\(\int \frac {\sec ^5(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [146]

Optimal result

Integrand size = 33, antiderivative size = 232 \[ \int \frac {\sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {(2 A+21 C) \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {32 (5 A+54 C) \tan (c+d x)}{105 a^4 d}+\frac {(2 A+21 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(10 A+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {16 (5 A+54 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3} \] Output:

1/2*(2*A+21*C)*arctanh(sin(d*x+c))/a^4/d-32/105*(5*A+54*C)*tan(d*x+c)/a^4/ 
d+1/2*(2*A+21*C)*sec(d*x+c)*tan(d*x+c)/a^4/d-1/105*(10*A+129*C)*sec(d*x+c) 
^3*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2-16/105*(5*A+54*C)*sec(d*x+c)^2*tan(d* 
x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A+C)*sec(d*x+c)^5*tan(d*x+c)/d/(a+a*sec(d*x 
+c))^4-2/5*C*sec(d*x+c)^4*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(746\) vs. \(2(232)=464\).

Time = 5.41 (sec) , antiderivative size = 746, normalized size of antiderivative = 3.22 \[ \int \frac {\sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/3360*(Cos[(c + d*x)/2]*Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2)*(53760*(2* 
A + 21*C)*Cos[(c + d*x)/2]^7*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - L 
og[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]^2* 
(-14*(1010*A + 5229*C)*Sin[(d*x)/2] + 4*(3790*A + 41667*C)*Sin[(3*d*x)/2] 
- 17220*A*Sin[c - (d*x)/2] - 183162*C*Sin[c - (d*x)/2] + 17220*A*Sin[c + ( 
d*x)/2] + 100842*C*Sin[c + (d*x)/2] - 14140*A*Sin[2*c + (d*x)/2] - 155526* 
C*Sin[2*c + (d*x)/2] - 9800*A*Sin[c + (3*d*x)/2] - 37380*C*Sin[c + (3*d*x) 
/2] + 15160*A*Sin[2*c + (3*d*x)/2] + 101148*C*Sin[2*c + (3*d*x)/2] - 9800* 
A*Sin[3*c + (3*d*x)/2] - 102900*C*Sin[3*c + (3*d*x)/2] + 10920*A*Sin[c + ( 
5*d*x)/2] + 119364*C*Sin[c + (5*d*x)/2] - 4760*A*Sin[2*c + (5*d*x)/2] - 88 
20*C*Sin[2*c + (5*d*x)/2] + 10920*A*Sin[3*c + (5*d*x)/2] + 78204*C*Sin[3*c 
 + (5*d*x)/2] - 4760*A*Sin[4*c + (5*d*x)/2] - 49980*C*Sin[4*c + (5*d*x)/2] 
 + 5890*A*Sin[2*c + (7*d*x)/2] + 64053*C*Sin[2*c + (7*d*x)/2] - 1470*A*Sin 
[3*c + (7*d*x)/2] + 3885*C*Sin[3*c + (7*d*x)/2] + 5890*A*Sin[4*c + (7*d*x) 
/2] + 44733*C*Sin[4*c + (7*d*x)/2] - 1470*A*Sin[5*c + (7*d*x)/2] - 15435*C 
*Sin[5*c + (7*d*x)/2] + 2030*A*Sin[3*c + (9*d*x)/2] + 21987*C*Sin[3*c + (9 
*d*x)/2] - 210*A*Sin[4*c + (9*d*x)/2] + 3675*C*Sin[4*c + (9*d*x)/2] + 2030 
*A*Sin[5*c + (9*d*x)/2] + 16107*C*Sin[5*c + (9*d*x)/2] - 210*A*Sin[6*c + ( 
9*d*x)/2] - 2205*C*Sin[6*c + (9*d*x)/2] + 320*A*Sin[4*c + (11*d*x)/2] + 34 
56*C*Sin[4*c + (11*d*x)/2] + 840*C*Sin[5*c + (11*d*x)/2] + 320*A*Sin[6*...
 

Rubi [A] (verified)

Time = 1.80 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.06, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4573, 25, 3042, 4507, 25, 3042, 4507, 3042, 4507, 3042, 4274, 3042, 4254, 24, 4255, 3042, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

\(\displaystyle -\frac {\int -\frac {\sec ^5(c+d x) (a (2 A-5 C)+a (2 A+9 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\sec ^5(c+d x) (a (2 A-5 C)+a (2 A+9 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (a (2 A-5 C)+a (2 A+9 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int -\frac {\sec ^4(c+d x) \left (56 a^2 C-a^2 (10 A+73 C) \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int \frac {\sec ^4(c+d x) \left (56 a^2 C-a^2 (10 A+73 C) \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (56 a^2 C-a^2 (10 A+73 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {-\frac {\frac {\int \frac {\sec ^3(c+d x) \left (3 a^3 (10 A+129 C)-a^3 (50 A+477 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a^3 (10 A+129 C)-a^3 (50 A+477 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {-\frac {\frac {\frac {\int \sec ^2(c+d x) \left (32 a^4 (5 A+54 C)-105 a^4 (2 A+21 C) \sec (c+d x)\right )dx}{a^2}+\frac {16 a^3 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (32 a^4 (5 A+54 C)-105 a^4 (2 A+21 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {16 a^3 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {-\frac {\frac {\frac {32 a^4 (5 A+54 C) \int \sec ^2(c+d x)dx-105 a^4 (2 A+21 C) \int \sec ^3(c+d x)dx}{a^2}+\frac {16 a^3 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\frac {32 a^4 (5 A+54 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-105 a^4 (2 A+21 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {16 a^3 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {-\frac {\frac {\frac {-\frac {32 a^4 (5 A+54 C) \int 1d(-\tan (c+d x))}{d}-105 a^4 (2 A+21 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {16 a^3 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {-\frac {\frac {\frac {\frac {32 a^4 (5 A+54 C) \tan (c+d x)}{d}-105 a^4 (2 A+21 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {16 a^3 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {-\frac {\frac {\frac {\frac {32 a^4 (5 A+54 C) \tan (c+d x)}{d}-105 a^4 (2 A+21 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {16 a^3 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\frac {\frac {32 a^4 (5 A+54 C) \tan (c+d x)}{d}-105 a^4 (2 A+21 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {16 a^3 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {-\frac {\frac {\frac {16 a^3 (5 A+54 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {32 a^4 (5 A+54 C) \tan (c+d x)}{d}-105 a^4 (2 A+21 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}}{3 a^2}+\frac {(10 A+129 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {14 a C \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

Input:

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

Output:

-1/7*((A + C)*Sec[c + d*x]^5*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + (( 
-14*a*C*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - (((10* 
A + 129*C)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(1 + Sec[c + d*x])^2) + ((16* 
a^3*(5*A + 54*C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) + ( 
(32*a^4*(5*A + 54*C)*Tan[c + d*x])/d - 105*a^4*(2*A + 21*C)*(ArcTanh[Sin[c 
 + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/a^2)/(3*a^2))/(5*a^2) 
)/(7*a^2)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

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 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] (verified)

Time = 0.55 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(\frac {-6720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {21 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {21 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-1070 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \left (\left (\frac {636 A}{107}+66 C \right ) \cos \left (2 d x +2 c \right )+\left (\frac {328 A}{107}+\frac {17964 C}{535}\right ) \cos \left (3 d x +3 c \right )+\left (A +\frac {11619 C}{1070}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {16 A}{107}+\frac {864 C}{535}\right ) \cos \left (5 d x +5 c \right )+\left (\frac {904 A}{107}+\frac {51252 C}{535}\right ) \cos \left (d x +c \right )+\frac {529 A}{107}+\frac {58161 C}{1070}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6720 d \,a^{4} \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(192\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A -\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-84 C -8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {36 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (84 C +8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {36 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{4}}\)	\(222\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A -\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-84 C -8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {36 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (84 C +8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {36 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{4}}\)	\(222\)
risch	\(-\frac {i \left (210 A \,{\mathrm e}^{10 i \left (d x +c \right )}+2205 C \,{\mathrm e}^{10 i \left (d x +c \right )}+1470 A \,{\mathrm e}^{9 i \left (d x +c \right )}+15435 C \,{\mathrm e}^{9 i \left (d x +c \right )}+4760 A \,{\mathrm e}^{8 i \left (d x +c \right )}+49980 C \,{\mathrm e}^{8 i \left (d x +c \right )}+9800 A \,{\mathrm e}^{7 i \left (d x +c \right )}+102900 C \,{\mathrm e}^{7 i \left (d x +c \right )}+14140 A \,{\mathrm e}^{6 i \left (d x +c \right )}+155526 C \,{\mathrm e}^{6 i \left (d x +c \right )}+17220 A \,{\mathrm e}^{5 i \left (d x +c \right )}+183162 C \,{\mathrm e}^{5 i \left (d x +c \right )}+15160 A \,{\mathrm e}^{4 i \left (d x +c \right )}+166668 C \,{\mathrm e}^{4 i \left (d x +c \right )}+10920 A \,{\mathrm e}^{3 i \left (d x +c \right )}+119364 C \,{\mathrm e}^{3 i \left (d x +c \right )}+5890 A \,{\mathrm e}^{2 i \left (d x +c \right )}+64053 C \,{\mathrm e}^{2 i \left (d x +c \right )}+2030 A \,{\mathrm e}^{i \left (d x +c \right )}+21987 C \,{\mathrm e}^{i \left (d x +c \right )}+320 A +3456 C \right )}{105 a^{4} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{4} d}-\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{4} d}+\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{4} d}\)	\(371\)

Input:

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

Output:

1/6720*(-6720*(1+cos(2*d*x+2*c))*(A+21/2*C)*ln(tan(1/2*d*x+1/2*c)-1)+6720* 
(1+cos(2*d*x+2*c))*(A+21/2*C)*ln(tan(1/2*d*x+1/2*c)+1)-1070*sec(1/2*d*x+1/ 
2*c)^6*((636/107*A+66*C)*cos(2*d*x+2*c)+(328/107*A+17964/535*C)*cos(3*d*x+ 
3*c)+(A+11619/1070*C)*cos(4*d*x+4*c)+(16/107*A+864/535*C)*cos(5*d*x+5*c)+( 
904/107*A+51252/535*C)*cos(d*x+c)+529/107*A+58161/1070*C)*tan(1/2*d*x+1/2* 
c))/d/a^4/(1+cos(2*d*x+2*c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.53 \[ \int \frac {\sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {105 \, {\left ({\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (64 \, {\left (5 \, A + 54 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (1070 \, A + 11619 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (310 \, A + 3411 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (130 \, A + 1509 \, C\right )} \cos \left (d x + c\right )^{2} + 420 \, C \cos \left (d x + c\right ) - 105 \, C\right )} \sin \left (d x + c\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \] Input:

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

Output:

1/420*(105*((2*A + 21*C)*cos(d*x + c)^6 + 4*(2*A + 21*C)*cos(d*x + c)^5 + 
6*(2*A + 21*C)*cos(d*x + c)^4 + 4*(2*A + 21*C)*cos(d*x + c)^3 + (2*A + 21* 
C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 105*((2*A + 21*C)*cos(d*x + c)^ 
6 + 4*(2*A + 21*C)*cos(d*x + c)^5 + 6*(2*A + 21*C)*cos(d*x + c)^4 + 4*(2*A 
 + 21*C)*cos(d*x + c)^3 + (2*A + 21*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 
 1) - 2*(64*(5*A + 54*C)*cos(d*x + c)^5 + (1070*A + 11619*C)*cos(d*x + c)^ 
4 + 4*(310*A + 3411*C)*cos(d*x + c)^3 + 4*(130*A + 1509*C)*cos(d*x + c)^2 
+ 420*C*cos(d*x + c) - 105*C)*sin(d*x + c))/(a^4*d*cos(d*x + c)^6 + 4*a^4* 
d*cos(d*x + c)^5 + 6*a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + a^4*d 
*cos(d*x + c)^2)
 

Sympy [F]

\[ \int \frac {\sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{7}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:

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

Output:

(Integral(A*sec(c + d*x)**5/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c 
 + d*x)**2 + 4*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**7/(sec(c + 
 d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) 
)/a**4
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {3 \, C {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} - \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + 5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )}}{840 \, d} \] Input:

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

Output:

-1/840*(3*C*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^3/(co 
s(d*x + c) + 1)^3)/(a^4 - 2*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4* 
sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c) + 
1) + 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x 
+ c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 2940*log(sin(d* 
x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 2940*log(sin(d*x + c)/(cos(d*x + c) + 
 1) - 1)/a^4) + 5*A*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c 
)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin( 
d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) 
+ 1) + 1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4))/d
 

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {420 \, {\left (2 \, A + 21 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {420 \, {\left (2 \, A + 21 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {840 \, {\left (9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 189 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1365 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11655 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \] Input:

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

Output:

1/840*(420*(2*A + 21*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420*(2*A 
+ 21*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 840*(9*C*tan(1/2*d*x + 1/ 
2*c)^3 - 7*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^4) - 
(15*A*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 + 105 
*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 189*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A 
*a^24*tan(1/2*d*x + 1/2*c)^3 + 1365*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 1575*A 
*a^24*tan(1/2*d*x + 1/2*c) + 11655*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d
 

Mupad [B] (verification not implemented)

Time = 13.18 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {21\,C}{2}\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {3\,\left (A+C\right )}{40\,a^4}+\frac {2\,A+6\,C}{40\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^4}-\frac {A-15\,C}{24\,a^4}+\frac {2\,A+6\,C}{8\,a^4}\right )}{d}-\frac {7\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A+C\right )}{4\,a^4}-\frac {3\,\left (A-15\,C\right )}{8\,a^4}+\frac {3\,\left (2\,A+6\,C\right )}{4\,a^4}-\frac {4\,A-20\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \] Input:

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

Output:

(2*atanh(tan(c/2 + (d*x)/2))*(A + (21*C)/2))/(a^4*d) - (tan(c/2 + (d*x)/2) 
^5*((3*(A + C))/(40*a^4) + (2*A + 6*C)/(40*a^4)))/d - (tan(c/2 + (d*x)/2)^ 
3*((A + C)/(4*a^4) - (A - 15*C)/(24*a^4) + (2*A + 6*C)/(8*a^4)))/d - (7*C* 
tan(c/2 + (d*x)/2) - 9*C*tan(c/2 + (d*x)/2)^3)/(d*(a^4*tan(c/2 + (d*x)/2)^ 
4 - 2*a^4*tan(c/2 + (d*x)/2)^2 + a^4)) - (tan(c/2 + (d*x)/2)*((5*(A + C))/ 
(4*a^4) - (3*(A - 15*C))/(8*a^4) + (3*(2*A + 6*C))/(4*a^4) - (4*A - 20*C)/ 
(8*a^4)))/d - (tan(c/2 + (d*x)/2)^7*(A + C))/(56*a^4*d)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.02 \[ \int \frac {\sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 840*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*a - 8820*log(tan((c 
+ d*x)/2) - 1)*tan((c + d*x)/2)**4*c + 1680*log(tan((c + d*x)/2) - 1)*tan( 
(c + d*x)/2)**2*a + 17640*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*c 
- 840*log(tan((c + d*x)/2) - 1)*a - 8820*log(tan((c + d*x)/2) - 1)*c + 840 
*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a + 8820*log(tan((c + d*x)/ 
2) + 1)*tan((c + d*x)/2)**4*c - 1680*log(tan((c + d*x)/2) + 1)*tan((c + d* 
x)/2)**2*a - 17640*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c + 840*l 
og(tan((c + d*x)/2) + 1)*a + 8820*log(tan((c + d*x)/2) + 1)*c - 15*tan((c 
+ d*x)/2)**11*a - 15*tan((c + d*x)/2)**11*c - 75*tan((c + d*x)/2)**9*a - 1 
59*tan((c + d*x)/2)**9*c - 190*tan((c + d*x)/2)**7*a - 1002*tan((c + d*x)/ 
2)**7*c - 910*tan((c + d*x)/2)**5*a - 9114*tan((c + d*x)/2)**5*c + 2765*ta 
n((c + d*x)/2)**3*a + 29505*tan((c + d*x)/2)**3*c - 1575*tan((c + d*x)/2)* 
a - 17535*tan((c + d*x)/2)*c)/(840*a**4*d*(tan((c + d*x)/2)**4 - 2*tan((c 
+ d*x)/2)**2 + 1))