Integrand size = 27, antiderivative size = 106 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a b x}{4}-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d} \]
1/4*a*b*x-1/30*(a^2+4*b^2)*cos(d*x+c)^3/d+1/4*a*b*cos(d*x+c)*sin(d*x+c)/d- 1/10*a*cos(d*x+c)^3*(a+b*sin(d*x+c))/d-1/5*cos(d*x+c)^3*(a+b*sin(d*x+c))^2 /d
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-30 \left (2 a^2+b^2\right ) \cos (c+d x)-5 \left (4 a^2+b^2\right ) \cos (3 (c+d x))+3 b (20 a (c+d x)+b \cos (5 (c+d x))-5 a \sin (4 (c+d x)))}{240 d} \]
(-30*(2*a^2 + b^2)*Cos[c + d*x] - 5*(4*a^2 + b^2)*Cos[3*(c + d*x)] + 3*b*( 20*a*(c + d*x) + b*Cos[5*(c + d*x)] - 5*a*Sin[4*(c + d*x)]))/(240*d)
Time = 0.54 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3341, 27, 3042, 3341, 3042, 3148, 3042, 3115, 24}
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.
| \(\displaystyle \int \sin (c+d x) \cos ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) \cos (c+d x)^2 (a+b \sin (c+d x))^2dx\) |
| \(\Big \downarrow \) 3341 |
| \(\displaystyle \frac {1}{5} \int 2 \cos ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))dx-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{5} \int \cos ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))dx-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \int \cos (c+d x)^2 (b+a \sin (c+d x)) (a+b \sin (c+d x))dx-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3341 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{4} \int \cos ^2(c+d x) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )dx-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{4} \int \cos (c+d x)^2 \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )dx-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{4} \left (5 a b \int \cos ^2(c+d x)dx-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{3 d}\right )-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{4} \left (5 a b \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{3 d}\right )-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{4} \left (5 a b \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{3 d}\right )-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{4} \left (5 a b \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{3 d}\right )-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{4 d}\right )-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\) |
-1/5*(Cos[c + d*x]^3*(a + b*Sin[c + d*x])^2)/d + (2*(-1/4*(a*Cos[c + d*x]^ 3*(a + b*Sin[c + d*x]))/d + (-1/3*((a^2 + 4*b^2)*Cos[c + d*x]^3)/d + 5*a*b *(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/5
3.11.61.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[1/(m + p + 1) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Sim p[a*c*(m + p + 1) + b*d*m + (a*d*m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && !LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && S implerQ[c + d*x, a + b*x])
Time = 0.32 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89
| method | result | size |
| derivativedivides | \(\frac {-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d}\) | \(94\) |
| default | \(\frac {-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d}\) | \(94\) |
| parallelrisch | \(\frac {60 a b x d -60 \cos \left (d x +c \right ) a^{2}-30 \cos \left (d x +c \right ) b^{2}+3 \cos \left (5 d x +5 c \right ) b^{2}-15 a b \sin \left (4 d x +4 c \right )-20 \cos \left (3 d x +3 c \right ) a^{2}-5 \cos \left (3 d x +3 c \right ) b^{2}-80 a^{2}-32 b^{2}}{240 d}\) | \(100\) |
| risch | \(\frac {a b x}{4}-\frac {a^{2} \cos \left (d x +c \right )}{4 d}-\frac {b^{2} \cos \left (d x +c \right )}{8 d}+\frac {\cos \left (5 d x +5 c \right ) b^{2}}{80 d}-\frac {a b \sin \left (4 d x +4 c \right )}{16 d}-\frac {\cos \left (3 d x +3 c \right ) a^{2}}{12 d}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{48 d}\) | \(102\) |
| norman | \(\frac {-\frac {10 a^{2}+4 b^{2}}{15 d}+\frac {a b x}{4}-\frac {2 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (2 a^{2}+2 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (4 a^{2}-2 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {3 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a b x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a b x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(289\) |
1/d*(-1/3*a^2*cos(d*x+c)^3+2*a*b*(-1/4*sin(d*x+c)*cos(d*x+c)^3+1/8*cos(d*x +c)*sin(d*x+c)+1/8*d*x+1/8*c)+b^2*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos (d*x+c)^3))
Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.69 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {12 \, b^{2} \cos \left (d x + c\right )^{5} + 15 \, a b d x - 20 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (2 \, a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]
1/60*(12*b^2*cos(d*x + c)^5 + 15*a*b*d*x - 20*(a^2 + b^2)*cos(d*x + c)^3 - 15*(2*a*b*cos(d*x + c)^3 - a*b*cos(d*x + c))*sin(d*x + c))/d
Time = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.62 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} - \frac {a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 b^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \sin {\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((-a**2*cos(c + d*x)**3/(3*d) + a*b*x*sin(c + d*x)**4/4 + a*b*x*s in(c + d*x)**2*cos(c + d*x)**2/2 + a*b*x*cos(c + d*x)**4/4 + a*b*sin(c + d *x)**3*cos(c + d*x)/(4*d) - a*b*sin(c + d*x)*cos(c + d*x)**3/(4*d) - b**2* sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 2*b**2*cos(c + d*x)**5/(15*d), Ne( d, 0)), (x*(a + b*sin(c))**2*sin(c)*cos(c)**2, True))
Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {80 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 16 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{2}}{240 \, d} \]
-1/240*(80*a^2*cos(d*x + c)^3 - 15*(4*d*x + 4*c - sin(4*d*x + 4*c))*a*b - 16*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*b^2)/d
Time = 0.43 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {1}{4} \, a b x + \frac {b^{2} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac {{\left (4 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )}{8 \, d} \]
1/4*a*b*x + 1/80*b^2*cos(5*d*x + 5*c)/d - 1/16*a*b*sin(4*d*x + 4*c)/d - 1/ 48*(4*a^2 + b^2)*cos(3*d*x + 3*c)/d - 1/8*(2*a^2 + b^2)*cos(d*x + c)/d
Time = 13.19 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.70 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a\,b\,x}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (4\,a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {4\,a^2}{3}+\frac {4\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {8\,a^2}{3}-\frac {4\,b^2}{3}\right )+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {2\,a^2}{3}+\frac {4\,b^2}{15}-3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]
(a*b*x)/4 - (tan(c/2 + (d*x)/2)^6*(4*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^2*( (4*a^2)/3 + (4*b^2)/3) + tan(c/2 + (d*x)/2)^4*((8*a^2)/3 - (4*b^2)/3) + 2* a^2*tan(c/2 + (d*x)/2)^8 + (2*a^2)/3 + (4*b^2)/15 - 3*a*b*tan(c/2 + (d*x)/ 2)^3 + 3*a*b*tan(c/2 + (d*x)/2)^7 - (a*b*tan(c/2 + (d*x)/2)^9)/2 + (a*b*ta n(c/2 + (d*x)/2))/2)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^5)