Integrand size = 31, antiderivative size = 78 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {2 (A-B) (a+a \sin (c+d x))^5}{5 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^6}{6 a^3 d}-\frac {B (a+a \sin (c+d x))^7}{7 a^4 d} \]
2/5*(A-B)*(a+a*sin(d*x+c))^5/a^2/d-1/6*(A-3*B)*(a+a*sin(d*x+c))^6/a^3/d-1/ 7*B*(a+a*sin(d*x+c))^7/a^4/d
Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.68 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {a^3 (1+\sin (c+d x))^5 \left (-49 A+9 B+5 (7 A-9 B) \sin (c+d x)+30 B \sin ^2(c+d x)\right )}{210 d} \]
-1/210*(a^3*(1 + Sin[c + d*x])^5*(-49*A + 9*B + 5*(7*A - 9*B)*Sin[c + d*x] + 30*B*Sin[c + d*x]^2))/d
Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 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.
| \(\displaystyle \int \cos ^3(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^3 (a \sin (c+d x)+a)^3 (A+B \sin (c+d x))dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \frac {(a-a \sin (c+d x)) (\sin (c+d x) a+a)^4 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (a-a \sin (c+d x)) (\sin (c+d x) a+a)^4 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^4 d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\int \left (-B (\sin (c+d x) a+a)^6-a (A-3 B) (\sin (c+d x) a+a)^5+2 a^2 (A-B) (\sin (c+d x) a+a)^4\right )d(a \sin (c+d x))}{a^4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2}{5} a^2 (A-B) (a \sin (c+d x)+a)^5-\frac {1}{6} a (A-3 B) (a \sin (c+d x)+a)^6-\frac {1}{7} B (a \sin (c+d x)+a)^7}{a^4 d}\) |
((2*a^2*(A - B)*(a + a*Sin[c + d*x])^5)/5 - (a*(A - 3*B)*(a + a*Sin[c + d* x])^6)/6 - (B*(a + a*Sin[c + d*x])^7)/7)/(a^4*d)
3.10.88.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.68 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.45
| method | result | size |
| derivativedivides | \(-\frac {a^{3} \left (\frac {\left (\sin ^{7}\left (d x +c \right )\right ) B}{7}+\frac {\left (A +3 B \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (3 A +2 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-2 B +2 A \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-3 B -2 A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-3 A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\) | \(113\) |
| default | \(-\frac {a^{3} \left (\frac {\left (\sin ^{7}\left (d x +c \right )\right ) B}{7}+\frac {\left (A +3 B \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (3 A +2 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-2 B +2 A \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-3 B -2 A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-3 A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\) | \(113\) |
| parallelrisch | \(\frac {9 \left (\left (-\frac {3 A}{8}-\frac {17 B}{72}\right ) \cos \left (2 d x +2 c \right )+\left (-\frac {A}{12}-\frac {B}{36}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {A}{216}+\frac {B}{72}\right ) \cos \left (6 d x +6 c \right )+\left (\frac {A}{54}-\frac {5 B}{72}\right ) \sin \left (3 d x +3 c \right )+\left (-\frac {A}{30}-\frac {13 B}{360}\right ) \sin \left (5 d x +5 c \right )+\frac {B \sin \left (7 d x +7 c \right )}{504}+\left (A +\frac {3 B}{8}\right ) \sin \left (d x +c \right )+\frac {49 A}{108}+\frac {B}{4}\right ) a^{3}}{8 d}\) | \(125\) |
| risch | \(\frac {9 \sin \left (d x +c \right ) A \,a^{3}}{8 d}+\frac {27 a^{3} B \sin \left (d x +c \right )}{64 d}+\frac {\sin \left (7 d x +7 c \right ) B \,a^{3}}{448 d}+\frac {a^{3} \cos \left (6 d x +6 c \right ) A}{192 d}+\frac {a^{3} \cos \left (6 d x +6 c \right ) B}{64 d}-\frac {3 \sin \left (5 d x +5 c \right ) A \,a^{3}}{80 d}-\frac {13 \sin \left (5 d x +5 c \right ) B \,a^{3}}{320 d}-\frac {3 a^{3} \cos \left (4 d x +4 c \right ) A}{32 d}-\frac {a^{3} \cos \left (4 d x +4 c \right ) B}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{3}}{48 d}-\frac {5 \sin \left (3 d x +3 c \right ) B \,a^{3}}{64 d}-\frac {27 a^{3} \cos \left (2 d x +2 c \right ) A}{64 d}-\frac {17 a^{3} \cos \left (2 d x +2 c \right ) B}{64 d}\) | \(230\) |
| norman | \(\frac {\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (22 A \,a^{3}+18 B \,a^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (22 A \,a^{3}+18 B \,a^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (76 A \,a^{3}+36 B \,a^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (76 A \,a^{3}+36 B \,a^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \left (13 A +6 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 a^{3} \left (13 A +6 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {24 a^{3} \left (49 A +6 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {2 a^{3} \left (241 A +144 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 a^{3} \left (241 A +144 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}\) | \(356\) |
-a^3/d*(1/7*sin(d*x+c)^7*B+1/6*(A+3*B)*sin(d*x+c)^6+1/5*(3*A+2*B)*sin(d*x+ c)^5+1/4*(-2*B+2*A)*sin(d*x+c)^4+1/3*(-3*B-2*A)*sin(d*x+c)^3+1/2*(-3*A-B)* sin(d*x+c)^2-A*sin(d*x+c))
Time = 0.30 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.47 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {35 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} - 210 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 2 \, {\left (15 \, B a^{3} \cos \left (d x + c\right )^{6} - 3 \, {\left (21 \, A + 29 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 8 \, {\left (7 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 16 \, {\left (7 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{210 \, d} \]
1/210*(35*(A + 3*B)*a^3*cos(d*x + c)^6 - 210*(A + B)*a^3*cos(d*x + c)^4 + 2*(15*B*a^3*cos(d*x + c)^6 - 3*(21*A + 29*B)*a^3*cos(d*x + c)^4 + 8*(7*A + 3*B)*a^3*cos(d*x + c)^2 + 16*(7*A + 3*B)*a^3)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (70) = 140\).
Time = 0.48 (sec) , antiderivative size = 313, normalized size of antiderivative = 4.01 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {2 A a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac {A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A a^{3} \cos ^{6}{\left (c + d x \right )}}{12 d} - \frac {3 A a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 B a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {3 B a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {B a^{3} \cos ^{6}{\left (c + d x \right )}}{4 d} - \frac {B a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((2*A*a**3*sin(c + d*x)**5/(5*d) + A*a**3*sin(c + d*x)**3*cos(c + d*x)**2/d + 2*A*a**3*sin(c + d*x)**3/(3*d) - A*a**3*sin(c + d*x)**2*cos(c + d*x)**4/(4*d) + A*a**3*sin(c + d*x)*cos(c + d*x)**2/d - A*a**3*cos(c + d*x)**6/(12*d) - 3*A*a**3*cos(c + d*x)**4/(4*d) + 2*B*a**3*sin(c + d*x)**7 /(35*d) + B*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 2*B*a**3*sin(c + d*x)**5/(5*d) + B*a**3*sin(c + d*x)**3*cos(c + d*x)**2/d - 3*B*a**3*sin(c + d*x)**2*cos(c + d*x)**4/(4*d) - B*a**3*cos(c + d*x)**6/(4*d) - B*a**3*co s(c + d*x)**4/(4*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**3*cos(c) **3, True))
Time = 0.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.62 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {30 \, B a^{3} \sin \left (d x + c\right )^{7} + 35 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{6} + 42 \, {\left (3 \, A + 2 \, B\right )} a^{3} \sin \left (d x + c\right )^{5} + 105 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{4} - 70 \, {\left (2 \, A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} - 105 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} - 210 \, A a^{3} \sin \left (d x + c\right )}{210 \, d} \]
-1/210*(30*B*a^3*sin(d*x + c)^7 + 35*(A + 3*B)*a^3*sin(d*x + c)^6 + 42*(3* A + 2*B)*a^3*sin(d*x + c)^5 + 105*(A - B)*a^3*sin(d*x + c)^4 - 70*(2*A + 3 *B)*a^3*sin(d*x + c)^3 - 105*(3*A + B)*a^3*sin(d*x + c)^2 - 210*A*a^3*sin( d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (72) = 144\).
Time = 0.41 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.21 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {30 \, B a^{3} \sin \left (d x + c\right )^{7} + 35 \, A a^{3} \sin \left (d x + c\right )^{6} + 105 \, B a^{3} \sin \left (d x + c\right )^{6} + 126 \, A a^{3} \sin \left (d x + c\right )^{5} + 84 \, B a^{3} \sin \left (d x + c\right )^{5} + 105 \, A a^{3} \sin \left (d x + c\right )^{4} - 105 \, B a^{3} \sin \left (d x + c\right )^{4} - 140 \, A a^{3} \sin \left (d x + c\right )^{3} - 210 \, B a^{3} \sin \left (d x + c\right )^{3} - 315 \, A a^{3} \sin \left (d x + c\right )^{2} - 105 \, B a^{3} \sin \left (d x + c\right )^{2} - 210 \, A a^{3} \sin \left (d x + c\right )}{210 \, d} \]
-1/210*(30*B*a^3*sin(d*x + c)^7 + 35*A*a^3*sin(d*x + c)^6 + 105*B*a^3*sin( d*x + c)^6 + 126*A*a^3*sin(d*x + c)^5 + 84*B*a^3*sin(d*x + c)^5 + 105*A*a^ 3*sin(d*x + c)^4 - 105*B*a^3*sin(d*x + c)^4 - 140*A*a^3*sin(d*x + c)^3 - 2 10*B*a^3*sin(d*x + c)^3 - 315*A*a^3*sin(d*x + c)^2 - 105*B*a^3*sin(d*x + c )^2 - 210*A*a^3*sin(d*x + c))/d
Time = 9.62 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.62 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^4\,\left (A-B\right )}{2}-\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}+\frac {a^3\,{\sin \left (c+d\,x\right )}^6\,\left (A+3\,B\right )}{6}+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a^3\,{\sin \left (c+d\,x\right )}^3\,\left (2\,A+3\,B\right )}{3}+\frac {a^3\,{\sin \left (c+d\,x\right )}^5\,\left (3\,A+2\,B\right )}{5}-A\,a^3\,\sin \left (c+d\,x\right )}{d} \]