Integrand size = 21, antiderivative size = 47 \[ \int \frac {\cos ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 (a-a \sin (c+d x))^3}{3 a^4 d}+\frac {(a-a \sin (c+d x))^4}{4 a^5 d} \]
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin (c+d x) \left (12-6 \sin (c+d x)-4 \sin ^2(c+d x)+3 \sin ^3(c+d x)\right )}{12 a d} \]
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3146, 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.
\(\displaystyle \int \frac {\cos ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (2 a (a-a \sin (c+d x))^2-(a-a \sin (c+d x))^3\right )d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} (a-a \sin (c+d x))^4-\frac {2}{3} a (a-a \sin (c+d x))^3}{a^5 d}\) |
3.1.52.3.1 Defintions of rubi rules used
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 _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )}{a d}\) | \(45\) |
default | \(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )}{a d}\) | \(45\) |
risch | \(\frac {3 \sin \left (d x +c \right )}{4 a d}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {\sin \left (3 d x +3 c \right )}{12 d a}+\frac {\cos \left (2 d x +2 c \right )}{8 a d}\) | \(67\) |
parallelrisch | \(\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a d}\) | \(101\) |
norman | \(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {2 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {10 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {10 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {10 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {14 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {14 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(181\) |
Time = 0.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, \cos \left (d x + c\right )^{4} + 4 \, {\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right )}{12 \, a d} \]
Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (37) = 74\).
Time = 6.87 (sec) , antiderivative size = 530, normalized size of antiderivative = 11.28 \[ \int \frac {\cos ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {6 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {10 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {10 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{5}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]
Piecewise((6*tan(c/2 + d*x/2)**7/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c /2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 6*tan(c/2 + d*x/2)**6/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/ 2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 10*tan(c/2 + d*x/2)**5/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/ 2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 10*tan(c/2 + d*x/2)**3/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/ 2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 6*tan(c/2 + d*x/2)**2/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3 *a*d) + 6*tan(c/2 + d*x/2)/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/2 + d *x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d ), Ne(d, 0)), (x*cos(c)**5/(a*sin(c) + a), True))
Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} + 12 \, \sin \left (d x + c\right )}{12 \, a d} \]
Time = 0.33 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} + 12 \, \sin \left (d x + c\right )}{12 \, a d} \]
Time = 6.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\sin \left (c+d\,x\right )}{a}-\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}+\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}}{d} \]