Integrand size = 29, antiderivative size = 109 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {4 (a+a \sin (c+d x))^5}{5 a^3 d}-\frac {2 (a+a \sin (c+d x))^6}{a^4 d}+\frac {13 (a+a \sin (c+d x))^7}{7 a^5 d}-\frac {3 (a+a \sin (c+d x))^8}{4 a^6 d}+\frac {(a+a \sin (c+d x))^9}{9 a^7 d} \]
4/5*(a+a*sin(d*x+c))^5/a^3/d-2*(a+a*sin(d*x+c))^6/a^4/d+13/7*(a+a*sin(d*x+ c))^7/a^5/d-3/4*(a+a*sin(d*x+c))^8/a^6/d+1/9*(a+a*sin(d*x+c))^9/a^7/d
Time = 0.39 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 (7560 \cos (2 (c+d x))+1260 \cos (4 (c+d x))-840 \cos (6 (c+d x))-315 \cos (8 (c+d x))-16380 \sin (c+d x)+1680 \sin (3 (c+d x))+2016 \sin (5 (c+d x))+270 \sin (7 (c+d x))-70 \sin (9 (c+d x)))}{161280 d} \]
-1/161280*(a^2*(7560*Cos[2*(c + d*x)] + 1260*Cos[4*(c + d*x)] - 840*Cos[6* (c + d*x)] - 315*Cos[8*(c + d*x)] - 16380*Sin[c + d*x] + 1680*Sin[3*(c + d *x)] + 2016*Sin[5*(c + d*x)] + 270*Sin[7*(c + d*x)] - 70*Sin[9*(c + d*x)]) )/d
Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 99, 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.
\(\displaystyle \int \sin ^2(c+d x) \cos ^5(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^5 (a \sin (c+d x)+a)^2dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^2 \sin ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left ((\sin (c+d x) a+a)^8-6 a (\sin (c+d x) a+a)^7+13 a^2 (\sin (c+d x) a+a)^6-12 a^3 (\sin (c+d x) a+a)^5+4 a^4 (\sin (c+d x) a+a)^4\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {4}{5} a^4 (a \sin (c+d x)+a)^5-2 a^3 (a \sin (c+d x)+a)^6+\frac {13}{7} a^2 (a \sin (c+d x)+a)^7+\frac {1}{9} (a \sin (c+d x)+a)^9-\frac {3}{4} a (a \sin (c+d x)+a)^8}{a^7 d}\) |
((4*a^4*(a + a*Sin[c + d*x])^5)/5 - 2*a^3*(a + a*Sin[c + d*x])^6 + (13*a^2 *(a + a*Sin[c + d*x])^7)/7 - (3*a*(a + a*Sin[c + d*x])^8)/4 + (a + a*Sin[c + d*x])^9/9)/(a^7*d)
3.6.12.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 0.43 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {2 \left (\sin ^{6}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\) | \(79\) |
default | \(\frac {a^{2} \left (\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {2 \left (\sin ^{6}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\) | \(79\) |
parallelrisch | \(\frac {a^{2} \left (-\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (2406 \cos \left (2 d x +2 c \right )+315 \sin \left (5 d x +5 c \right )-70 \cos \left (6 d x +6 c \right )+3150 \sin \left (d x +c \right )+1785 \sin \left (3 d x +3 c \right )+60 \cos \left (4 d x +4 c \right )+4324\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40320 d}\) | \(118\) |
risch | \(\frac {13 a^{2} \sin \left (d x +c \right )}{128 d}+\frac {a^{2} \sin \left (9 d x +9 c \right )}{2304 d}+\frac {a^{2} \cos \left (8 d x +8 c \right )}{512 d}-\frac {3 a^{2} \sin \left (7 d x +7 c \right )}{1792 d}+\frac {a^{2} \cos \left (6 d x +6 c \right )}{192 d}-\frac {a^{2} \sin \left (5 d x +5 c \right )}{80 d}-\frac {a^{2} \cos \left (4 d x +4 c \right )}{128 d}-\frac {a^{2} \sin \left (3 d x +3 c \right )}{96 d}-\frac {3 a^{2} \cos \left (2 d x +2 c \right )}{64 d}\) | \(152\) |
norman | \(\frac {\frac {8 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {48 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {136 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {11104 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 d}-\frac {136 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {48 a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {8 a^{2} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}\) | \(265\) |
a^2/d*(1/9*sin(d*x+c)^9+1/4*sin(d*x+c)^8-1/7*sin(d*x+c)^7-2/3*sin(d*x+c)^6 -1/5*sin(d*x+c)^5+1/2*sin(d*x+c)^4+1/3*sin(d*x+c)^3)
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {315 \, a^{2} \cos \left (d x + c\right )^{8} - 420 \, a^{2} \cos \left (d x + c\right )^{6} + 4 \, {\left (35 \, a^{2} \cos \left (d x + c\right )^{8} - 95 \, a^{2} \cos \left (d x + c\right )^{6} + 12 \, a^{2} \cos \left (d x + c\right )^{4} + 16 \, a^{2} \cos \left (d x + c\right )^{2} + 32 \, a^{2}\right )} \sin \left (d x + c\right )}{1260 \, d} \]
1/1260*(315*a^2*cos(d*x + c)^8 - 420*a^2*cos(d*x + c)^6 + 4*(35*a^2*cos(d* x + c)^8 - 95*a^2*cos(d*x + c)^6 + 12*a^2*cos(d*x + c)^4 + 16*a^2*cos(d*x + c)^2 + 32*a^2)*sin(d*x + c))/d
Time = 0.91 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.74 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {8 a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {4 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {8 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {4 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \cos ^{8}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((8*a**2*sin(c + d*x)**9/(315*d) + 4*a**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 8*a**2*sin(c + d*x)**7/(105*d) + a**2*sin(c + d*x)**5*co s(c + d*x)**4/(5*d) + 4*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + a**2 *sin(c + d*x)**3*cos(c + d*x)**4/(3*d) - a**2*sin(c + d*x)**2*cos(c + d*x) **6/(3*d) - a**2*cos(c + d*x)**8/(12*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*s in(c)**2*cos(c)**5, True))
Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {140 \, a^{2} \sin \left (d x + c\right )^{9} + 315 \, a^{2} \sin \left (d x + c\right )^{8} - 180 \, a^{2} \sin \left (d x + c\right )^{7} - 840 \, a^{2} \sin \left (d x + c\right )^{6} - 252 \, a^{2} \sin \left (d x + c\right )^{5} + 630 \, a^{2} \sin \left (d x + c\right )^{4} + 420 \, a^{2} \sin \left (d x + c\right )^{3}}{1260 \, d} \]
1/1260*(140*a^2*sin(d*x + c)^9 + 315*a^2*sin(d*x + c)^8 - 180*a^2*sin(d*x + c)^7 - 840*a^2*sin(d*x + c)^6 - 252*a^2*sin(d*x + c)^5 + 630*a^2*sin(d*x + c)^4 + 420*a^2*sin(d*x + c)^3)/d
Time = 0.40 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.39 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \cos \left (8 \, d x + 8 \, c\right )}{512 \, d} + \frac {a^{2} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a^{2} \cos \left (4 \, d x + 4 \, c\right )}{128 \, d} - \frac {3 \, a^{2} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {a^{2} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {3 \, a^{2} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {a^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a^{2} \sin \left (3 \, d x + 3 \, c\right )}{96 \, d} + \frac {13 \, a^{2} \sin \left (d x + c\right )}{128 \, d} \]
1/512*a^2*cos(8*d*x + 8*c)/d + 1/192*a^2*cos(6*d*x + 6*c)/d - 1/128*a^2*co s(4*d*x + 4*c)/d - 3/64*a^2*cos(2*d*x + 2*c)/d + 1/2304*a^2*sin(9*d*x + 9* c)/d - 3/1792*a^2*sin(7*d*x + 7*c)/d - 1/80*a^2*sin(5*d*x + 5*c)/d - 1/96* a^2*sin(3*d*x + 3*c)/d + 13/128*a^2*sin(d*x + c)/d
Time = 9.47 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {a^2\,{\sin \left (c+d\,x\right )}^8}{4}-\frac {a^2\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^6}{3}-\frac {a^2\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a^2\,{\sin \left (c+d\,x\right )}^4}{2}+\frac {a^2\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]