Integrand size = 27, antiderivative size = 89 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {4 (a+a \sin (c+d x))^5}{5 a^3 d}+\frac {4 (a+a \sin (c+d x))^6}{3 a^4 d}-\frac {5 (a+a \sin (c+d x))^7}{7 a^5 d}+\frac {(a+a \sin (c+d x))^8}{8 a^6 d} \]
-4/5*(a+a*sin(d*x+c))^5/a^3/d+4/3*(a+a*sin(d*x+c))^6/a^4/d-5/7*(a+a*sin(d* x+c))^7/a^5/d+1/8*(a+a*sin(d*x+c))^8/a^6/d
Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 (-2590+10920 \cos (2 (c+d x))+3780 \cos (4 (c+d x))+280 \cos (6 (c+d x))-105 \cos (8 (c+d x))-16800 \sin (c+d x)+1120 \sin (3 (c+d x))+2016 \sin (5 (c+d x))+480 \sin (7 (c+d x)))}{107520 d} \]
-1/107520*(a^2*(-2590 + 10920*Cos[2*(c + d*x)] + 3780*Cos[4*(c + d*x)] + 2 80*Cos[6*(c + d*x)] - 105*Cos[8*(c + d*x)] - 16800*Sin[c + d*x] + 1120*Sin [3*(c + d*x)] + 2016*Sin[5*(c + d*x)] + 480*Sin[7*(c + d*x)]))/d
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 86, 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.
\(\displaystyle \int \sin (c+d x) \cos ^5(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x) \cos (c+d x)^5 (a \sin (c+d x)+a)^2dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin (c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a \sin (c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left ((\sin (c+d x) a+a)^7-5 a (\sin (c+d x) a+a)^6+8 a^2 (\sin (c+d x) a+a)^5-4 a^3 (\sin (c+d x) a+a)^4\right )d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {4}{5} a^3 (a \sin (c+d x)+a)^5+\frac {4}{3} a^2 (a \sin (c+d x)+a)^6+\frac {1}{8} (a \sin (c+d x)+a)^8-\frac {5}{7} a (a \sin (c+d x)+a)^7}{a^6 d}\) |
((-4*a^3*(a + a*Sin[c + d*x])^5)/5 + (4*a^2*(a + a*Sin[c + d*x])^6)/3 - (5 *a*(a + a*Sin[c + d*x])^7)/7 + (a + a*Sin[c + d*x])^8/8)/(a^6*d)
3.6.13.3.1 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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 0.34 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {2 \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {4 \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\) | \(79\) |
default | \(\frac {a^{2} \left (\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {2 \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {4 \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\) | \(79\) |
parallelrisch | \(-\frac {a^{2} \left (3780 \cos \left (4 d x +4 c \right )-14875+10920 \cos \left (2 d x +2 c \right )+1120 \sin \left (3 d x +3 c \right )-16800 \sin \left (d x +c \right )+480 \sin \left (7 d x +7 c \right )+2016 \sin \left (5 d x +5 c \right )-105 \cos \left (8 d x +8 c \right )+280 \cos \left (6 d x +6 c \right )\right )}{107520 d}\) | \(96\) |
risch | \(\frac {5 a^{2} \sin \left (d x +c \right )}{32 d}+\frac {a^{2} \cos \left (8 d x +8 c \right )}{1024 d}-\frac {a^{2} \sin \left (7 d x +7 c \right )}{224 d}-\frac {a^{2} \cos \left (6 d x +6 c \right )}{384 d}-\frac {3 a^{2} \sin \left (5 d x +5 c \right )}{160 d}-\frac {9 a^{2} \cos \left (4 d x +4 c \right )}{256 d}-\frac {a^{2} \sin \left (3 d x +3 c \right )}{96 d}-\frac {13 a^{2} \cos \left (2 d x +2 c \right )}{128 d}\) | \(135\) |
norman | \(\frac {\frac {16 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {16 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {1376 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {1376 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {16 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {16 a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {10 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {80 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}\) | \(265\) |
a^2/d*(1/8*sin(d*x+c)^8+2/7*sin(d*x+c)^7-1/6*sin(d*x+c)^6-4/5*sin(d*x+c)^5 -1/4*sin(d*x+c)^4+2/3*sin(d*x+c)^3+1/2*sin(d*x+c)^2)
Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {105 \, a^{2} \cos \left (d x + c\right )^{8} - 280 \, a^{2} \cos \left (d x + c\right )^{6} - 16 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{6} - 3 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a^{2}\right )} \sin \left (d x + c\right )}{840 \, d} \]
1/840*(105*a^2*cos(d*x + c)^8 - 280*a^2*cos(d*x + c)^6 - 16*(15*a^2*cos(d* x + c)^6 - 3*a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^2 - 8*a^2)*sin(d*x + c))/d
Time = 0.65 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.56 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {16 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {8 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {a^{2} \cos ^{8}{\left (c + d x \right )}}{24 d} - \frac {a^{2} \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin {\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((16*a**2*sin(c + d*x)**7/(105*d) + 8*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + 2*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/(6*d) - a**2*cos(c + d*x)**8/(24*d) - a**2*co s(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)*cos(c)**5, Tru e))
Time = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.09 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {105 \, a^{2} \sin \left (d x + c\right )^{8} + 240 \, a^{2} \sin \left (d x + c\right )^{7} - 140 \, a^{2} \sin \left (d x + c\right )^{6} - 672 \, a^{2} \sin \left (d x + c\right )^{5} - 210 \, a^{2} \sin \left (d x + c\right )^{4} + 560 \, a^{2} \sin \left (d x + c\right )^{3} + 420 \, a^{2} \sin \left (d x + c\right )^{2}}{840 \, d} \]
1/840*(105*a^2*sin(d*x + c)^8 + 240*a^2*sin(d*x + c)^7 - 140*a^2*sin(d*x + c)^6 - 672*a^2*sin(d*x + c)^5 - 210*a^2*sin(d*x + c)^4 + 560*a^2*sin(d*x + c)^3 + 420*a^2*sin(d*x + c)^2)/d
Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.51 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a^{2} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {9 \, a^{2} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {13 \, a^{2} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac {a^{2} \sin \left (7 \, d x + 7 \, c\right )}{224 \, d} - \frac {3 \, a^{2} \sin \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac {a^{2} \sin \left (3 \, d x + 3 \, c\right )}{96 \, d} + \frac {5 \, a^{2} \sin \left (d x + c\right )}{32 \, d} \]
1/1024*a^2*cos(8*d*x + 8*c)/d - 1/384*a^2*cos(6*d*x + 6*c)/d - 9/256*a^2*c os(4*d*x + 4*c)/d - 13/128*a^2*cos(2*d*x + 2*c)/d - 1/224*a^2*sin(7*d*x + 7*c)/d - 3/160*a^2*sin(5*d*x + 5*c)/d - 1/96*a^2*sin(3*d*x + 3*c)/d + 5/32 *a^2*sin(d*x + c)/d
Time = 9.63 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.08 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a^2\,{\sin \left (c+d\,x\right )}^6}{6}-\frac {4\,a^2\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {a^2\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {a^2\,{\sin \left (c+d\,x\right )}^2}{2}}{d} \]