Integrand size = 27, antiderivative size = 55 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cos ^4(c+d x)}{4 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^5(c+d x)}{5 a d} \]
Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^2(c+d x) \left (30-20 \sin (c+d x)-15 \sin ^2(c+d x)+12 \sin ^3(c+d x)\right )}{60 a d} \]
Time = 0.38 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3314, 3042, 3044, 244, 2009, 3045, 15}
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.
\(\displaystyle \int \frac {\sin (c+d x) \cos ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^5}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \cos ^3(c+d x) \sin (c+d x)dx}{a}-\frac {\int \cos ^3(c+d x) \sin ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x)^3 \sin (c+d x)dx}{a}-\frac {\int \cos (c+d x)^3 \sin (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \cos (c+d x)^3 \sin (c+d x)dx}{a}-\frac {\int \sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )d\sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \cos (c+d x)^3 \sin (c+d x)dx}{a}-\frac {\int \left (\sin ^2(c+d x)-\sin ^4(c+d x)\right )d\sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \cos (c+d x)^3 \sin (c+d x)dx}{a}-\frac {\frac {1}{3} \sin ^3(c+d x)-\frac {1}{5} \sin ^5(c+d x)}{a d}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \cos ^3(c+d x)d\cos (c+d x)}{a d}-\frac {\frac {1}{3} \sin ^3(c+d x)-\frac {1}{5} \sin ^5(c+d x)}{a d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {\frac {1}{3} \sin ^3(c+d x)-\frac {1}{5} \sin ^5(c+d x)}{a d}-\frac {\cos ^4(c+d x)}{4 a d}\) |
3.6.35.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d a}\) | \(49\) |
default | \(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d a}\) | \(49\) |
parallelrisch | \(\frac {-15 \cos \left (4 d x +4 c \right )+75-60 \cos \left (2 d x +2 c \right )+10 \sin \left (3 d x +3 c \right )-60 \sin \left (d x +c \right )+6 \sin \left (5 d x +5 c \right )}{480 d a}\) | \(63\) |
risch | \(-\frac {\sin \left (d x +c \right )}{8 a d}+\frac {\sin \left (5 d x +5 c \right )}{80 d a}-\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {\sin \left (3 d x +3 c \right )}{48 d a}-\frac {\cos \left (2 d x +2 c \right )}{8 a d}\) | \(84\) |
norman | \(\frac {\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {2 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {12 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {12 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {12 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {12 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {4 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(221\) |
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {15 \, \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right )}{60 \, a d} \]
Leaf count of result is larger than twice the leaf count of optimal. 741 vs. \(2 (39) = 78\).
Time = 11.43 (sec) , antiderivative size = 741, normalized size of antiderivative = 13.47 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {30 \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} - \frac {40 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} + \frac {30 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} + \frac {16 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} + \frac {30 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} - \frac {40 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} + \frac {30 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )} \cos ^{5}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]
Piecewise((30*tan(c/2 + d*x/2)**8/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*ta n(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2) **4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 40*tan(c/2 + d*x/2)**7/(15*a* d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d* x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a* d) + 30*tan(c/2 + d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 16*tan(c/2 + d*x/2)**5/(15*a*d*tan( c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)** 6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 3 0*tan(c/2 + d*x/2)**4/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/ 2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d *tan(c/2 + d*x/2)**2 + 15*a*d) - 40*tan(c/2 + d*x/2)**3/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 15 0*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*tan( c/2 + d*x/2)**2/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c /2 + d*x/2)**2 + 15*a*d), Ne(d, 0)), (x*sin(c)*cos(c)**5/(a*sin(c) + a), T rue))
Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2}}{60 \, a d} \]
Time = 0.42 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2}}{60 \, a d} \]
Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}-\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}+\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}}{d} \]