\(\int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [108]

Optimal result

Integrand size = 31, antiderivative size = 238 \[ \int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {(8 A-21 B) \text {arctanh}(\sin (c+d x))}{2 a^4 d}+\frac {8 (83 A-216 B) \tan (c+d x)}{105 a^4 d}-\frac {(8 A-21 B) \sec (c+d x) \tan (c+d x)}{2 a^4 d}+\frac {(52 A-129 B) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {4 (83 A-216 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(A-2 B) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3} \] Output:

-1/2*(8*A-21*B)*arctanh(sin(d*x+c))/a^4/d+8/105*(83*A-216*B)*tan(d*x+c)/a^ 
4/d-1/2*(8*A-21*B)*sec(d*x+c)*tan(d*x+c)/a^4/d+1/105*(52*A-129*B)*sec(d*x+ 
c)^3*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2+4/105*(83*A-216*B)*sec(d*x+c)^2*tan 
(d*x+c)/a^4/d/(1+sec(d*x+c))+1/7*(A-B)*sec(d*x+c)^5*tan(d*x+c)/d/(a+a*sec( 
d*x+c))^4+1/5*(A-2*B)*sec(d*x+c)^4*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(880\) vs. \(2(238)=476\).

Time = 8.37 (sec) , antiderivative size = 880, normalized size of antiderivative = 3.70 \[ \int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

Integrate[(Sec[c + d*x]^6*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]
 

Output:

(-8*(-8*A + 21*B)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + 
(d*x)/2]]*Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(d*(B + A*Cos[c + d*x])*(a 
+ a*Sec[c + d*x])^4) + (8*(-8*A + 21*B)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + 
 (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(d*(B 
 + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^4) + (Cos[c/2 + (d*x)/2]*Sec[c/2]* 
Sec[c]*Sec[c + d*x]^5*(A + B*Sec[c + d*x])*(-38668*A*Sin[(d*x)/2] + 73206* 
B*Sin[(d*x)/2] + 64384*A*Sin[(3*d*x)/2] - 166668*B*Sin[(3*d*x)/2] - 70896* 
A*Sin[c - (d*x)/2] + 183162*B*Sin[c - (d*x)/2] + 50316*A*Sin[c + (d*x)/2] 
- 100842*B*Sin[c + (d*x)/2] - 59248*A*Sin[2*c + (d*x)/2] + 155526*B*Sin[2* 
c + (d*x)/2] - 22820*A*Sin[c + (3*d*x)/2] + 37380*B*Sin[c + (3*d*x)/2] + 4 
8004*A*Sin[2*c + (3*d*x)/2] - 101148*B*Sin[2*c + (3*d*x)/2] - 39200*A*Sin[ 
3*c + (3*d*x)/2] + 102900*B*Sin[3*c + (3*d*x)/2] + 46032*A*Sin[c + (5*d*x) 
/2] - 119364*B*Sin[c + (5*d*x)/2] - 8750*A*Sin[2*c + (5*d*x)/2] + 8820*B*S 
in[2*c + (5*d*x)/2] + 35742*A*Sin[3*c + (5*d*x)/2] - 78204*B*Sin[3*c + (5* 
d*x)/2] - 19040*A*Sin[4*c + (5*d*x)/2] + 49980*B*Sin[4*c + (5*d*x)/2] + 24 
664*A*Sin[2*c + (7*d*x)/2] - 64053*B*Sin[2*c + (7*d*x)/2] - 1050*A*Sin[3*c 
 + (7*d*x)/2] - 3885*B*Sin[3*c + (7*d*x)/2] + 19834*A*Sin[4*c + (7*d*x)/2] 
 - 44733*B*Sin[4*c + (7*d*x)/2] - 5880*A*Sin[5*c + (7*d*x)/2] + 15435*B*Si 
n[5*c + (7*d*x)/2] + 8456*A*Sin[3*c + (9*d*x)/2] - 21987*B*Sin[3*c + (9*d* 
x)/2] + 630*A*Sin[4*c + (9*d*x)/2] - 3675*B*Sin[4*c + (9*d*x)/2] + 6986...
 

Rubi [A] (verified)

Time = 1.81 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {3042, 4507, 3042, 4507, 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.

\(\displaystyle \int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^6 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\)

\(\Big \downarrow \) 4507

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int \frac {\sec ^4(c+d x) \left (28 a^2 (A-2 B)-a^2 (24 A-73 B) \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (28 a^2 (A-2 B)-a^2 (24 A-73 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\frac {\int \frac {\sec ^3(c+d x) \left (3 a^3 (52 A-129 B)-a^3 (176 A-477 B) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a^3 (52 A-129 B)-a^3 (176 A-477 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\frac {\frac {\int \sec ^2(c+d x) \left (8 a^4 (83 A-216 B)-105 a^4 (8 A-21 B) \sec (c+d x)\right )dx}{a^2}+\frac {4 a^3 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (8 a^4 (83 A-216 B)-105 a^4 (8 A-21 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {4 a^3 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {\frac {\frac {8 a^4 (83 A-216 B) \int \sec ^2(c+d x)dx-105 a^4 (8 A-21 B) \int \sec ^3(c+d x)dx}{a^2}+\frac {4 a^3 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {8 a^4 (83 A-216 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-105 a^4 (8 A-21 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^3 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {\frac {\frac {-\frac {8 a^4 (83 A-216 B) \int 1d(-\tan (c+d x))}{d}-105 a^4 (8 A-21 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^3 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {\frac {\frac {\frac {8 a^4 (83 A-216 B) \tan (c+d x)}{d}-105 a^4 (8 A-21 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^3 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {\frac {\frac {\frac {8 a^4 (83 A-216 B) \tan (c+d x)}{d}-105 a^4 (8 A-21 B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {4 a^3 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {\frac {8 a^4 (83 A-216 B) \tan (c+d x)}{d}-105 a^4 (8 A-21 B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {4 a^3 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {\frac {\frac {4 a^3 (83 A-216 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {8 a^4 (83 A-216 B) \tan (c+d x)}{d}-105 a^4 (8 A-21 B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}}{3 a^2}+\frac {(52 A-129 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}+\frac {7 a (A-2 B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}+\frac {(A-B) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

Input:

Int[(Sec[c + d*x]^6*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]
 

Output:

((A - B)*Sec[c + d*x]^5*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((7*a 
*(A - 2*B)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((( 
52*A - 129*B)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(1 + Sec[c + d*x])^2) + (( 
4*a^3*(83*A - 216*B)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) 
 + ((8*a^4*(83*A - 216*B)*Tan[c + d*x])/d - 105*a^4*(8*A - 21*B)*(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 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]
 

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(\frac {3360 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {21 B}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3360 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {21 B}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+559 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {15 \left (228 A -\frac {1177 B}{2}\right ) \cos \left (2 d x +2 c \right )}{559}+\frac {\left (1733 A -4491 B \right ) \cos \left (3 d x +3 c \right )}{559}+\left (A -\frac {11619 B}{4472}\right ) \cos \left (4 d x +4 c \right )+\frac {\left (83 A -216 B \right ) \cos \left (5 d x +5 c \right )}{559}+\frac {\left (4994 A -12813 B \right ) \cos \left (d x +c \right )}{559}+\frac {2861 A}{559}-\frac {58161 B}{4472}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{840 d \,a^{4} \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(196\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {-36 B +8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (84 B -32 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-84 B +32 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-36 B +8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {4 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{8 d \,a^{4}}\)	\(234\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {-36 B +8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (84 B -32 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-84 B +32 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-36 B +8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {4 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{8 d \,a^{4}}\)	\(234\)
risch	\(\frac {i \left (840 A \,{\mathrm e}^{10 i \left (d x +c \right )}-2205 \,{\mathrm e}^{10 i \left (d x +c \right )} B +5880 A \,{\mathrm e}^{9 i \left (d x +c \right )}-15435 B \,{\mathrm e}^{9 i \left (d x +c \right )}+19040 A \,{\mathrm e}^{8 i \left (d x +c \right )}-49980 B \,{\mathrm e}^{8 i \left (d x +c \right )}+39200 A \,{\mathrm e}^{7 i \left (d x +c \right )}-102900 B \,{\mathrm e}^{7 i \left (d x +c \right )}+59248 A \,{\mathrm e}^{6 i \left (d x +c \right )}-155526 B \,{\mathrm e}^{6 i \left (d x +c \right )}+70896 A \,{\mathrm e}^{5 i \left (d x +c \right )}-183162 B \,{\mathrm e}^{5 i \left (d x +c \right )}+64384 A \,{\mathrm e}^{4 i \left (d x +c \right )}-166668 B \,{\mathrm e}^{4 i \left (d x +c \right )}+46032 A \,{\mathrm e}^{3 i \left (d x +c \right )}-119364 B \,{\mathrm e}^{3 i \left (d x +c \right )}+24664 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64053 B \,{\mathrm e}^{2 i \left (d x +c \right )}+8456 \,{\mathrm e}^{i \left (d x +c \right )} A -21987 B \,{\mathrm e}^{i \left (d x +c \right )}+1328 A -3456 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{4} d}-\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a^{4} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{4} d}+\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a^{4} d}\)	\(372\)

Input:

int(sec(d*x+c)^6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBO 
SE)
 

Output:

1/840*(3360*(1+cos(2*d*x+2*c))*(A-21/8*B)*ln(tan(1/2*d*x+1/2*c)-1)-3360*(1 
+cos(2*d*x+2*c))*(A-21/8*B)*ln(tan(1/2*d*x+1/2*c)+1)+559*tan(1/2*d*x+1/2*c 
)*(15/559*(228*A-1177/2*B)*cos(2*d*x+2*c)+1/559*(1733*A-4491*B)*cos(3*d*x+ 
3*c)+(A-11619/4472*B)*cos(4*d*x+4*c)+1/559*(83*A-216*B)*cos(5*d*x+5*c)+1/5 
59*(4994*A-12813*B)*cos(d*x+c)+2861/559*A-58161/4472*B)*sec(1/2*d*x+1/2*c) 
^6)/d/a^4/(1+cos(2*d*x+2*c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {105 \, {\left ({\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A - 21 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (83 \, A - 216 \, B\right )} \cos \left (d x + c\right )^{5} + {\left (4472 \, A - 11619 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (1318 \, A - 3411 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (592 \, A - 1509 \, B\right )} \cos \left (d x + c\right )^{2} + 210 \, {\left (A - 2 \, B\right )} \cos \left (d x + c\right ) + 105 \, B\right )} \sin \left (d x + c\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \] Input:

integrate(sec(d*x+c)^6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="f 
ricas")
 

Output:

-1/420*(105*((8*A - 21*B)*cos(d*x + c)^6 + 4*(8*A - 21*B)*cos(d*x + c)^5 + 
 6*(8*A - 21*B)*cos(d*x + c)^4 + 4*(8*A - 21*B)*cos(d*x + c)^3 + (8*A - 21 
*B)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 105*((8*A - 21*B)*cos(d*x + c) 
^6 + 4*(8*A - 21*B)*cos(d*x + c)^5 + 6*(8*A - 21*B)*cos(d*x + c)^4 + 4*(8* 
A - 21*B)*cos(d*x + c)^3 + (8*A - 21*B)*cos(d*x + c)^2)*log(-sin(d*x + c) 
+ 1) - 2*(16*(83*A - 216*B)*cos(d*x + c)^5 + (4472*A - 11619*B)*cos(d*x + 
c)^4 + 4*(1318*A - 3411*B)*cos(d*x + c)^3 + 4*(592*A - 1509*B)*cos(d*x + c 
)^2 + 210*(A - 2*B)*cos(d*x + c) + 105*B)*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]

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

integrate(sec(d*x+c)**6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**4,x)
 

Output:

(Integral(A*sec(c + d*x)**6/(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(B*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 = 0.05 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {3 \, B {\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 )} - A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \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)^6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="m 
axima")
 

Output:

-1/840*(3*B*(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) - A*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + 
 c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d*x + c)/(cos(d*x + c) + 1) + 
805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) 
 + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + 
 c)/(cos(d*x + c) + 1) + 1)/a^4 + 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) 
 - 1)/a^4))/d
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.12 \[ \int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {420 \, {\left (8 \, A - 21 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {420 \, {\left (8 \, A - 21 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {840 \, {\left (2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 189 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1365 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 11655 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \] Input:

integrate(sec(d*x+c)^6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="g 
iac")
 

Output:

-1/840*(420*(8*A - 21*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420*(8*A 
 - 21*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 840*(2*A*tan(1/2*d*x + 1 
/2*c)^3 - 9*B*tan(1/2*d*x + 1/2*c)^3 - 2*A*tan(1/2*d*x + 1/2*c) + 7*B*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*B*a^24*tan(1/2*d*x + 1/2*c)^7 + 147*A*a^24*tan(1/2*d 
*x + 1/2*c)^5 - 189*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 805*A*a^24*tan(1/2*d*x 
 + 1/2*c)^3 - 1365*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 5145*A*a^24*tan(1/2*d*x 
 + 1/2*c) - 11655*B*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d
 

Mupad [B] (verification not implemented)

Time = 10.99 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{4\,a^4}+\frac {4\,A-6\,B}{8\,a^4}+\frac {5\,A-15\,B}{24\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A-B\right )}{4\,a^4}-\frac {5\,B}{2\,a^4}+\frac {3\,\left (4\,A-6\,B\right )}{4\,a^4}+\frac {3\,\left (5\,A-15\,B\right )}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-9\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A-7\,B\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {3\,\left (A-B\right )}{40\,a^4}+\frac {4\,A-6\,B}{40\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (8\,A-21\,B\right )}{a^4\,d} \] Input:

int((A + B/cos(c + d*x))/(cos(c + d*x)^6*(a + a/cos(c + d*x))^4),x)
                                                                                    
                                                                                    
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.97 \[ \int \frac {\sec ^6(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)^6*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x)
 

Output:

(3360*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*b - 6720*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*b + 
3360*log(tan((c + d*x)/2) - 1)*a - 8820*log(tan((c + d*x)/2) - 1)*b - 3360 
*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*b + 6720*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*b - 3360* 
log(tan((c + d*x)/2) + 1)*a + 8820*log(tan((c + d*x)/2) + 1)*b + 15*tan((c 
 + d*x)/2)**11*a - 15*tan((c + d*x)/2)**11*b + 117*tan((c + d*x)/2)**9*a - 
 159*tan((c + d*x)/2)**9*b + 526*tan((c + d*x)/2)**7*a - 1002*tan((c + d*x 
)/2)**7*b + 3682*tan((c + d*x)/2)**5*a - 9114*tan((c + d*x)/2)**5*b - 1116 
5*tan((c + d*x)/2)**3*a + 29505*tan((c + d*x)/2)**3*b + 6825*tan((c + d*x) 
/2)*a - 17535*tan((c + d*x)/2)*b)/(840*a**4*d*(tan((c + d*x)/2)**4 - 2*tan 
((c + d*x)/2)**2 + 1))