Integrand size = 29, antiderivative size = 97 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a A \sin (c+d x)}{d}+\frac {(A b+a B) \sin ^2(c+d x)}{2 d}-\frac {(a A-b B) \sin ^3(c+d x)}{3 d}-\frac {(A b+a B) \sin ^4(c+d x)}{4 d}-\frac {b B \sin ^5(c+d x)}{5 d} \]
a*A*sin(d*x+c)/d+1/2*(A*b+B*a)*sin(d*x+c)^2/d-1/3*(A*a-B*b)*sin(d*x+c)^3/d -1/4*(A*b+B*a)*sin(d*x+c)^4/d-1/5*b*B*sin(d*x+c)^5/d
Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\sin (c+d x) \left (60 a A+30 (A b+a B) \sin (c+d x)-20 (a A-b B) \sin ^2(c+d x)-15 (A b+a B) \sin ^3(c+d x)-12 b B \sin ^4(c+d x)\right )}{60 d} \]
(Sin[c + d*x]*(60*a*A + 30*(A*b + a*B)*Sin[c + d*x] - 20*(a*A - b*B)*Sin[c + d*x]^2 - 15*(A*b + a*B)*Sin[c + d*x]^3 - 12*b*B*Sin[c + d*x]^4))/(60*d)
Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3316, 27, 652, 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.
\(\displaystyle \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^3 (a+b \sin (c+d x)) (A+B \sin (c+d x))dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \frac {(a+b \sin (c+d x)) (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )}{b}d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a+b \sin (c+d x)) (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )d(b \sin (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \frac {\int \left (-b^4 B \sin ^4(c+d x)-b^3 (A b+a B) \sin ^3(c+d x)+b^3 (b B-a A) \sin ^2(c+d x)+b^3 (A b+a B) \sin (c+d x)+a A b^3\right )d(b \sin (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{4} b^4 (a B+A b) \sin ^4(c+d x)-\frac {1}{3} b^4 (a A-b B) \sin ^3(c+d x)+\frac {1}{2} b^4 (a B+A b) \sin ^2(c+d x)+a A b^4 \sin (c+d x)-\frac {1}{5} b^5 B \sin ^5(c+d x)}{b^4 d}\) |
(a*A*b^4*Sin[c + d*x] + (b^4*(A*b + a*B)*Sin[c + d*x]^2)/2 - (b^4*(a*A - b *B)*Sin[c + d*x]^3)/3 - (b^4*(A*b + a*B)*Sin[c + d*x]^4)/4 - (b^5*B*Sin[c + d*x]^5)/5)/(b^4*d)
3.16.30.3.1 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[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c *x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.53 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(-\frac {\frac {B b \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (A b +B a \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (a A -B b \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A b -B a \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right ) a}{d}\) | \(83\) |
default | \(-\frac {\frac {B b \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (A b +B a \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (a A -B b \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A b -B a \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right ) a}{d}\) | \(83\) |
parallelrisch | \(\frac {-60 A \cos \left (2 d x +2 c \right ) b -15 A \cos \left (4 d x +4 c \right ) b +40 a A \sin \left (3 d x +3 c \right )+360 A \sin \left (d x +c \right ) a -60 B \cos \left (2 d x +2 c \right ) a -6 B \sin \left (5 d x +5 c \right ) b -15 B \cos \left (4 d x +4 c \right ) a -10 B \sin \left (3 d x +3 c \right ) b +60 B b \sin \left (d x +c \right )+75 A b +75 B a}{480 d}\) | \(126\) |
risch | \(\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {b B \sin \left (d x +c \right )}{8 d}-\frac {\sin \left (5 d x +5 c \right ) B b}{80 d}-\frac {\cos \left (4 d x +4 c \right ) A b}{32 d}-\frac {\cos \left (4 d x +4 c \right ) B a}{32 d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (3 d x +3 c \right ) B b}{48 d}-\frac {\cos \left (2 d x +2 c \right ) A b}{8 d}-\frac {\cos \left (2 d x +2 c \right ) B a}{8 d}\) | \(140\) |
norman | \(\frac {\frac {\left (2 A b +2 B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 A b +2 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 \left (2 a A +B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 \left (2 a A +B b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 \left (25 a A -4 B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a A \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(219\) |
-1/d*(1/5*B*b*sin(d*x+c)^5+1/4*(A*b+B*a)*sin(d*x+c)^4+1/3*(A*a-B*b)*sin(d* x+c)^3+1/2*(-A*b-B*a)*sin(d*x+c)^2-A*sin(d*x+c)*a)
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {15 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, B b \cos \left (d x + c\right )^{4} - {\left (5 \, A a + B b\right )} \cos \left (d x + c\right )^{2} - 10 \, A a - 2 \, B b\right )} \sin \left (d x + c\right )}{60 \, d} \]
-1/60*(15*(B*a + A*b)*cos(d*x + c)^4 + 4*(3*B*b*cos(d*x + c)^4 - (5*A*a + B*b)*cos(d*x + c)^2 - 10*A*a - 2*B*b)*sin(d*x + c))/d
Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.32 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A b \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {B a \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {B b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a + b \sin {\left (c \right )}\right ) \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((2*A*a*sin(c + d*x)**3/(3*d) + A*a*sin(c + d*x)*cos(c + d*x)**2/ d - A*b*cos(c + d*x)**4/(4*d) - B*a*cos(c + d*x)**4/(4*d) + 2*B*b*sin(c + d*x)**5/(15*d) + B*b*sin(c + d*x)**3*cos(c + d*x)**2/(3*d), Ne(d, 0)), (x* (A + B*sin(c))*(a + b*sin(c))*cos(c)**3, True))
Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {12 \, B b \sin \left (d x + c\right )^{5} + 15 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{4} + 20 \, {\left (A a - B b\right )} \sin \left (d x + c\right )^{3} - 60 \, A a \sin \left (d x + c\right ) - 30 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{2}}{60 \, d} \]
-1/60*(12*B*b*sin(d*x + c)^5 + 15*(B*a + A*b)*sin(d*x + c)^4 + 20*(A*a - B *b)*sin(d*x + c)^3 - 60*A*a*sin(d*x + c) - 30*(B*a + A*b)*sin(d*x + c)^2)/ d
Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {12 \, B b \sin \left (d x + c\right )^{5} + 15 \, B a \sin \left (d x + c\right )^{4} + 15 \, A b \sin \left (d x + c\right )^{4} + 20 \, A a \sin \left (d x + c\right )^{3} - 20 \, B b \sin \left (d x + c\right )^{3} - 30 \, B a \sin \left (d x + c\right )^{2} - 30 \, A b \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \]
-1/60*(12*B*b*sin(d*x + c)^5 + 15*B*a*sin(d*x + c)^4 + 15*A*b*sin(d*x + c) ^4 + 20*A*a*sin(d*x + c)^3 - 20*B*b*sin(d*x + c)^3 - 30*B*a*sin(d*x + c)^2 - 30*A*b*sin(d*x + c)^2 - 60*A*a*sin(d*x + c))/d
Time = 12.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {\frac {B\,b\,{\sin \left (c+d\,x\right )}^5}{5}+\left (\frac {A\,b}{4}+\frac {B\,a}{4}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {A\,a}{3}-\frac {B\,b}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (-\frac {A\,b}{2}-\frac {B\,a}{2}\right )\,{\sin \left (c+d\,x\right )}^2-A\,a\,\sin \left (c+d\,x\right )}{d} \]