Integrand size = 31, antiderivative size = 134 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {1}{8} a^2 (5 A+2 B) x-\frac {a^2 (5 A+2 B) \cos ^3(c+d x)}{12 d}+\frac {a^2 (5 A+2 B) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {(5 A+2 B) \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{20 d} \]
1/8*a^2*(5*A+2*B)*x-1/12*a^2*(5*A+2*B)*cos(d*x+c)^3/d+1/8*a^2*(5*A+2*B)*co s(d*x+c)*sin(d*x+c)/d-1/5*B*cos(d*x+c)^3*(a+a*sin(d*x+c))^2/d-1/20*(5*A+2* B)*cos(d*x+c)^3*(a^2+a^2*sin(d*x+c))/d
Time = 0.72 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.99 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {a^2 \cos (c+d x) \left (80 A+62 B+\frac {60 (5 A+2 B) \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )}{\sqrt {\cos ^2(c+d x)}}+8 (10 A+7 B) \cos (2 (c+d x))-6 B \cos (4 (c+d x))-135 A \sin (c+d x)-30 B \sin (c+d x)+15 A \sin (3 (c+d x))+30 B \sin (3 (c+d x))\right )}{240 d} \]
-1/240*(a^2*Cos[c + d*x]*(80*A + 62*B + (60*(5*A + 2*B)*ArcSin[Sqrt[1 - Si n[c + d*x]]/Sqrt[2]])/Sqrt[Cos[c + d*x]^2] + 8*(10*A + 7*B)*Cos[2*(c + d*x )] - 6*B*Cos[4*(c + d*x)] - 135*A*Sin[c + d*x] - 30*B*Sin[c + d*x] + 15*A* Sin[3*(c + d*x)] + 30*B*Sin[3*(c + d*x)]))/d
Time = 0.53 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 3339, 3042, 3157, 3042, 3148, 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.
| \(\displaystyle \int \cos ^2(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^2 (a \sin (c+d x)+a)^2 (A+B \sin (c+d x))dx\) |
| \(\Big \downarrow \) 3339 |
| \(\displaystyle \frac {1}{5} (5 A+2 B) \int \cos ^2(c+d x) (\sin (c+d x) a+a)^2dx-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+2 B) \int \cos (c+d x)^2 (\sin (c+d x) a+a)^2dx-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3157 |
| \(\displaystyle \frac {1}{5} (5 A+2 B) \left (\frac {5}{4} a \int \cos ^2(c+d x) (\sin (c+d x) a+a)dx-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+2 B) \left (\frac {5}{4} a \int \cos (c+d x)^2 (\sin (c+d x) a+a)dx-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {1}{5} (5 A+2 B) \left (\frac {5}{4} a \left (a \int \cos ^2(c+d x)dx-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+2 B) \left (\frac {5}{4} a \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} (5 A+2 B) \left (\frac {5}{4} a \left (a \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} (5 A+2 B) \left (\frac {5}{4} a \left (a \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {a \cos ^3(c+d x)}{3 d}\right )-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}\right )-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
-1/5*(B*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^2)/d + ((5*A + 2*B)*(-1/4*(Cos [c + d*x]^3*(a^2 + a^2*Sin[c + d*x]))/d + (5*a*(-1/3*(a*Cos[c + d*x]^3)/d + a*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d))))/4))/5
3.10.79.3.1 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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && NeQ[m + p, 0] && Integers Q[2*m, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ a^2 - b^2, 0] && NeQ[m + p + 1, 0]
Time = 0.56 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.72
| method | result | size |
| parallelrisch | \(\frac {\left (\left (-\frac {2 A}{3}-\frac {5 B}{12}\right ) \cos \left (3 d x +3 c \right )+\left (-\frac {A}{8}-\frac {B}{4}\right ) \sin \left (4 d x +4 c \right )+\frac {B \cos \left (5 d x +5 c \right )}{20}+A \sin \left (2 d x +2 c \right )+\left (-2 A -\frac {3 B}{2}\right ) \cos \left (d x +c \right )+\frac {5 d x A}{2}+d x B -\frac {8 A}{3}-\frac {28 B}{15}\right ) a^{2}}{4 d}\) | \(96\) |
| risch | \(\frac {5 a^{2} x A}{8}+\frac {a^{2} x B}{4}-\frac {A \,a^{2} \cos \left (d x +c \right )}{2 d}-\frac {3 a^{2} \cos \left (d x +c \right ) B}{8 d}+\frac {a^{2} \cos \left (5 d x +5 c \right ) B}{80 d}-\frac {\sin \left (4 d x +4 c \right ) A \,a^{2}}{32 d}-\frac {\sin \left (4 d x +4 c \right ) B \,a^{2}}{16 d}-\frac {a^{2} \cos \left (3 d x +3 c \right ) A}{6 d}-\frac {5 a^{2} \cos \left (3 d x +3 c \right ) B}{48 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{2}}{4 d}\) | \(154\) |
| derivativedivides | \(\frac {A \,a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+B \,a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-\frac {2 A \,a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+2 B \,a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {B \,a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}}{d}\) | \(182\) |
| default | \(\frac {A \,a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+B \,a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-\frac {2 A \,a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+2 B \,a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {B \,a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}}{d}\) | \(182\) |
| norman | \(\frac {\left (\frac {5}{8} A \,a^{2}+\frac {1}{4} B \,a^{2}\right ) x +\left (\frac {5}{8} A \,a^{2}+\frac {1}{4} B \,a^{2}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {25}{4} A \,a^{2}+\frac {5}{2} B \,a^{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {25}{4} A \,a^{2}+\frac {5}{2} B \,a^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {25}{8} A \,a^{2}+\frac {5}{4} B \,a^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {25}{8} A \,a^{2}+\frac {5}{4} B \,a^{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {20 A \,a^{2}+14 B \,a^{2}}{15 d}-\frac {\left (4 A \,a^{2}+2 B \,a^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (4 A \,a^{2}+4 B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (8 A \,a^{2}+2 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (8 A \,a^{2}+8 B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{2} \left (3 A -2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a^{2} \left (3 A -2 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{2} \left (6 B +7 A \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a^{2} \left (6 B +7 A \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(399\) |
1/4*((-2/3*A-5/12*B)*cos(3*d*x+3*c)+(-1/8*A-1/4*B)*sin(4*d*x+4*c)+1/20*B*c os(5*d*x+5*c)+A*sin(2*d*x+2*c)+(-2*A-3/2*B)*cos(d*x+c)+5/2*d*x*A+d*x*B-8/3 *A-28/15*B)*a^2/d
Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.71 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {24 \, B a^{2} \cos \left (d x + c\right )^{5} - 80 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{3} + 15 \, {\left (5 \, A + 2 \, B\right )} a^{2} d x - 15 \, {\left (2 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} - {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(24*B*a^2*cos(d*x + c)^5 - 80*(A + B)*a^2*cos(d*x + c)^3 + 15*(5*A + 2*B)*a^2*d*x - 15*(2*(A + 2*B)*a^2*cos(d*x + c)^3 - (5*A + 2*B)*a^2*cos(d *x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (117) = 234\).
Time = 0.28 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.77 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {A a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {A a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {A a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {A a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {2 A a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {B a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {B a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {2 B a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {B a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((A*a**2*x*sin(c + d*x)**4/8 + A*a**2*x*sin(c + d*x)**2*cos(c + d *x)**2/4 + A*a**2*x*sin(c + d*x)**2/2 + A*a**2*x*cos(c + d*x)**4/8 + A*a** 2*x*cos(c + d*x)**2/2 + A*a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - A*a**2 *sin(c + d*x)*cos(c + d*x)**3/(8*d) + A*a**2*sin(c + d*x)*cos(c + d*x)/(2* d) - 2*A*a**2*cos(c + d*x)**3/(3*d) + B*a**2*x*sin(c + d*x)**4/4 + B*a**2* x*sin(c + d*x)**2*cos(c + d*x)**2/2 + B*a**2*x*cos(c + d*x)**4/4 + B*a**2* sin(c + d*x)**3*cos(c + d*x)/(4*d) - B*a**2*sin(c + d*x)**2*cos(c + d*x)** 3/(3*d) - B*a**2*sin(c + d*x)*cos(c + d*x)**3/(4*d) - 2*B*a**2*cos(c + d*x )**5/(15*d) - B*a**2*cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(A + B*sin(c))*( a*sin(c) + a)**2*cos(c)**2, True))
Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {320 \, A a^{2} \cos \left (d x + c\right )^{3} + 160 \, B a^{2} \cos \left (d x + c\right )^{3} - 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} A a^{2} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 32 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} B a^{2} - 30 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{2}}{480 \, d} \]
-1/480*(320*A*a^2*cos(d*x + c)^3 + 160*B*a^2*cos(d*x + c)^3 - 15*(4*d*x + 4*c - sin(4*d*x + 4*c))*A*a^2 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 - 32*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*B*a^2 - 30*(4*d*x + 4*c - sin( 4*d*x + 4*c))*B*a^2)/d
Time = 0.35 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.97 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B a^{2} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {A a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {1}{8} \, {\left (5 \, A a^{2} + 2 \, B a^{2}\right )} x - \frac {{\left (8 \, A a^{2} + 5 \, B a^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (4 \, A a^{2} + 3 \, B a^{2}\right )} \cos \left (d x + c\right )}{8 \, d} - \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \]
1/80*B*a^2*cos(5*d*x + 5*c)/d + 1/4*A*a^2*sin(2*d*x + 2*c)/d + 1/8*(5*A*a^ 2 + 2*B*a^2)*x - 1/48*(8*A*a^2 + 5*B*a^2)*cos(3*d*x + 3*c)/d - 1/8*(4*A*a^ 2 + 3*B*a^2)*cos(d*x + c)/d - 1/32*(A*a^2 + 2*B*a^2)*sin(4*d*x + 4*c)/d
Time = 11.17 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.74 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,A+2\,B\right )}{4\,\left (\frac {5\,A\,a^2}{4}+\frac {B\,a^2}{2}\right )}\right )\,\left (5\,A+2\,B\right )}{4\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d}-\frac {\frac {4\,A\,a^2}{3}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,a^2}{4}-\frac {B\,a^2}{2}\right )+\frac {14\,B\,a^2}{15}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (4\,A\,a^2+2\,B\,a^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {7\,A\,a^2}{2}+3\,B\,a^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {7\,A\,a^2}{2}+3\,B\,a^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {3\,A\,a^2}{4}-\frac {B\,a^2}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (8\,A\,a^2+8\,B\,a^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {8\,A\,a^2}{3}+\frac {8\,B\,a^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {16\,A\,a^2}{3}+\frac {4\,B\,a^2}{3}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(a^2*atan((a^2*tan(c/2 + (d*x)/2)*(5*A + 2*B))/(4*((5*A*a^2)/4 + (B*a^2)/2 )))*(5*A + 2*B))/(4*d) - (a^2*(5*A + 2*B)*(atan(tan(c/2 + (d*x)/2)) - (d*x )/2))/(4*d) - ((4*A*a^2)/3 - tan(c/2 + (d*x)/2)*((3*A*a^2)/4 - (B*a^2)/2) + (14*B*a^2)/15 + tan(c/2 + (d*x)/2)^8*(4*A*a^2 + 2*B*a^2) - tan(c/2 + (d* x)/2)^3*((7*A*a^2)/2 + 3*B*a^2) + tan(c/2 + (d*x)/2)^7*((7*A*a^2)/2 + 3*B* a^2) + tan(c/2 + (d*x)/2)^9*((3*A*a^2)/4 - (B*a^2)/2) + tan(c/2 + (d*x)/2) ^6*(8*A*a^2 + 8*B*a^2) + tan(c/2 + (d*x)/2)^2*((8*A*a^2)/3 + (8*B*a^2)/3) + tan(c/2 + (d*x)/2)^4*((16*A*a^2)/3 + (4*B*a^2)/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))