Integrand size = 29, antiderivative size = 143 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a A \sin (c+d x)}{d}+\frac {(A b+a B) \sin ^2(c+d x)}{2 d}-\frac {(2 a A-b B) \sin ^3(c+d x)}{3 d}-\frac {(A b+a B) \sin ^4(c+d x)}{2 d}+\frac {(a A-2 b B) \sin ^5(c+d x)}{5 d}+\frac {(A b+a B) \sin ^6(c+d x)}{6 d}+\frac {b B \sin ^7(c+d x)}{7 d} \]
a*A*sin(d*x+c)/d+1/2*(A*b+B*a)*sin(d*x+c)^2/d-1/3*(2*A*a-B*b)*sin(d*x+c)^3 /d-1/2*(A*b+B*a)*sin(d*x+c)^4/d+1/5*(A*a-2*B*b)*sin(d*x+c)^5/d+1/6*(A*b+B* a)*sin(d*x+c)^6/d+1/7*b*B*sin(d*x+c)^7/d
Time = 0.22 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.81 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\sin (c+d x) \left (210 a A+105 (A b+a B) \sin (c+d x)-70 (2 a A-b B) \sin ^2(c+d x)-105 (A b+a B) \sin ^3(c+d x)+42 (a A-2 b B) \sin ^4(c+d x)+35 (A b+a B) \sin ^5(c+d x)+30 b B \sin ^6(c+d x)\right )}{210 d} \]
(Sin[c + d*x]*(210*a*A + 105*(A*b + a*B)*Sin[c + d*x] - 70*(2*a*A - b*B)*S in[c + d*x]^2 - 105*(A*b + a*B)*Sin[c + d*x]^3 + 42*(a*A - 2*b*B)*Sin[c + d*x]^4 + 35*(A*b + a*B)*Sin[c + d*x]^5 + 30*b*B*Sin[c + d*x]^6))/(210*d)
Time = 0.37 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.04, 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.
| \(\displaystyle \int \cos ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^5 (a+b \sin (c+d x)) (A+B \sin (c+d x))dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {(a+b \sin (c+d x)) (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (a+b \sin (c+d x)) (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^6 d}\) |
| \(\Big \downarrow \) 652 |
| \(\displaystyle \frac {\int \left (b^6 B \sin ^6(c+d x)+b^5 (A b+a B) \sin ^5(c+d x)-b^5 (2 b B-a A) \sin ^4(c+d x)-2 b^5 (A b+a B) \sin ^3(c+d x)+b^5 (b B-2 a A) \sin ^2(c+d x)+b^5 (A b+a B) \sin (c+d x)+a A b^5\right )d(b \sin (c+d x))}{b^6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{6} b^6 (a B+A b) \sin ^6(c+d x)+\frac {1}{5} b^6 (a A-2 b B) \sin ^5(c+d x)-\frac {1}{2} b^6 (a B+A b) \sin ^4(c+d x)-\frac {1}{3} b^6 (2 a A-b B) \sin ^3(c+d x)+\frac {1}{2} b^6 (a B+A b) \sin ^2(c+d x)+a A b^6 \sin (c+d x)+\frac {1}{7} b^7 B \sin ^7(c+d x)}{b^6 d}\) |
(a*A*b^6*Sin[c + d*x] + (b^6*(A*b + a*B)*Sin[c + d*x]^2)/2 - (b^6*(2*a*A - b*B)*Sin[c + d*x]^3)/3 - (b^6*(A*b + a*B)*Sin[c + d*x]^4)/2 + (b^6*(a*A - 2*b*B)*Sin[c + d*x]^5)/5 + (b^6*(A*b + a*B)*Sin[c + d*x]^6)/6 + (b^7*B*Si n[c + d*x]^7)/7)/(b^6*d)
3.16.29.3.1 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 = 0.84 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {\frac {B b \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (A b +B a \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (a A -2 B b \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-2 A b -2 B a \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-2 a A +B b \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (A b +B a \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right ) a}{d}\) | \(116\) |
| default | \(\frac {\frac {B b \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (A b +B a \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (a A -2 B b \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-2 A b -2 B a \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-2 a A +B b \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (A b +B a \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right ) a}{d}\) | \(116\) |
| parallelrisch | \(\frac {-210 A \cos \left (4 d x +4 c \right ) b -525 A \cos \left (2 d x +2 c \right ) b +4200 A \sin \left (d x +c \right ) a -35 A \cos \left (6 d x +6 c \right ) b +84 A \sin \left (5 d x +5 c \right ) a +700 a A \sin \left (3 d x +3 c \right )-210 B \cos \left (4 d x +4 c \right ) a -525 B \cos \left (2 d x +2 c \right ) a +525 B b \sin \left (d x +c \right )-15 B \sin \left (7 d x +7 c \right ) b -35 B \cos \left (6 d x +6 c \right ) a -63 B \sin \left (5 d x +5 c \right ) b -35 B \sin \left (3 d x +3 c \right ) b +770 A b +770 B a}{6720 d}\) | \(178\) |
| risch | \(\frac {5 a A \sin \left (d x +c \right )}{8 d}+\frac {5 b B \sin \left (d x +c \right )}{64 d}-\frac {\sin \left (7 d x +7 c \right ) B b}{448 d}-\frac {\cos \left (6 d x +6 c \right ) A b}{192 d}-\frac {\cos \left (6 d x +6 c \right ) B a}{192 d}+\frac {\sin \left (5 d x +5 c \right ) a A}{80 d}-\frac {3 \sin \left (5 d x +5 c \right ) B b}{320 d}-\frac {\cos \left (4 d x +4 c \right ) A b}{32 d}-\frac {\cos \left (4 d x +4 c \right ) B a}{32 d}+\frac {5 a A \sin \left (3 d x +3 c \right )}{48 d}-\frac {\sin \left (3 d x +3 c \right ) B b}{192 d}-\frac {5 \cos \left (2 d x +2 c \right ) A b}{64 d}-\frac {5 \cos \left (2 d x +2 c \right ) B a}{64 d}\) | \(204\) |
| norman | \(\frac {\frac {\left (2 A b +2 B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 A b +2 B a \right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 A b +2 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 A b +2 B a \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (5 a A +2 B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 \left (5 a A +2 B b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 \left (91 a A +38 B b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {2 \left (113 a A -16 B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 \left (113 a A -16 B b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {5 \left (4 A b +4 B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {5 \left (4 A b +4 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a A \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}\) | \(323\) |
1/d*(1/7*B*b*sin(d*x+c)^7+1/6*(A*b+B*a)*sin(d*x+c)^6+1/5*(A*a-2*B*b)*sin(d *x+c)^5+1/4*(-2*A*b-2*B*a)*sin(d*x+c)^4+1/3*(-2*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 = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.62 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {35 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{6} + 2 \, {\left (15 \, B b \cos \left (d x + c\right )^{6} - 3 \, {\left (7 \, A a + B b\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, A a + B b\right )} \cos \left (d x + c\right )^{2} - 56 \, A a - 8 \, B b\right )} \sin \left (d x + c\right )}{210 \, d} \]
-1/210*(35*(B*a + A*b)*cos(d*x + c)^6 + 2*(15*B*b*cos(d*x + c)^6 - 3*(7*A* a + B*b)*cos(d*x + c)^4 - 4*(7*A*a + B*b)*cos(d*x + c)^2 - 56*A*a - 8*B*b) *sin(d*x + c))/d
Time = 0.46 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.24 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {8 A a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 A a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A b \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {B a \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac {8 B b \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 B b \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {B b \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a + b \sin {\left (c \right )}\right ) \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((8*A*a*sin(c + d*x)**5/(15*d) + 4*A*a*sin(c + d*x)**3*cos(c + d* x)**2/(3*d) + A*a*sin(c + d*x)*cos(c + d*x)**4/d - A*b*cos(c + d*x)**6/(6* d) - B*a*cos(c + d*x)**6/(6*d) + 8*B*b*sin(c + d*x)**7/(105*d) + 4*B*b*sin (c + d*x)**5*cos(c + d*x)**2/(15*d) + B*b*sin(c + d*x)**3*cos(c + d*x)**4/ (3*d), Ne(d, 0)), (x*(A + B*sin(c))*(a + b*sin(c))*cos(c)**5, True))
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.81 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {30 \, B b \sin \left (d x + c\right )^{7} + 35 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{6} + 42 \, {\left (A a - 2 \, B b\right )} \sin \left (d x + c\right )^{5} - 105 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{4} - 70 \, {\left (2 \, A a - B b\right )} \sin \left (d x + c\right )^{3} + 210 \, A a \sin \left (d x + c\right ) + 105 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{2}}{210 \, d} \]
1/210*(30*B*b*sin(d*x + c)^7 + 35*(B*a + A*b)*sin(d*x + c)^6 + 42*(A*a - 2 *B*b)*sin(d*x + c)^5 - 105*(B*a + A*b)*sin(d*x + c)^4 - 70*(2*A*a - B*b)*s in(d*x + c)^3 + 210*A*a*sin(d*x + c) + 105*(B*a + A*b)*sin(d*x + c)^2)/d
Time = 0.39 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.01 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {B b \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {{\left (B a + A b\right )} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {{\left (B a + A b\right )} \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {5 \, {\left (B a + A b\right )} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (4 \, A a - 3 \, B b\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (20 \, A a - B b\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (8 \, A a + B b\right )} \sin \left (d x + c\right )}{64 \, d} \]
-1/448*B*b*sin(7*d*x + 7*c)/d - 1/192*(B*a + A*b)*cos(6*d*x + 6*c)/d - 1/3 2*(B*a + A*b)*cos(4*d*x + 4*c)/d - 5/64*(B*a + A*b)*cos(2*d*x + 2*c)/d + 1 /320*(4*A*a - 3*B*b)*sin(5*d*x + 5*c)/d + 1/192*(20*A*a - B*b)*sin(3*d*x + 3*c)/d + 5/64*(8*A*a + B*b)*sin(d*x + c)/d
Time = 12.50 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.83 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\frac {B\,b\,{\sin \left (c+d\,x\right )}^7}{7}+\left (\frac {A\,b}{6}+\frac {B\,a}{6}\right )\,{\sin \left (c+d\,x\right )}^6+\left (\frac {A\,a}{5}-\frac {2\,B\,b}{5}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-\frac {A\,b}{2}-\frac {B\,a}{2}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {B\,b}{3}-\frac {2\,A\,a}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {A\,b}{2}+\frac {B\,a}{2}\right )\,{\sin \left (c+d\,x\right )}^2+A\,a\,\sin \left (c+d\,x\right )}{d} \]