Integrand size = 31, antiderivative size = 131 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} a (4 A+3 C) x+\frac {b (5 A+4 C) \sin (c+d x)}{5 d}+\frac {a (4 A+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a C \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {b C \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {b (5 A+4 C) \sin ^3(c+d x)}{15 d} \]
1/8*a*(4*A+3*C)*x+1/5*b*(5*A+4*C)*sin(d*x+c)/d+1/8*a*(4*A+3*C)*cos(d*x+c)* sin(d*x+c)/d+1/4*a*C*cos(d*x+c)^3*sin(d*x+c)/d+1/5*b*C*cos(d*x+c)^4*sin(d* x+c)/d-1/15*b*(5*A+4*C)*sin(d*x+c)^3/d
Time = 0.32 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {480 b (A+C) \sin (c+d x)-160 b (A+2 C) \sin ^3(c+d x)+96 b C \sin ^5(c+d x)+15 a (4 (4 A+3 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+C \sin (4 (c+d x)))}{480 d} \]
(480*b*(A + C)*Sin[c + d*x] - 160*b*(A + 2*C)*Sin[c + d*x]^3 + 96*b*C*Sin[ c + d*x]^5 + 15*a*(4*(4*A + 3*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d*x)] + C*Sin[4*(c + d*x)]))/(480*d)
Time = 0.65 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3513, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 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.
| \(\displaystyle \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3513 |
| \(\displaystyle \frac {1}{5} \int \cos ^2(c+d x) \left (5 a C \cos ^2(c+d x)+b (5 A+4 C) \cos (c+d x)+5 a A\right )dx+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (5 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a A\right )dx+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \cos ^2(c+d x) (5 a (4 A+3 C)+4 b (5 A+4 C) \cos (c+d x))dx+\frac {5 a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (5 a (4 A+3 C)+4 b (5 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {5 a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a (4 A+3 C) \int \cos ^2(c+d x)dx+4 b (5 A+4 C) \int \cos ^3(c+d x)dx\right )+\frac {5 a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a (4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+4 b (5 A+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {5 a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a (4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 b (5 A+4 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {5 a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a (4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 b (5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {5 a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a (4 A+3 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 b (5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {5 a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a (4 A+3 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {4 b (5 A+4 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {5 a C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
(b*C*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) + ((5*a*C*Cos[c + d*x]^3*Sin[c + d *x])/(4*d) + (5*a*(4*A + 3*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (4*b*(5*A + 4*C)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d)/4)/5
3.6.22.3.1 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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ (-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3) )), x] + Simp[1/(b*(m + 3)) Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c *(m + 3) + b*d*(C*(m + 2) + A*(m + 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Time = 5.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68
| method | result | size |
| parallelrisch | \(\frac {120 a \left (A +C \right ) \sin \left (2 d x +2 c \right )+40 b \left (A +\frac {5 C}{4}\right ) \sin \left (3 d x +3 c \right )+15 \sin \left (4 d x +4 c \right ) a C +6 C b \sin \left (5 d x +5 c \right )+360 b \left (A +\frac {5 C}{6}\right ) \sin \left (d x +c \right )+240 x \left (A +\frac {3 C}{4}\right ) d a}{480 d}\) | \(89\) |
| derivativedivides | \(\frac {\frac {C b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(117\) |
| default | \(\frac {\frac {C b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(117\) |
| parts | \(\frac {a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {A b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {C b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}\) | \(125\) |
| risch | \(\frac {a x A}{2}+\frac {3 a C x}{8}+\frac {3 \sin \left (d x +c \right ) A b}{4 d}+\frac {5 b C \sin \left (d x +c \right )}{8 d}+\frac {C b \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) a C}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A b}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) C b}{48 d}+\frac {\sin \left (2 d x +2 c \right ) a A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a C}{4 d}\) | \(134\) |
| norman | \(\frac {\left (\frac {1}{2} a A +\frac {3}{8} a C \right ) x +\left (5 a A +\frac {15}{4} a C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (5 a A +\frac {15}{4} a C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a A +\frac {3}{8} a C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} a A +\frac {15}{8} a C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} a A +\frac {15}{8} a C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\left (4 a A -8 A b +5 a C -8 C b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 a A +8 A b +5 a C +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 a A -32 A b +3 a C -16 C b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (12 a A +32 A b +3 a C +16 C b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {4 b \left (25 A +29 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(293\) |
1/480*(120*a*(A+C)*sin(2*d*x+2*c)+40*b*(A+5/4*C)*sin(3*d*x+3*c)+15*sin(4*d *x+4*c)*a*C+6*C*b*sin(5*d*x+5*c)+360*b*(A+5/6*C)*sin(d*x+c)+240*x*(A+3/4*C )*d*a)/d
Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A + 3 \, C\right )} a d x + {\left (24 \, C b \cos \left (d x + c\right )^{4} + 30 \, C a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right ) + 16 \, {\left (5 \, A + 4 \, C\right )} b\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(15*(4*A + 3*C)*a*d*x + (24*C*b*cos(d*x + c)^4 + 30*C*a*cos(d*x + c) ^3 + 8*(5*A + 4*C)*b*cos(d*x + c)^2 + 15*(4*A + 3*C)*a*cos(d*x + c) + 16*( 5*A + 4*C)*b)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (119) = 238\).
Time = 0.25 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.13 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 C b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C b \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((A*a*x*sin(c + d*x)**2/2 + A*a*x*cos(c + d*x)**2/2 + A*a*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*A*b*sin(c + d*x)**3/(3*d) + A*b*sin(c + d*x)* cos(c + d*x)**2/d + 3*C*a*x*sin(c + d*x)**4/8 + 3*C*a*x*sin(c + d*x)**2*co s(c + d*x)**2/4 + 3*C*a*x*cos(c + d*x)**4/8 + 3*C*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*C*a*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 8*C*b*sin(c + d* x)**5/(15*d) + 4*C*b*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + C*b*sin(c + d *x)*cos(c + d*x)**4/d, Ne(d, 0)), (x*(A + C*cos(c)**2)*(a + b*cos(c))*cos( c)**2, True))
Time = 0.20 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b}{480 \, d} \]
1/480*(120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a + 15*(12*d*x + 12*c + sin( 4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b + 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*b )/d
Time = 0.33 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.83 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} \, {\left (4 \, A a + 3 \, C a\right )} x + \frac {C b \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {C a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, A b + 5 \, C b\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a + C a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (6 \, A b + 5 \, C b\right )} \sin \left (d x + c\right )}{8 \, d} \]
1/8*(4*A*a + 3*C*a)*x + 1/80*C*b*sin(5*d*x + 5*c)/d + 1/32*C*a*sin(4*d*x + 4*c)/d + 1/48*(4*A*b + 5*C*b)*sin(3*d*x + 3*c)/d + 1/4*(A*a + C*a)*sin(2* d*x + 2*c)/d + 1/8*(6*A*b + 5*C*b)*sin(d*x + c)/d
Time = 2.77 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.13 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\left (2\,A\,b-A\,a-\frac {5\,C\,a}{4}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {16\,A\,b}{3}-2\,A\,a-\frac {C\,a}{2}+\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,b}{3}+\frac {116\,C\,b}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (2\,A\,a+\frac {16\,A\,b}{3}+\frac {C\,a}{2}+\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a+2\,A\,b+\frac {5\,C\,a}{4}+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+3\,C\right )}{4\,\left (A\,a+\frac {3\,C\,a}{4}\right )}\right )\,\left (4\,A+3\,C\right )}{4\,d}-\frac {a\,\left (4\,A+3\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*(A*a + 2*A*b + (5*C*a)/4 + 2*C*b) + tan(c/2 + (d*x)/2) ^5*((20*A*b)/3 + (116*C*b)/15) - tan(c/2 + (d*x)/2)^9*(A*a - 2*A*b + (5*C* a)/4 - 2*C*b) + tan(c/2 + (d*x)/2)^3*(2*A*a + (16*A*b)/3 + (C*a)/2 + (8*C* b)/3) - tan(c/2 + (d*x)/2)^7*(2*A*a - (16*A*b)/3 + (C*a)/2 - (8*C*b)/3))/( 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)) + (a*atan((a*ta n(c/2 + (d*x)/2)*(4*A + 3*C))/(4*(A*a + (3*C*a)/4)))*(4*A + 3*C))/(4*d) - (a*(4*A + 3*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)