Integrand size = 31, antiderivative size = 132 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\left (a^2-b^2\right ) (A b-a B) (a+b \sin (c+d x))^3}{3 b^4 d}+\frac {\left (2 a A b-3 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^4 d}-\frac {(A b-3 a B) (a+b \sin (c+d x))^5}{5 b^4 d}-\frac {B (a+b \sin (c+d x))^6}{6 b^4 d} \]
[Out]
Time = 0.13 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2916, 786} \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^4 d}-\frac {\left (a^2-b^2\right ) (A b-a B) (a+b \sin (c+d x))^3}{3 b^4 d}-\frac {(A b-3 a B) (a+b \sin (c+d x))^5}{5 b^4 d}-\frac {B (a+b \sin (c+d x))^6}{6 b^4 d} \]
[In]
[Out]
Rule 786
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a+x)^2 \left (A+\frac {B x}{b}\right ) \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {\left (-a^2+b^2\right ) (A b-a B) (a+x)^2}{b}+\frac {\left (2 a A b-3 a^2 B+b^2 B\right ) (a+x)^3}{b}+\frac {(-A b+3 a B) (a+x)^4}{b}-\frac {B (a+x)^5}{b}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = -\frac {\left (a^2-b^2\right ) (A b-a B) (a+b \sin (c+d x))^3}{3 b^4 d}+\frac {\left (2 a A b-3 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^4 d}-\frac {(A b-3 a B) (a+b \sin (c+d x))^5}{5 b^4 d}-\frac {B (a+b \sin (c+d x))^6}{6 b^4 d} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.84 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {(a+b \sin (c+d x))^3 \left (-2 a^2 A b+20 A b^3+a^3 B-5 a b^2 B+3 b \left (2 a A b-a^2 B+5 b^2 B\right ) \sin (c+d x)-6 b^2 (2 A b-a B) \sin ^2(c+d x)-10 b^3 B \sin ^3(c+d x)\right )}{60 b^4 d} \]
[In]
[Out]
Time = 0.84 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(-\frac {\frac {B \,b^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (A \,b^{2}+2 B a b \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\left (a^{2}-b^{2}\right ) B +2 A a b \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-2 B a b +\left (a^{2}-b^{2}\right ) A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-2 A a b -B \,a^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right ) a^{2}}{d}\) | \(130\) |
default | \(-\frac {\frac {B \,b^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (A \,b^{2}+2 B a b \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\left (a^{2}-b^{2}\right ) B +2 A a b \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-2 B a b +\left (a^{2}-b^{2}\right ) A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-2 A a b -B \,a^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right ) a^{2}}{d}\) | \(130\) |
parallelrisch | \(\frac {\left (-240 A a b -120 B \,a^{2}-45 B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (80 A \,a^{2}-20 A \,b^{2}-40 B a b \right ) \sin \left (3 d x +3 c \right )+\left (-60 A a b -30 B \,a^{2}\right ) \cos \left (4 d x +4 c \right )+\left (-12 A \,b^{2}-24 B a b \right ) \sin \left (5 d x +5 c \right )+5 B \cos \left (6 d x +6 c \right ) b^{2}+\left (720 A \,a^{2}+120 A \,b^{2}+240 B a b \right ) \sin \left (d x +c \right )+300 A a b +150 B \,a^{2}+40 B \,b^{2}}{960 d}\) | \(164\) |
risch | \(\frac {3 \sin \left (d x +c \right ) A \,a^{2}}{4 d}+\frac {\sin \left (d x +c \right ) A \,b^{2}}{8 d}+\frac {\sin \left (d x +c \right ) B a b}{4 d}+\frac {\cos \left (6 d x +6 c \right ) B \,b^{2}}{192 d}-\frac {\sin \left (5 d x +5 c \right ) A \,b^{2}}{80 d}-\frac {\sin \left (5 d x +5 c \right ) B a b}{40 d}-\frac {\cos \left (4 d x +4 c \right ) A a b}{16 d}-\frac {\cos \left (4 d x +4 c \right ) B \,a^{2}}{32 d}+\frac {A \,a^{2} \sin \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (3 d x +3 c \right ) A \,b^{2}}{48 d}-\frac {\sin \left (3 d x +3 c \right ) B a b}{24 d}-\frac {\cos \left (2 d x +2 c \right ) A a b}{4 d}-\frac {\cos \left (2 d x +2 c \right ) B \,a^{2}}{8 d}-\frac {3 \cos \left (2 d x +2 c \right ) B \,b^{2}}{64 d}\) | \(240\) |
norman | \(\frac {\frac {\left (8 A a b +4 B \,a^{2}+4 B \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (8 A a b +4 B \,a^{2}+4 B \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (11 A \,a^{2}+4 A \,b^{2}+8 B a b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (11 A \,a^{2}+4 A \,b^{2}+8 B a b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 \left (15 A \,a^{2}+2 A \,b^{2}+4 B a b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {4 \left (15 A \,a^{2}+2 A \,b^{2}+4 B a b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {2 \left (2 A a b +B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 A a b +B \,a^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (6 A a b +3 B \,a^{2}-2 B \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(346\) |
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {10 \, B b^{2} \cos \left (d x + c\right )^{6} - 15 \, {\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} - 10 \, A a^{2} - 4 \, B a b - 2 \, A b^{2} - {\left (5 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \]
[In]
[Out]
Time = 0.35 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.73 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A a b \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {2 A b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {4 B a b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {2 B a b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {B b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {B b^{2} \cos ^{6}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a + b \sin {\left (c \right )}\right )^{2} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]
[In]
[Out]
none
Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.97 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {10 \, B b^{2} \sin \left (d x + c\right )^{6} + 12 \, {\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{5} + 15 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{4} - 60 \, A a^{2} \sin \left (d x + c\right ) + 20 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{60 \, d} \]
[In]
[Out]
none
Time = 0.53 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.27 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {10 \, B b^{2} \sin \left (d x + c\right )^{6} + 24 \, B a b \sin \left (d x + c\right )^{5} + 12 \, A b^{2} \sin \left (d x + c\right )^{5} + 15 \, B a^{2} \sin \left (d x + c\right )^{4} + 30 \, A a b \sin \left (d x + c\right )^{4} - 15 \, B b^{2} \sin \left (d x + c\right )^{4} + 20 \, A a^{2} \sin \left (d x + c\right )^{3} - 40 \, B a b \sin \left (d x + c\right )^{3} - 20 \, A b^{2} \sin \left (d x + c\right )^{3} - 30 \, B a^{2} \sin \left (d x + c\right )^{2} - 60 \, A a b \sin \left (d x + c\right )^{2} - 60 \, A a^{2} \sin \left (d x + c\right )}{60 \, d} \]
[In]
[Out]
Time = 12.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.96 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )-{\sin \left (c+d\,x\right )}^5\,\left (\frac {A\,b^2}{5}+\frac {2\,B\,a\,b}{5}\right )+{\sin \left (c+d\,x\right )}^3\,\left (-\frac {A\,a^2}{3}+\frac {2\,B\,a\,b}{3}+\frac {A\,b^2}{3}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {B\,a^2}{4}+\frac {A\,a\,b}{2}-\frac {B\,b^2}{4}\right )-\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^6}{6}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \]
[In]
[Out]