Integrand size = 39, antiderivative size = 102 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a (3 A+4 (B+C)) x+\frac {a (A+B+C) \sin (c+d x)}{d}+\frac {a (3 A+4 (B+C)) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a (A+B) \sin ^3(c+d x)}{3 d} \]
1/8*a*(3*A+4*B+4*C)*x+a*(A+B+C)*sin(d*x+c)/d+1/8*a*(3*A+4*B+4*C)*cos(d*x+c )*sin(d*x+c)/d+1/4*a*A*cos(d*x+c)^3*sin(d*x+c)/d-1/3*a*(A+B)*sin(d*x+c)^3/ d
Time = 0.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (24 A c+48 B c+36 A d x+48 B d x+48 C d x+24 (3 A+3 B+4 C) \sin (c+d x)+24 (A+B+C) \sin (2 (c+d x))+8 A \sin (3 (c+d x))+8 B \sin (3 (c+d x))+3 A \sin (4 (c+d x)))}{96 d} \]
(a*(24*A*c + 48*B*c + 36*A*d*x + 48*B*d*x + 48*C*d*x + 24*(3*A + 3*B + 4*C )*Sin[c + d*x] + 24*(A + B + C)*Sin[2*(c + d*x)] + 8*A*Sin[3*(c + d*x)] + 8*B*Sin[3*(c + d*x)] + 3*A*Sin[4*(c + d*x)]))/(96*d)
Time = 0.69 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 4562, 25, 3042, 4535, 3042, 3115, 24, 4532, 3042, 3492, 2009}
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.
| \(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {1}{4} \int -\cos ^3(c+d x) \left (4 a C \sec ^2(c+d x)+a (3 A+4 (B+C)) \sec (c+d x)+4 a (A+B)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \int \cos ^3(c+d x) \left (4 a C \sec ^2(c+d x)+a (3 A+4 (B+C)) \sec (c+d x)+4 a (A+B)\right )dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {4 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (3 A+4 (B+C)) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a (A+B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 (B+C)) \int \cos ^2(c+d x)dx+\int \cos ^3(c+d x) \left (4 a C \sec ^2(c+d x)+4 a (A+B)\right )dx\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 (B+C)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\int \frac {4 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a (A+B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {4 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a (A+B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+a (3 A+4 (B+C)) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {4 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a (A+B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+a (3 A+4 (B+C)) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4532 |
| \(\displaystyle \frac {1}{4} \left (\int \cos (c+d x) \left (4 a (A+B) \cos ^2(c+d x)+4 a C\right )dx+a (3 A+4 (B+C)) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 a (A+B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+4 a C\right )dx+a (3 A+4 (B+C)) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3492 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 (B+C)) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\int \left (4 a (A+B+C)-4 a (A+B) \sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 (B+C)) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\frac {4}{3} a (A+B) \sin ^3(c+d x)-4 a (A+B+C) \sin (c+d x)}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
(a*A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*(3*A + 4*(B + C))*(x/2 + (Cos [c + d*x]*Sin[c + d*x])/(2*d)) - (-4*a*(A + B + C)*Sin[c + d*x] + (4*a*(A + B)*Sin[c + d*x]^3)/3)/d)/4
3.5.15.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 ), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[ {e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
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 _)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si mp[1/(d*n) Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
Time = 0.38 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76
| method | result | size |
| parallelrisch | \(\frac {\left (3 \left (A +B +C \right ) \sin \left (2 d x +2 c \right )+\left (A +B \right ) \sin \left (3 d x +3 c \right )+\frac {3 A \sin \left (4 d x +4 c \right )}{8}+3 \left (3 A +3 B +4 C \right ) \sin \left (d x +c \right )+\frac {9 x d \left (A +\frac {4 B}{3}+\frac {4 C}{3}\right )}{2}\right ) a}{12 d}\) | \(78\) |
| derivativedivides | \(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {a B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \sin \left (d x +c \right )}{d}\) | \(141\) |
| default | \(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {a B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \sin \left (d x +c \right )}{d}\) | \(141\) |
| risch | \(\frac {3 a A x}{8}+\frac {a B x}{2}+\frac {a x C}{2}+\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) C a}{d}+\frac {a A \sin \left (4 d x +4 c \right )}{32 d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}+\frac {\sin \left (3 d x +3 c \right ) a B}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a B}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a}{4 d}\) | \(151\) |
| norman | \(\frac {\left (\frac {3}{8} a A +\frac {1}{2} a B +\frac {1}{2} C a \right ) x +\left (-\frac {3}{2} a A -2 a B -2 C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {3}{8} a A -\frac {1}{2} a B -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {3}{8} a A -\frac {1}{2} a B -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{4} a A +a B +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{4} a A +a B +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {3}{8} a A +\frac {1}{2} a B +\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\frac {a \left (3 A +4 B +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {a \left (13 A -20 B -36 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}+\frac {a \left (13 A +12 B +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a \left (29 A -4 B +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {a \left (31 A +4 B +36 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {a \left (47 A +20 B -12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\) | \(360\) |
1/12*(3*(A+B+C)*sin(2*d*x+2*c)+(A+B)*sin(3*d*x+3*c)+3/8*A*sin(4*d*x+4*c)+3 *(3*A+3*B+4*C)*sin(d*x+c)+9/2*x*d*(A+4/3*B+4/3*C))*a/d
Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.85 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A + 4 \, B + 4 \, C\right )} a d x + {\left (6 \, A a \cos \left (d x + c\right )^{3} + 8 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 4 \, B + 4 \, C\right )} a \cos \left (d x + c\right ) + 8 \, {\left (2 \, A + 2 \, B + 3 \, C\right )} a\right )} \sin \left (d x + c\right )}{24 \, d} \]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
1/24*(3*(3*A + 4*B + 4*C)*a*d*x + (6*A*a*cos(d*x + c)^3 + 8*(A + B)*a*cos( d*x + c)^2 + 3*(3*A + 4*B + 4*C)*a*cos(d*x + c) + 8*(2*A + 2*B + 3*C)*a)*s in(d*x + c))/d
\[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a \left (\int A \cos ^{4}{\left (c + d x \right )}\, dx + \int A \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
a*(Integral(A*cos(c + d*x)**4, x) + Integral(A*cos(c + d*x)**4*sec(c + d*x ), x) + Integral(B*cos(c + d*x)**4*sec(c + d*x), x) + Integral(B*cos(c + d *x)**4*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**2, x ) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**3, x))
Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.29 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 96 \, C a \sin \left (d x + c\right )}{96 \, d} \]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 3*(12*d*x + 12*c + sin(4 *d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a + 32*(sin(d*x + c)^3 - 3*sin(d*x + c ))*B*a - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a - 24*(2*d*x + 2*c + sin(2 *d*x + 2*c))*C*a - 96*C*a*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (96) = 192\).
Time = 0.31 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.14 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A a + 4 \, B a + 4 \, C a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 49 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 28 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 60 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 31 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 52 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 84 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 39 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
1/24*(3*(3*A*a + 4*B*a + 4*C*a)*(d*x + c) + 2*(9*A*a*tan(1/2*d*x + 1/2*c)^ 7 + 12*B*a*tan(1/2*d*x + 1/2*c)^7 + 12*C*a*tan(1/2*d*x + 1/2*c)^7 + 49*A*a *tan(1/2*d*x + 1/2*c)^5 + 28*B*a*tan(1/2*d*x + 1/2*c)^5 + 60*C*a*tan(1/2*d *x + 1/2*c)^5 + 31*A*a*tan(1/2*d*x + 1/2*c)^3 + 52*B*a*tan(1/2*d*x + 1/2*c )^3 + 84*C*a*tan(1/2*d*x + 1/2*c)^3 + 39*A*a*tan(1/2*d*x + 1/2*c) + 36*B*a *tan(1/2*d*x + 1/2*c) + 36*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c) ^2 + 1)^4)/d
Time = 19.29 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.05 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {3\,A\,a}{4}+B\,a+C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {49\,A\,a}{12}+\frac {7\,B\,a}{3}+5\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {31\,A\,a}{12}+\frac {13\,B\,a}{3}+7\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+3\,B\,a+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,B+4\,C\right )}{4\,\left (\frac {3\,A\,a}{4}+B\,a+C\,a\right )}\right )\,\left (3\,A+4\,B+4\,C\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*((13*A*a)/4 + 3*B*a + 3*C*a) + tan(c/2 + (d*x)/2)^7*(( 3*A*a)/4 + B*a + C*a) + tan(c/2 + (d*x)/2)^3*((31*A*a)/12 + (13*B*a)/3 + 7 *C*a) + tan(c/2 + (d*x)/2)^5*((49*A*a)/12 + (7*B*a)/3 + 5*C*a))/(d*(4*tan( c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c /2 + (d*x)/2)^8 + 1)) + (a*atan((a*tan(c/2 + (d*x)/2)*(3*A + 4*B + 4*C))/( 4*((3*A*a)/4 + B*a + C*a)))*(3*A + 4*B + 4*C))/(4*d)