Integrand size = 29, antiderivative size = 125 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {3}{8} a (A+B) x+\frac {a (5 A+4 B) \sin (c+d x)}{5 d}+\frac {3 a (A+B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (5 A+4 B) \sin ^3(c+d x)}{15 d} \]
3/8*a*(A+B)*x+1/5*a*(5*A+4*B)*sin(d*x+c)/d+3/8*a*(A+B)*cos(d*x+c)*sin(d*x+ c)/d+1/4*a*(A+B)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*a*B*cos(d*x+c)^4*sin(d*x+c) /d-1/15*a*(5*A+4*B)*sin(d*x+c)^3/d
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {a \left (480 (A+B) \sin (c+d x)-160 (A+2 B) \sin ^3(c+d x)+96 B \sin ^5(c+d x)+15 (A+B) (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))\right )}{480 d} \]
(a*(480*(A + B)*Sin[c + d*x] - 160*(A + 2*B)*Sin[c + d*x]^3 + 96*B*Sin[c + d*x]^5 + 15*(A + B)*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)] )))/(480*d)
Time = 0.69 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.96, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 3447, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 3115, 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.
| \(\displaystyle \int \cos ^3(c+d x) (a \cos (c+d x)+a) (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int \cos ^3(c+d x) \left ((a A+a B) \cos (c+d x)+a A+a B \cos ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left ((a A+a B) \sin \left (c+d x+\frac {\pi }{2}\right )+a A+a B \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{5} \int \cos ^3(c+d x) (a (5 A+4 B)+5 a (A+B) \cos (c+d x))dx+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (5 A+4 B)+5 a (A+B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (5 a (A+B) \int \cos ^4(c+d x)dx+a (5 A+4 B) \int \cos ^3(c+d x)dx\right )+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (a (5 A+4 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+5 a (A+B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{5} \left (5 a (A+B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {a (5 A+4 B) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (5 a (A+B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {a (5 A+4 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a (5 A+4 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a (5 A+4 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a (5 A+4 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {a (5 A+4 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
(a*B*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) + (-((a*(5*A + 4*B)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d) + 5*a*(A + B)*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d ) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/5
3.1.1.3.1 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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
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]
Time = 3.46 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.70
| method | result | size |
| parallelrisch | \(\frac {\left (8 \left (A +B \right ) \sin \left (2 d x +2 c \right )+\frac {2 \left (4 A +5 B \right ) \sin \left (3 d x +3 c \right )}{3}+\left (A +B \right ) \sin \left (4 d x +4 c \right )+\frac {2 B \sin \left (5 d x +5 c \right )}{5}+4 \left (6 A +5 B \right ) \sin \left (d x +c \right )+12 \left (A +B \right ) x d \right ) a}{32 d}\) | \(87\) |
| parts | \(\frac {\left (a A +B a \right ) \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {a A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {B a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}\) | \(102\) |
| derivativedivides | \(\frac {\frac {B a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(128\) |
| default | \(\frac {\frac {B a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(128\) |
| risch | \(\frac {3 a x A}{8}+\frac {3 a B x}{8}+\frac {3 \sin \left (d x +c \right ) a A}{4 d}+\frac {5 a B \sin \left (d x +c \right )}{8 d}+\frac {B a \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) a A}{32 d}+\frac {\sin \left (4 d x +4 c \right ) B a}{32 d}+\frac {\sin \left (3 d x +3 c \right ) a A}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) B a}{48 d}+\frac {\sin \left (2 d x +2 c \right ) a A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\) | \(150\) |
| norman | \(\frac {\frac {3 a \left (A +B \right ) x}{8}+\frac {13 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {3 a \left (A +B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {15 a \left (A +B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {15 a \left (A +B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a \left (A +B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a \left (A +B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {3 a \left (A +B \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {4 a \left (25 A +29 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {a \left (29 A +13 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {a \left (35 A +19 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(225\) |
1/32*(8*(A+B)*sin(2*d*x+2*c)+2/3*(4*A+5*B)*sin(3*d*x+3*c)+(A+B)*sin(4*d*x+ 4*c)+2/5*B*sin(5*d*x+5*c)+4*(6*A+5*B)*sin(d*x+c)+12*(A+B)*x*d)*a/d
Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.70 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {45 \, {\left (A + B\right )} a d x + {\left (24 \, B a \cos \left (d x + c\right )^{4} + 30 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{2} + 45 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 16 \, {\left (5 \, A + 4 \, B\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(45*(A + B)*a*d*x + (24*B*a*cos(d*x + c)^4 + 30*(A + B)*a*cos(d*x + c)^3 + 8*(5*A + 4*B)*a*cos(d*x + c)^2 + 45*(A + B)*a*cos(d*x + c) + 16*(5* A + 4*B)*a)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (117) = 234\).
Time = 0.27 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.66 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\begin {cases} \frac {3 A a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 A a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {8 B a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 B a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 B a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right ) \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((3*A*a*x*sin(c + d*x)**4/8 + 3*A*a*x*sin(c + d*x)**2*cos(c + d*x )**2/4 + 3*A*a*x*cos(c + d*x)**4/8 + 3*A*a*sin(c + d*x)**3*cos(c + d*x)/(8 *d) + 2*A*a*sin(c + d*x)**3/(3*d) + 5*A*a*sin(c + d*x)*cos(c + d*x)**3/(8* d) + A*a*sin(c + d*x)*cos(c + d*x)**2/d + 3*B*a*x*sin(c + d*x)**4/8 + 3*B* a*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*B*a*x*cos(c + d*x)**4/8 + 8*B*a* sin(c + d*x)**5/(15*d) + 4*B*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 3*B *a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + B*a*sin(c + d*x)*cos(c + d*x)**4/d + 5*B*a*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(A + B*cos(c))* (a*cos(c) + a)*cos(c)**3, True))
Time = 0.21 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.99 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=-\frac {160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a}{480 \, d} \]
-1/480*(160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 15*(12*d*x + 12*c + si n(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a - 32*(3*sin(d*x + c)^5 - 10*sin(d *x + c)^3 + 15*sin(d*x + c))*B*a - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a)/d
Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {3}{8} \, {\left (A a + B a\right )} x + \frac {B a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (A a + B a\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, A a + 5 \, B a\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a + B a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (6 \, A a + 5 \, B a\right )} \sin \left (d x + c\right )}{8 \, d} \]
3/8*(A*a + B*a)*x + 1/80*B*a*sin(5*d*x + 5*c)/d + 1/32*(A*a + B*a)*sin(4*d *x + 4*c)/d + 1/48*(4*A*a + 5*B*a)*sin(3*d*x + 3*c)/d + 1/4*(A*a + B*a)*si n(2*d*x + 2*c)/d + 1/8*(6*A*a + 5*B*a)*sin(d*x + c)/d
Time = 0.97 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.89 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {29\,A\,a}{6}+\frac {13\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,a}{3}+\frac {116\,B\,a}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {35\,A\,a}{6}+\frac {19\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+\frac {13\,B\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,a\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,\left (A+B\right )}{4\,d}+\frac {3\,a\,\mathrm {atan}\left (\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{4\,\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}\right )}\right )\,\left (A+B\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*((13*A*a)/4 + (13*B*a)/4) + tan(c/2 + (d*x)/2)^9*((3*A *a)/4 + (3*B*a)/4) + tan(c/2 + (d*x)/2)^7*((29*A*a)/6 + (13*B*a)/6) + tan( c/2 + (d*x)/2)^3*((35*A*a)/6 + (19*B*a)/6) + tan(c/2 + (d*x)/2)^5*((20*A*a )/3 + (116*B*a)/15))/(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)) - (3*a*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(A + B))/(4*d) + (3*a*a tan((3*a*tan(c/2 + (d*x)/2)*(A + B))/(4*((3*A*a)/4 + (3*B*a)/4)))*(A + B)) /(4*d)