Integrand size = 31, antiderivative size = 131 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} b (3 A+4 C) x+\frac {a (4 A+5 C) \sin (c+d x)}{5 d}+\frac {b (3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A b \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 C) \sin ^3(c+d x)}{15 d} \]
1/8*b*(3*A+4*C)*x+1/5*a*(4*A+5*C)*sin(d*x+c)/d+1/8*b*(3*A+4*C)*cos(d*x+c)* sin(d*x+c)/d+1/4*A*b*cos(d*x+c)^3*sin(d*x+c)/d+1/5*a*A*cos(d*x+c)^4*sin(d* x+c)/d-1/15*a*(4*A+5*C)*sin(d*x+c)^3/d
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {480 a (A+C) \sin (c+d x)-160 a (2 A+C) \sin ^3(c+d x)+96 a A \sin ^5(c+d x)+15 b (4 (3 A+4 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+A \sin (4 (c+d x)))}{480 d} \]
(480*a*(A + C)*Sin[c + d*x] - 160*a*(2*A + C)*Sin[c + d*x]^3 + 96*a*A*Sin[ c + d*x]^5 + 15*b*(4*(3*A + 4*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d*x)] + A*Sin[4*(c + d*x)]))/(480*d)
Time = 0.63 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.94, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4563, 25, 3042, 4535, 3042, 3113, 2009, 4533, 3042, 3115, 24}
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.
\(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
\(\Big \downarrow \) 4563 |
\(\displaystyle \frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}-\frac {1}{5} \int -\cos ^4(c+d x) \left (5 b C \sec ^2(c+d x)+a (4 A+5 C) \sec (c+d x)+5 A b\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) \left (5 b C \sec ^2(c+d x)+a (4 A+5 C) \sec (c+d x)+5 A b\right )dx+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {5 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )+5 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \frac {1}{5} \left (a (4 A+5 C) \int \cos ^3(c+d x)dx+\int \cos ^4(c+d x) \left (5 b C \sec ^2(c+d x)+5 A b\right )dx\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (a (4 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+\int \frac {5 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \frac {1}{5} \left (\int \frac {5 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {a (4 A+5 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\int \frac {5 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle \frac {1}{5} \left (\frac {5}{4} b (3 A+4 C) \int \cos ^2(c+d x)dx-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 A b \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {5}{4} b (3 A+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 A b \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{5} \left (\frac {5}{4} b (3 A+4 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 A b \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{5} \left (-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5}{4} b (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )+\frac {5 A b \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
(a*A*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) + ((5*A*b*Cos[c + d*x]^3*Sin[c + d *x])/(4*d) + (5*b*(3*A + 4*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4 - (a*(4*A + 5*C)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d)/5
3.7.45.3.1 Defintions of rubi rules used
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
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[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
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_)]^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] + Simp[1/(d*n) Int[(d*Csc[e + f*x] )^(n + 1)*Simp[A*b*n + a*(C*n + 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, C}, x] && LtQ[n, -1]
Time = 0.65 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68
method | result | size |
parallelrisch | \(\frac {120 b \left (A +C \right ) \sin \left (2 d x +2 c \right )+50 a \left (A +\frac {4 C}{5}\right ) \sin \left (3 d x +3 c \right )+15 A b \sin \left (4 d x +4 c \right )+6 a A \sin \left (5 d x +5 c \right )+300 \left (A +\frac {6 C}{5}\right ) a \sin \left (d x +c \right )+180 \left (A +\frac {4 C}{3}\right ) x b d}{480 d}\) | \(89\) |
derivativedivides | \(\frac {\frac {a A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+A b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(117\) |
default | \(\frac {\frac {a A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+A b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(117\) |
risch | \(\frac {3 A b x}{8}+\frac {b x C}{2}+\frac {5 a A \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) C a}{4 d}+\frac {a A \sin \left (5 d x +5 c \right )}{80 d}+\frac {A b \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 a A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) C a}{12 d}+\frac {A b \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C b}{4 d}\) | \(134\) |
norman | \(\frac {\left (\frac {3}{8} A b +\frac {1}{2} C b \right ) x +\left (-\frac {15}{8} A b -\frac {5}{2} C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {15}{8} A b -\frac {5}{2} C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{8} A b +\frac {1}{2} C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {3}{8} A b +\frac {1}{2} C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {3}{8} A b +\frac {1}{2} C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {9}{8} A b +\frac {3}{2} C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {9}{8} A b +\frac {3}{2} C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-\frac {2 \left (2 a A -3 A b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {2 \left (2 a A +3 A b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {\left (8 a A -5 A b +8 C a -4 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {\left (8 a A +5 A b +8 C a +4 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (88 a A -5 A b -40 C a +60 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 d}+\frac {\left (88 a A +5 A b -40 C a -60 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}-\frac {8 a \left (19 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\) | \(410\) |
1/480*(120*b*(A+C)*sin(2*d*x+2*c)+50*a*(A+4/5*C)*sin(3*d*x+3*c)+15*A*b*sin (4*d*x+4*c)+6*a*A*sin(5*d*x+5*c)+300*(A+6/5*C)*a*sin(d*x+c)+180*(A+4/3*C)* x*b*d)/d
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, A + 4 \, C\right )} b d x + {\left (24 \, A a \cos \left (d x + c\right )^{4} + 30 \, A b \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A + 4 \, C\right )} b \cos \left (d x + c\right ) + 16 \, {\left (4 \, A + 5 \, C\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(15*(3*A + 4*C)*b*d*x + (24*A*a*cos(d*x + c)^4 + 30*A*b*cos(d*x + c) ^3 + 8*(4*A + 5*C)*a*cos(d*x + c)^2 + 15*(3*A + 4*C)*b*cos(d*x + c) + 16*( 4*A + 5*C)*a)*sin(d*x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b}{480 \, d} \]
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a - 1 60*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*b + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*b )/d
Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (119) = 238\).
Time = 0.32 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.31 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, A b + 4 \, C b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]
1/120*(15*(3*A*b + 4*C*b)*(d*x + c) + 2*(120*A*a*tan(1/2*d*x + 1/2*c)^9 + 120*C*a*tan(1/2*d*x + 1/2*c)^9 - 75*A*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*b*ta n(1/2*d*x + 1/2*c)^9 + 160*A*a*tan(1/2*d*x + 1/2*c)^7 + 320*C*a*tan(1/2*d* x + 1/2*c)^7 - 30*A*b*tan(1/2*d*x + 1/2*c)^7 - 120*C*b*tan(1/2*d*x + 1/2*c )^7 + 464*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*C*a*tan(1/2*d*x + 1/2*c)^5 + 16 0*A*a*tan(1/2*d*x + 1/2*c)^3 + 320*C*a*tan(1/2*d*x + 1/2*c)^3 + 30*A*b*tan (1/2*d*x + 1/2*c)^3 + 120*C*b*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*tan(1/2*d*x + 1/2*c) + 120*C*a*tan(1/2*d*x + 1/2*c) + 75*A*b*tan(1/2*d*x + 1/2*c) + 6 0*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 18.18 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.91 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (2\,A\,a-\frac {5\,A\,b}{4}+2\,C\,a-C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,A\,a}{3}-\frac {A\,b}{2}+\frac {16\,C\,a}{3}-2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,A\,a}{3}+\frac {A\,b}{2}+\frac {16\,C\,a}{3}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+\frac {5\,A\,b}{4}+2\,C\,a+C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,C\right )}{4\,\left (\frac {3\,A\,b}{4}+C\,b\right )}\right )\,\left (3\,A+4\,C\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*(2*A*a + (5*A*b)/4 + 2*C*a + C*b) + tan(c/2 + (d*x)/2) ^5*((116*A*a)/15 + (20*C*a)/3) + tan(c/2 + (d*x)/2)^9*(2*A*a - (5*A*b)/4 + 2*C*a - C*b) + tan(c/2 + (d*x)/2)^3*((8*A*a)/3 + (A*b)/2 + (16*C*a)/3 + 2 *C*b) + tan(c/2 + (d*x)/2)^7*((8*A*a)/3 - (A*b)/2 + (16*C*a)/3 - 2*C*b))/( d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2 )^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (b*atan((b*ta n(c/2 + (d*x)/2)*(3*A + 4*C))/(4*((3*A*b)/4 + C*b)))*(3*A + 4*C))/(4*d)