\(\int \frac {\sec ^4(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [200]

Optimal result

Integrand size = 35, antiderivative size = 259 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(75 A+283 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(465 A+1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(45 A+157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(195 A+787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d} \] Output:

-1/32*(75*A+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+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+ 
c))^(5/2)-1/16*(5*A+21*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/ 
2)+1/120*(465*A+1729*C)*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)+1/80*(45*A 
+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+ 
787*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^3/d
 

Mathematica [A] (verified)

Time = 5.76 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left (4125 A+15053 C+50 (153 A+521 C) \cos (c+d x)+108 (45 A+157 C) \cos (2 (c+d x))+2550 A \cos (3 (c+d x))+9110 C \cos (3 (c+d x))+735 A \cos (4 (c+d x))+2671 C \cos (4 (c+d x))+\frac {60 \sqrt {2} (75 A+283 C) \arctan \left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^3(c+d x) (1+\cos (c+d x))^2 \sqrt {-1+\sec (c+d x)}}{-1+\cos (c+d x)}\right ) \sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right ) \tan (c+d x)}{960 d (A+2 C+A \cos (2 (c+d x))) (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

((4125*A + 15053*C + 50*(153*A + 521*C)*Cos[c + d*x] + 108*(45*A + 157*C)* 
Cos[2*(c + d*x)] + 2550*A*Cos[3*(c + d*x)] + 9110*C*Cos[3*(c + d*x)] + 735 
*A*Cos[4*(c + d*x)] + 2671*C*Cos[4*(c + d*x)] + (60*Sqrt[2]*(75*A + 283*C) 
*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Cos[c + d*x]^3*(1 + Cos[c + d*x]) 
^2*Sqrt[-1 + Sec[c + d*x]])/(-1 + Cos[c + d*x]))*Sec[c + d*x]^2*(A + C*Sec 
[c + d*x]^2)*Tan[c + d*x])/(960*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a*(1 + S 
ec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.79 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.10, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4573, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4509

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4498

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4489

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4282

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

-1/4*((A + C)*Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) 
- ((a*(5*A + 21*C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^ 
(3/2)) + ((-2*a^2*(45*A + 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 + 787*C)*Sqrt[a + a*Sec[c + d*x]]*Tan 
[c + d*x])/(3*d) - ((15*Sqrt[2]*a^(7/2)*(75*A + 283*C)*ArcTan[(Sqrt[a]*Tan 
[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (4*a^4*(465*A + 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 4573

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) 
*(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*C 
sc[e + f*x])^n*Simp[b*C*n + A*b*(2*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, C, n}, x] && EqQ[ 
a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

Maple [A] (warning: unable to verify)

Time = 1.70 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.22

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (1125 \cos \left (d x +c \right )^{3}+3375 \cos \left (d x +c \right )^{2}+3375 \cos \left (d x +c \right )+1125\right ) \sqrt {2}\, A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (4245 \cos \left (d x +c \right )^{3}+12735 \cos \left (d x +c \right )^{2}+12735 \cos \left (d x +c \right )+4245\right ) \sqrt {2}\, C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \left (-1470 \cos \left (d x +c \right )^{2}-2550 \cos \left (d x +c \right )-960\right ) A +\left (-5342 \cos \left (d x +c \right )^{4}-9110 \cos \left (d x +c \right )^{3}-3136 \cos \left (d x +c \right )^{2}+320 \cos \left (d x +c \right )-192\right ) C \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{480 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\)	\(316\)
parts	\(\frac {A \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}}{16}-\frac {17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{32}-\frac {75 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}+\frac {83 \csc \left (d x +c \right )}{32}-\frac {83 \cot \left (d x +c \right )}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3}}-\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-5342 \cos \left (d x +c \right )^{4}-9110 \cos \left (d x +c \right )^{3}-3136 \cos \left (d x +c \right )^{2}+320 \cos \left (d x +c \right )-192\right ) \sec \left (d x +c \right ) \tan \left (d x +c \right )+\left (4245 \cos \left (d x +c \right )^{3}+12735 \cos \left (d x +c \right )^{2}+12735 \cos \left (d x +c \right )+4245\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{480 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\)	\(336\)

Input:

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

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))+(4245*cos(d*x+c)^3+12735*cos(d*x+c)^2+12735*c 
os(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+(-5342*cos(d*x+c)^4-9110*cos(d*x+c)^3-3136*c 
os(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 = 588, normalized size of antiderivative = 2.27 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/960*(15*sqrt(2)*((75*A + 283*C)*cos(d*x + c)^5 + 3*(75*A + 283*C)*cos( 
d*x + c)^4 + 3*(75*A + 283*C)*cos(d*x + c)^3 + (75*A + 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 + 2671*C)*cos(d*x + c) 
^4 + 5*(255*A + 911*C)*cos(d*x + c)^3 + 32*(15*A + 49*C)*cos(d*x + c)^2 - 
160*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 + 283*C)*cos(d*x + 
c)^5 + 3*(75*A + 283*C)*cos(d*x + c)^4 + 3*(75*A + 283*C)*cos(d*x + c)^3 + 
 (75*A + 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 + 26 
71*C)*cos(d*x + c)^4 + 5*(255*A + 911*C)*cos(d*x + c)^3 + 32*(15*A + 49*C) 
*cos(d*x + c)^2 - 160*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]

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

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

Output:

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

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

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\frac {{\left ({\left ({\left (15 \, {\left (\frac {2 \, {\left (\sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2}} + \frac {13 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 29 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {1725 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 6733 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, {\left (549 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1973 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, {\left (83 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 291 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {15 \, {\left (75 \, \sqrt {2} A + 283 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{480 \, d} \] Input:

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

Output:

-1/480*((((15*(2*(sqrt(2)*A*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) 
) + 29*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - (172 
5*sqrt(2)*A*a^2*sgn(cos(d*x + c)) + 6733*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/ 
a^2)*tan(1/2*d*x + 1/2*c)^2 + 5*(549*sqrt(2)*A*a^2*sgn(cos(d*x + c)) + 197 
3*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15*(83*sq 
rt(2)*A*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 + 283*sqrt(2)*C)*log(abs(-sqrt(-a)*ta 
n(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2*s 
gn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)^4*(A+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)**4)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 
1),x)*a))/a**3