Integrand size = 29, antiderivative size = 55 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^3(c+d x)}{3 a^2 d}-\frac {\sin ^4(c+d x)}{2 a^2 d}+\frac {\sin ^5(c+d x)}{5 a^2 d} \]
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^3(c+d x) \left (10-15 \sin (c+d x)+6 \sin ^2(c+d x)\right )}{30 a^2 d} \]
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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^5(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^5}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^2(c+d x) (a-a \sin (c+d x))^2d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^2 \sin ^2(c+d x) (a-a \sin (c+d x))^2d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\sin ^4(c+d x) a^4-2 \sin ^3(c+d x) a^4+\sin ^2(c+d x) a^4\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} a^5 \sin ^5(c+d x)-\frac {1}{2} a^5 \sin ^4(c+d x)+\frac {1}{3} a^5 \sin ^3(c+d x)}{a^7 d}\) |
3.6.45.3.1 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 = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d \,a^{2}}\) | \(39\) |
default | \(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d \,a^{2}}\) | \(39\) |
parallelrisch | \(\frac {\left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (-13+3 \cos \left (2 d x +2 c \right )+15 \sin \left (d x +c \right )\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d \,a^{2}}\) | \(72\) |
risch | \(\frac {3 \sin \left (d x +c \right )}{8 a^{2} d}+\frac {\sin \left (5 d x +5 c \right )}{80 d \,a^{2}}-\frac {\cos \left (4 d x +4 c \right )}{16 d \,a^{2}}-\frac {7 \sin \left (3 d x +3 c \right )}{48 d \,a^{2}}+\frac {\cos \left (2 d x +2 c \right )}{4 d \,a^{2}}\) | \(84\) |
norman | \(\frac {\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {8 \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {112 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {112 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {16 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {16 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {88 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {88 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) | \(186\) |
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {15 \, \cos \left (d x + c\right )^{4} - 30 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 11 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right )}{30 \, a^{2} d} \]
-1/30*(15*cos(d*x + c)^4 - 30*cos(d*x + c)^2 - 2*(3*cos(d*x + c)^4 - 11*co s(d*x + c)^2 + 8)*sin(d*x + c))/(a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 588 vs. \(2 (44) = 88\).
Time = 34.25 (sec) , antiderivative size = 588, normalized size of antiderivative = 10.69 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {40 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} - \frac {120 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} + \frac {176 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} - \frac {120 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} + \frac {40 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 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 ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]
Piecewise((40*tan(c/2 + d*x/2)**7/(15*a**2*d*tan(c/2 + d*x/2)**10 + 75*a** 2*d*tan(c/2 + d*x/2)**8 + 150*a**2*d*tan(c/2 + d*x/2)**6 + 150*a**2*d*tan( c/2 + d*x/2)**4 + 75*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 120*tan(c/2 + d*x/2)**6/(15*a**2*d*tan(c/2 + d*x/2)**10 + 75*a**2*d*tan(c/2 + d*x/2)* *8 + 150*a**2*d*tan(c/2 + d*x/2)**6 + 150*a**2*d*tan(c/2 + d*x/2)**4 + 75* a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 176*tan(c/2 + d*x/2)**5/(15*a**2 *d*tan(c/2 + d*x/2)**10 + 75*a**2*d*tan(c/2 + d*x/2)**8 + 150*a**2*d*tan(c /2 + d*x/2)**6 + 150*a**2*d*tan(c/2 + d*x/2)**4 + 75*a**2*d*tan(c/2 + d*x/ 2)**2 + 15*a**2*d) - 120*tan(c/2 + d*x/2)**4/(15*a**2*d*tan(c/2 + d*x/2)** 10 + 75*a**2*d*tan(c/2 + d*x/2)**8 + 150*a**2*d*tan(c/2 + d*x/2)**6 + 150* a**2*d*tan(c/2 + d*x/2)**4 + 75*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 40*tan(c/2 + d*x/2)**3/(15*a**2*d*tan(c/2 + d*x/2)**10 + 75*a**2*d*tan(c/2 + d*x/2)**8 + 150*a**2*d*tan(c/2 + d*x/2)**6 + 150*a**2*d*tan(c/2 + d*x/2 )**4 + 75*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d), Ne(d, 0)), (x*sin(c)**2 *cos(c)**5/(a*sin(c) + a)**2, True))
Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 10 \, \sin \left (d x + c\right )^{3}}{30 \, a^{2} d} \]
Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 10 \, \sin \left (d x + c\right )^{3}}{30 \, a^{2} d} \]
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^3\,\left (6\,{\sin \left (c+d\,x\right )}^2-15\,\sin \left (c+d\,x\right )+10\right )}{30\,a^2\,d} \]