Integrand size = 31, antiderivative size = 105 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {4 (A-B) (a+a \sin (c+d x))^5}{5 a^3 d}-\frac {2 (A-2 B) (a+a \sin (c+d x))^6}{3 a^4 d}+\frac {(A-5 B) (a+a \sin (c+d x))^7}{7 a^5 d}+\frac {B (a+a \sin (c+d x))^8}{8 a^6 d} \]
4/5*(A-B)*(a+a*sin(d*x+c))^5/a^3/d-2/3*(A-2*B)*(a+a*sin(d*x+c))^6/a^4/d+1/ 7*(A-5*B)*(a+a*sin(d*x+c))^7/a^5/d+1/8*B*(a+a*sin(d*x+c))^8/a^6/d
Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (1+\sin (c+d x))^5 \left (232 A-47 B-5 (64 A-47 B) \sin (c+d x)+15 (8 A-19 B) \sin ^2(c+d x)+105 B \sin ^3(c+d x)\right )}{840 d} \]
(a^2*(1 + Sin[c + d*x])^5*(232*A - 47*B - 5*(64*A - 47*B)*Sin[c + d*x] + 1 5*(8*A - 19*B)*Sin[c + d*x]^2 + 105*B*Sin[c + d*x]^3))/(840*d)
Time = 0.33 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90, 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.
| \(\displaystyle \int \cos ^5(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^5 (a \sin (c+d x)+a)^2 (A+B \sin (c+d x))dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^6 d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\int \left (B (\sin (c+d x) a+a)^7+a (A-5 B) (\sin (c+d x) a+a)^6-4 a^2 (A-2 B) (\sin (c+d x) a+a)^5+4 a^3 (A-B) (\sin (c+d x) a+a)^4\right )d(a \sin (c+d x))}{a^6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {4}{5} a^3 (A-B) (a \sin (c+d x)+a)^5-\frac {2}{3} a^2 (A-2 B) (a \sin (c+d x)+a)^6+\frac {1}{7} a (A-5 B) (a \sin (c+d x)+a)^7+\frac {1}{8} B (a \sin (c+d x)+a)^8}{a^6 d}\) |
((4*a^3*(A - B)*(a + a*Sin[c + d*x])^5)/5 - (2*a^2*(A - 2*B)*(a + a*Sin[c + d*x])^6)/3 + (a*(A - 5*B)*(a + a*Sin[c + d*x])^7)/7 + (B*(a + a*Sin[c + d*x])^8)/8)/(a^6*d)
3.10.70.3.1 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 = 0.74 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.20
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\left (\sin ^{8}\left (d x +c \right )\right ) B}{8}+\frac {\left (A +2 B \right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (2 A -B \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (-4 B -A \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-4 A -B \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (2 B -A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (2 A +B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )\right )}{d}\) | \(126\) |
| default | \(\frac {a^{2} \left (\frac {\left (\sin ^{8}\left (d x +c \right )\right ) B}{8}+\frac {\left (A +2 B \right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (2 A -B \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (-4 B -A \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-4 A -B \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (2 B -A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (2 A +B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )\right )}{d}\) | \(126\) |
| parallelrisch | \(\frac {45 a^{2} \left (\left (-\frac {2 A}{9}-\frac {13 B}{90}\right ) \cos \left (2 d x +2 c \right )+\left (-\frac {4 A}{45}-\frac {B}{20}\right ) \cos \left (4 d x +4 c \right )+\left (-\frac {2 A}{135}-\frac {B}{270}\right ) \cos \left (6 d x +6 c \right )+\left (\frac {19 A}{135}-\frac {2 B}{135}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {A}{225}-\frac {2 B}{75}\right ) \sin \left (5 d x +5 c \right )+\left (-\frac {A}{315}-\frac {2 B}{315}\right ) \sin \left (7 d x +7 c \right )+\frac {B \cos \left (8 d x +8 c \right )}{720}+\left (A +\frac {2 B}{9}\right ) \sin \left (d x +c \right )+\frac {44 A}{135}+\frac {85 B}{432}\right )}{64 d}\) | \(142\) |
| risch | \(\frac {45 \sin \left (d x +c \right ) A \,a^{2}}{64 d}+\frac {5 \sin \left (d x +c \right ) B \,a^{2}}{32 d}+\frac {B \,a^{2} \cos \left (8 d x +8 c \right )}{1024 d}-\frac {\sin \left (7 d x +7 c \right ) A \,a^{2}}{448 d}-\frac {\sin \left (7 d x +7 c \right ) B \,a^{2}}{224 d}-\frac {a^{2} \cos \left (6 d x +6 c \right ) A}{96 d}-\frac {a^{2} \cos \left (6 d x +6 c \right ) B}{384 d}+\frac {\sin \left (5 d x +5 c \right ) A \,a^{2}}{320 d}-\frac {3 \sin \left (5 d x +5 c \right ) B \,a^{2}}{160 d}-\frac {a^{2} \cos \left (4 d x +4 c \right ) A}{16 d}-\frac {9 a^{2} \cos \left (4 d x +4 c \right ) B}{256 d}+\frac {19 A \,a^{2} \sin \left (3 d x +3 c \right )}{192 d}-\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{96 d}-\frac {5 a^{2} \cos \left (2 d x +2 c \right ) A}{32 d}-\frac {13 a^{2} \cos \left (2 d x +2 c \right ) B}{128 d}\) | \(266\) |
| norman | \(\frac {\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (2 A \,a^{2}+2 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (2 A \,a^{2}+2 B \,a^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 \left (8 A \,a^{2}+8 B \,a^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (26 A \,a^{2}+5 B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (26 A \,a^{2}+5 B \,a^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{2} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (17 A +8 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{2} \left (17 A +8 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{2} \left (167 A +8 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 a^{2} \left (167 A +8 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 a^{2} \left (2227 A +688 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {2 a^{2} \left (2227 A +688 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}\) | \(413\) |
a^2/d*(1/8*sin(d*x+c)^8*B+1/7*(A+2*B)*sin(d*x+c)^7+1/6*(2*A-B)*sin(d*x+c)^ 6+1/5*(-4*B-A)*sin(d*x+c)^5+1/4*(-4*A-B)*sin(d*x+c)^4+1/3*(2*B-A)*sin(d*x+ c)^3+1/2*(2*A+B)*sin(d*x+c)^2+A*sin(d*x+c))
Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.04 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {105 \, B a^{2} \cos \left (d x + c\right )^{8} - 280 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{6} - 8 \, {\left (15 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 6 \, {\left (4 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{4} - 8 \, {\left (4 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 16 \, {\left (4 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{840 \, d} \]
1/840*(105*B*a^2*cos(d*x + c)^8 - 280*(A + B)*a^2*cos(d*x + c)^6 - 8*(15*( A + 2*B)*a^2*cos(d*x + c)^6 - 6*(4*A + B)*a^2*cos(d*x + c)^4 - 8*(4*A + B) *a^2*cos(d*x + c)^2 - 16*(4*A + B)*a^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (99) = 198\).
Time = 0.67 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.94 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {8 A a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} + \frac {4 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A a^{2} \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {16 B a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {8 B a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {B a^{2} \cos ^{8}{\left (c + d x \right )}}{24 d} - \frac {B a^{2} \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((8*A*a**2*sin(c + d*x)**7/(105*d) + 4*A*a**2*sin(c + d*x)**5*cos (c + d*x)**2/(15*d) + 8*A*a**2*sin(c + d*x)**5/(15*d) + A*a**2*sin(c + d*x )**3*cos(c + d*x)**4/(3*d) + 4*A*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d ) + A*a**2*sin(c + d*x)*cos(c + d*x)**4/d - A*a**2*cos(c + d*x)**6/(3*d) + 16*B*a**2*sin(c + d*x)**7/(105*d) + 8*B*a**2*sin(c + d*x)**5*cos(c + d*x) **2/(15*d) + 2*B*a**2*sin(c + d*x)**3*cos(c + d*x)**4/(3*d) - B*a**2*sin(c + d*x)**2*cos(c + d*x)**6/(6*d) - B*a**2*cos(c + d*x)**8/(24*d) - B*a**2* cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**2*cos( c)**5, True))
Time = 0.22 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.35 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {105 \, B a^{2} \sin \left (d x + c\right )^{8} + 120 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{7} + 140 \, {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{6} - 168 \, {\left (A + 4 \, B\right )} a^{2} \sin \left (d x + c\right )^{5} - 210 \, {\left (4 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{4} - 280 \, {\left (A - 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{3} + 420 \, {\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} + 840 \, A a^{2} \sin \left (d x + c\right )}{840 \, d} \]
1/840*(105*B*a^2*sin(d*x + c)^8 + 120*(A + 2*B)*a^2*sin(d*x + c)^7 + 140*( 2*A - B)*a^2*sin(d*x + c)^6 - 168*(A + 4*B)*a^2*sin(d*x + c)^5 - 210*(4*A + B)*a^2*sin(d*x + c)^4 - 280*(A - 2*B)*a^2*sin(d*x + c)^3 + 420*(2*A + B) *a^2*sin(d*x + c)^2 + 840*A*a^2*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (97) = 194\).
Time = 0.53 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.92 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B a^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {{\left (4 \, A a^{2} + B a^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {{\left (16 \, A a^{2} + 9 \, B a^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {{\left (20 \, A a^{2} + 13 \, B a^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {{\left (A a^{2} - 6 \, B a^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (19 \, A a^{2} - 2 \, B a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (9 \, A a^{2} + 2 \, B a^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \]
1/1024*B*a^2*cos(8*d*x + 8*c)/d - 1/384*(4*A*a^2 + B*a^2)*cos(6*d*x + 6*c) /d - 1/256*(16*A*a^2 + 9*B*a^2)*cos(4*d*x + 4*c)/d - 1/128*(20*A*a^2 + 13* B*a^2)*cos(2*d*x + 2*c)/d - 1/448*(A*a^2 + 2*B*a^2)*sin(7*d*x + 7*c)/d + 1 /320*(A*a^2 - 6*B*a^2)*sin(5*d*x + 5*c)/d + 1/192*(19*A*a^2 - 2*B*a^2)*sin (3*d*x + 3*c)/d + 5/64*(9*A*a^2 + 2*B*a^2)*sin(d*x + c)/d
Time = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.33 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^2\,\left (2\,A+B\right )}{2}-\frac {a^2\,{\sin \left (c+d\,x\right )}^3\,\left (A-2\,B\right )}{3}-\frac {a^2\,{\sin \left (c+d\,x\right )}^4\,\left (4\,A+B\right )}{4}-\frac {a^2\,{\sin \left (c+d\,x\right )}^5\,\left (A+4\,B\right )}{5}+\frac {a^2\,{\sin \left (c+d\,x\right )}^7\,\left (A+2\,B\right )}{7}+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {a^2\,{\sin \left (c+d\,x\right )}^6\,\left (2\,A-B\right )}{6}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \]