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-15*Sqrt[2]*( 
75*A - 163*B + 283*C)*ArcSin[Tan[(c + d*x)/2]]*Cos[(c + d*x)/2]^4*Sqrt[Cos 
[c + d*x]/(1 + Cos[c + d*x])^2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]] 
+ ((4125*A - 7685*B + 15053*C + 10*(765*A - 1381*B + 2605*C)*Cos[c + d*x] 
+ 108*(45*A - 85*B + 157*C)*Cos[2*(c + d*x)] + 2550*A*Cos[3*(c + d*x)] - 5 
030*B*Cos[3*(c + d*x)] + 9110*C*Cos[3*(c + d*x)] + 735*A*Cos[4*(c + d*x)] 
- 1495*B*Cos[4*(c + d*x)] + 2671*C*Cos[4*(c + d*x)])*Sec[c + d*x]^3*Tan[(c 
 + d*x)/2])/16))/(30*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*( 
a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.87 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.09, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4572, 27, 3042, 4507, 27, 3042, 4509, 27, 3042, 4498, 27, 3042, 4489, 3042, 4282, 216}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4572

\(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a (B-C)+a (5 A-5 B+13 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a (B-C)+a (5 A-5 B+13 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (8 a (B-C)+a (5 A-5 B+13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int -\frac {\sec ^3(c+d x) \left (6 a^2 (5 A-13 B+21 C)-a^2 (45 A-85 B+157 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\sec ^3(c+d x) \left (6 a^2 (5 A-13 B+21 C)-a^2 (45 A-85 B+157 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (6 a^2 (5 A-13 B+21 C)-a^2 (45 A-85 B+157 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4509

\(\displaystyle \frac {-\frac {\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^3 (45 A-85 B+157 C)-a^3 (195 A-475 B+787 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^3 (45 A-85 B+157 C)-a^3 (195 A-475 B+787 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^3 (45 A-85 B+157 C)-a^3 (195 A-475 B+787 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4498

\(\displaystyle \frac {-\frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^4 (195 A-475 B+787 C)-2 a^4 (465 A-985 B+1729 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (195 A-475 B+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^4 (195 A-475 B+787 C)-2 a^4 (465 A-985 B+1729 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (195 A-475 B+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^4 (195 A-475 B+787 C)-2 a^4 (465 A-985 B+1729 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a^2 (195 A-475 B+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {-\frac {-\frac {-\frac {15 a^4 (75 A-163 B+283 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^4 (465 A-985 B+1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (195 A-475 B+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {-\frac {15 a^4 (75 A-163 B+283 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^4 (465 A-985 B+1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (195 A-475 B+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {-\frac {-\frac {-\frac {-\frac {30 a^4 (75 A-163 B+283 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^4 (465 A-985 B+1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (195 A-475 B+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {-\frac {2 a^2 (45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {-\frac {2 a^2 (195 A-475 B+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}-\frac {\frac {15 \sqrt {2} a^{7/2} (75 A-163 B+283 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^4 (465 A-985 B+1729 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{4 a^2}-\frac {a (5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/ 
2)) + (-1/2*(a*(5*A - 13*B + 21*C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a* 
Sec[c + d*x])^(3/2)) - ((-2*a^2*(45*A - 85*B + 157*C)*Sec[c + d*x]^2*Tan[c 
 + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) - ((-2*a^2*(195*A - 475*B + 787*C) 
*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) - ((15*Sqrt[2]*a^(7/2)*(75*A 
 - 163*B + 283*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c 
+ d*x]])])/d - (4*a^4*(465*A - 985*B + 1729*C)*Tan[c + d*x])/(d*Sqrt[a + a 
*Sec[c + d*x]]))/(3*a))/(5*a))/(4*a^2))/(8*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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4282

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2/f   Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ 
a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4489

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 
 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B 
, e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b 
*(m + 1), 0] &&  !LtQ[m, -2^(-1)]
 

rule 4498

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( 
csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* 
((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int 
[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) 
*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a 
*B, 0] &&  !LtQ[m, -1]
 

rule 4507

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b 
- a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 
2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)* 
(d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m 
 - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, 
A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G 
tQ[n, 0]
 

rule 4509

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), 
 x] + Simp[d/(b*(m + n))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 
 1)*Simp[b*B*(n - 1) + (A*b*(m + n) + a*B*m)*Csc[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] 
&& GtQ[n, 1]
 

rule 4572

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

Maple [A] (warning: unable to verify)

Time = 3.83 (sec) , antiderivative size = 445, normalized size of antiderivative = 1.61

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (1125 \cos \left (d x +c \right )^{3}+3375 \cos \left (d x +c \right )^{2}+3375 \cos \left (d x +c \right )+1125\right ) \sqrt {2}\, A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-2445 \cos \left (d x +c \right )^{3}-7335 \cos \left (d x +c \right )^{2}-7335 \cos \left (d x +c \right )-2445\right ) \sqrt {2}\, B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (4245 \cos \left (d x +c \right )^{3}+12735 \cos \left (d x +c \right )^{2}+12735 \cos \left (d x +c \right )+4245\right ) \sqrt {2}\, C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \left (-1470 \cos \left (d x +c \right )^{2}-2550 \cos \left (d x +c \right )-960\right ) A +\left (2990 \cos \left (d x +c \right )^{3}+5030 \cos \left (d x +c \right )^{2}+1600 \cos \left (d x +c \right )-320\right ) B \tan \left (d x +c \right )+\left (-5342 \cos \left (d x +c \right )^{4}-9110 \cos \left (d x +c \right )^{3}-3136 \cos \left (d x +c \right )^{2}+320 \cos \left (d x +c \right )-192\right ) C \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{480 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\)	\(445\)
parts	\(\frac {A \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}}{16}-\frac {17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{32}-\frac {75 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}+\frac {83 \csc \left (d x +c \right )}{32}-\frac {83 \cot \left (d x +c \right )}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3}}+\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-598 \cos \left (d x +c \right )^{3}-1006 \cos \left (d x +c \right )^{2}-320 \cos \left (d x +c \right )+64\right ) \tan \left (d x +c \right )+\left (489 \cos \left (d x +c \right )^{3}+1467 \cos \left (d x +c \right )^{2}+1467 \cos \left (d x +c \right )+489\right ) \sqrt {2}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )}{96 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}-\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-5342 \cos \left (d x +c \right )^{4}-9110 \cos \left (d x +c \right )^{3}-3136 \cos \left (d x +c \right )^{2}+320 \cos \left (d x +c \right )-192\right ) \sec \left (d x +c \right ) \tan \left (d x +c \right )+\left (4245 \cos \left (d x +c \right )^{3}+12735 \cos \left (d x +c \right )^{2}+12735 \cos \left (d x +c \right )+4245\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{480 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\)	\(515\)

Input:

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

Output:

-1/480/d/a^3*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d 
*x+c)+1)*((1125*cos(d*x+c)^3+3375*cos(d*x+c)^2+3375*cos(d*x+c)+1125)*2^(1/ 
2)*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^ 
(1/2)-cot(d*x+c)+csc(d*x+c))+(-2445*cos(d*x+c)^3-7335*cos(d*x+c)^2-7335*co 
s(d*x+c)-2445)*2^(1/2)*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x 
+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+(4245*cos(d*x+c)^3+12735* 
cos(d*x+c)^2+12735*cos(d*x+c)+4245)*2^(1/2)*C*(-cos(d*x+c)/(cos(d*x+c)+1)) 
^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+sin( 
d*x+c)*(-1470*cos(d*x+c)^2-2550*cos(d*x+c)-960)*A+(2990*cos(d*x+c)^3+5030* 
cos(d*x+c)^2+1600*cos(d*x+c)-320)*B*tan(d*x+c)+(-5342*cos(d*x+c)^4-9110*co 
s(d*x+c)^3-3136*cos(d*x+c)^2+320*cos(d*x+c)-192)*C*sec(d*x+c)*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 638, normalized size of antiderivative = 2.30 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/960*(15*sqrt(2)*((75*A - 163*B + 283*C)*cos(d*x + c)^5 + 3*(75*A - 163 
*B + 283*C)*cos(d*x + c)^4 + 3*(75*A - 163*B + 283*C)*cos(d*x + c)^3 + (75 
*A - 163*B + 283*C)*cos(d*x + c)^2)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt 
((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d* 
x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 
4*((735*A - 1495*B + 2671*C)*cos(d*x + c)^4 + 5*(255*A - 503*B + 911*C)*co 
s(d*x + c)^3 + 32*(15*A - 25*B + 49*C)*cos(d*x + c)^2 + 160*(B - C)*cos(d* 
x + c) + 96*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*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), 1/480*(15*sqrt(2)*((75*A - 163*B + 283*C)*cos(d*x + c)^5 
 + 3*(75*A - 163*B + 283*C)*cos(d*x + c)^4 + 3*(75*A - 163*B + 283*C)*cos( 
d*x + c)^3 + (75*A - 163*B + 283*C)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2) 
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c 
))) + 2*((735*A - 1495*B + 2671*C)*cos(d*x + c)^4 + 5*(255*A - 503*B + 911 
*C)*cos(d*x + c)^3 + 32*(15*A - 25*B + 49*C)*cos(d*x + c)^2 + 160*(B - C)* 
cos(d*x + c) + 96*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*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]

\[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4/(a*(sec( 
c + d*x) + 1))**(5/2), x)
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 447, normalized size of antiderivative = 1.61 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/480*((((15*(2*(sqrt(2)*A*a^2*sgn(cos(d*x + c)) - sqrt(2)*B*a^2*sgn(cos( 
d*x + c)) + sqrt(2)*C*a^2*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/a^2 + 
(13*sqrt(2)*A*a^2*sgn(cos(d*x + c)) - 21*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + 
 29*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - (1725*s 
qrt(2)*A*a^2*sgn(cos(d*x + c)) - 3685*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + 67 
33*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 + 5*(549*s 
qrt(2)*A*a^2*sgn(cos(d*x + c)) - 1133*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + 19 
73*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15*(83*s 
qrt(2)*A*a^2*sgn(cos(d*x + c)) - 155*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + 291 
*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d* 
x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) - 15*(75*sqrt(2)* 
A - 163*sqrt(2)*B + 283*sqrt(2)*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) 
+ sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2*sgn(cos(d*x + c))))/ 
d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{6}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**6)/(sec(c + d*x)**3 + 
3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1) 
*sec(c + d*x)**5)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 
1),x)*b + int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4)/(sec(c + d*x)**3 + 
3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*a))/a**3