Integrand size = 29, antiderivative size = 55 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^4(c+d x)}{4 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {\sin ^6(c+d x)}{6 a^2 d} \]
Time = 0.37 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^4(c+d x) \left (15-24 \sin (c+d x)+10 \sin ^2(c+d x)\right )}{60 a^2 d} \]
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96, 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, 49, 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.
\(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^5(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^5}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^3(c+d x) (a-a \sin (c+d x))^2d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^3 \sin ^3(c+d x) (a-a \sin (c+d x))^2d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\sin ^5(c+d x) a^5-2 \sin ^4(c+d x) a^5+\sin ^3(c+d x) a^5\right )d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{6} a^6 \sin ^6(c+d x)-\frac {2}{5} a^6 \sin ^5(c+d x)+\frac {1}{4} a^6 \sin ^4(c+d x)}{a^8 d}\) |
3.6.44.3.1 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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 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 + (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 = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}}{d \,a^{2}}\) | \(39\) |
default | \(\frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}}{d \,a^{2}}\) | \(39\) |
parallelrisch | \(\frac {-5 \cos \left (6 d x +6 c \right )-195 \cos \left (2 d x +2 c \right )-24 \sin \left (5 d x +5 c \right )+120 \sin \left (3 d x +3 c \right )-240 \sin \left (d x +c \right )+60 \cos \left (4 d x +4 c \right )+140}{960 d \,a^{2}}\) | \(74\) |
risch | \(-\frac {\sin \left (d x +c \right )}{4 a^{2} d}-\frac {\cos \left (6 d x +6 c \right )}{192 d \,a^{2}}-\frac {\sin \left (5 d x +5 c \right )}{40 d \,a^{2}}+\frac {\cos \left (4 d x +4 c \right )}{16 d \,a^{2}}+\frac {\sin \left (3 d x +3 c \right )}{8 d \,a^{2}}-\frac {13 \cos \left (2 d x +2 c \right )}{64 d \,a^{2}}\) | \(101\) |
Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {10 \, \cos \left (d x + c\right )^{6} - 45 \, \cos \left (d x + c\right )^{4} + 60 \, \cos \left (d x + c\right )^{2} + 24 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right )}{60 \, a^{2} d} \]
-1/60*(10*cos(d*x + c)^6 - 45*cos(d*x + c)^4 + 60*cos(d*x + c)^2 + 24*(cos (d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*sin(d*x + c))/(a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 682 vs. \(2 (46) = 92\).
Time = 53.78 (sec) , antiderivative size = 682, normalized size of antiderivative = 12.40 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {60 \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} - \frac {192 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} + \frac {280 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} - \frac {192 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} + \frac {60 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{3}{\left (c \right )} \cos ^{5}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]
Piecewise((60*tan(c/2 + d*x/2)**8/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a** 2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan (c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d* x/2)**2 + 15*a**2*d) - 192*tan(c/2 + d*x/2)**7/(15*a**2*d*tan(c/2 + d*x/2) **12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 3 00*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d *tan(c/2 + d*x/2)**2 + 15*a**2*d) + 280*tan(c/2 + d*x/2)**6/(15*a**2*d*tan (c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)** 4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 192*tan(c/2 + d*x/2)**5/( 15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2 *d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c /2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 60*tan(c/2 + d*x/2)**4/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**1 0 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225* a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d), N e(d, 0)), (x*sin(c)**3*cos(c)**5/(a*sin(c) + a)**2, True))
Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {10 \, \sin \left (d x + c\right )^{6} - 24 \, \sin \left (d x + c\right )^{5} + 15 \, \sin \left (d x + c\right )^{4}}{60 \, a^{2} d} \]
Time = 0.34 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {10 \, \sin \left (d x + c\right )^{6} - 24 \, \sin \left (d x + c\right )^{5} + 15 \, \sin \left (d x + c\right )^{4}}{60 \, a^{2} d} \]
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^4\,\left (10\,{\sin \left (c+d\,x\right )}^2-24\,\sin \left (c+d\,x\right )+15\right )}{60\,a^2\,d} \]