Integrand size = 27, antiderivative size = 97 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^6(c+d x)}{6 d}+\frac {a \sin ^7(c+d x)}{7 d}-\frac {a \sin ^8(c+d x)}{4 d}-\frac {2 a \sin ^9(c+d x)}{9 d}+\frac {a \sin ^{10}(c+d x)}{10 d}+\frac {a \sin ^{11}(c+d x)}{11 d} \]
1/6*a*sin(d*x+c)^6/d+1/7*a*sin(d*x+c)^7/d-1/4*a*sin(d*x+c)^8/d-2/9*a*sin(d *x+c)^9/d+1/10*a*sin(d*x+c)^10/d+1/11*a*sin(d*x+c)^11/d
Time = 0.37 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a (34650 \cos (2 (c+d x))-5775 \cos (6 (c+d x))+693 \cos (10 (c+d x))-34650 \sin (c+d x)+11550 \sin (3 (c+d x))+3465 \sin (5 (c+d x))-2475 \sin (7 (c+d x))-385 \sin (9 (c+d x))+315 \sin (11 (c+d x)))}{3548160 d} \]
-1/3548160*(a*(34650*Cos[2*(c + d*x)] - 5775*Cos[6*(c + d*x)] + 693*Cos[10 *(c + d*x)] - 34650*Sin[c + d*x] + 11550*Sin[3*(c + d*x)] + 3465*Sin[5*(c + d*x)] - 2475*Sin[7*(c + d*x)] - 385*Sin[9*(c + d*x)] + 315*Sin[11*(c + d *x)]))/d
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.01, 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, 99, 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.
\(\displaystyle \int \sin ^5(c+d x) \cos ^5(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^5 \cos (c+d x)^5 (a \sin (c+d x)+a)dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^5(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^5 \sin ^5(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a^{10} d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\sin ^{10}(c+d x) a^{10}+\sin ^9(c+d x) a^{10}-2 \sin ^8(c+d x) a^{10}-2 \sin ^7(c+d x) a^{10}+\sin ^6(c+d x) a^{10}+\sin ^5(c+d x) a^{10}\right )d(a \sin (c+d x))}{a^{10} d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{11} a^{11} \sin ^{11}(c+d x)+\frac {1}{10} a^{11} \sin ^{10}(c+d x)-\frac {2}{9} a^{11} \sin ^9(c+d x)-\frac {1}{4} a^{11} \sin ^8(c+d x)+\frac {1}{7} a^{11} \sin ^7(c+d x)+\frac {1}{6} a^{11} \sin ^6(c+d x)}{a^{10} d}\) |
((a^11*Sin[c + d*x]^6)/6 + (a^11*Sin[c + d*x]^7)/7 - (a^11*Sin[c + d*x]^8) /4 - (2*a^11*Sin[c + d*x]^9)/9 + (a^11*Sin[c + d*x]^10)/10 + (a^11*Sin[c + d*x]^11)/11)/(a^10*d)
3.5.94.3.1 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_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 0.56 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\left (\sin ^{11}\left (d x +c \right )\right )}{11}+\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}-\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}+\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}\right )}{d}\) | \(67\) |
default | \(\frac {a \left (\frac {\left (\sin ^{11}\left (d x +c \right )\right )}{11}+\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}-\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}+\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}\right )}{d}\) | \(67\) |
parallelrisch | \(-\frac {a \left (-10+\cos \left (3 d x +3 c \right )-6 \cos \left (2 d x +2 c \right )+15 \cos \left (d x +c \right )\right ) \left (4158 \cos \left (2 d x +2 c \right )+315 \sin \left (5 d x +5 c \right )+1830 \sin \left (d x +c \right )+1505 \sin \left (3 d x +3 c \right )+693 \cos \left (4 d x +4 c \right )+4389\right ) \left (\cos \left (3 d x +3 c \right )+6 \cos \left (2 d x +2 c \right )+15 \cos \left (d x +c \right )+10\right )}{887040 d}\) | \(121\) |
risch | \(\frac {5 a \sin \left (d x +c \right )}{512 d}-\frac {a \sin \left (11 d x +11 c \right )}{11264 d}-\frac {a \cos \left (10 d x +10 c \right )}{5120 d}+\frac {a \sin \left (9 d x +9 c \right )}{9216 d}+\frac {5 a \sin \left (7 d x +7 c \right )}{7168 d}+\frac {5 a \cos \left (6 d x +6 c \right )}{3072 d}-\frac {a \sin \left (5 d x +5 c \right )}{1024 d}-\frac {5 a \sin \left (3 d x +3 c \right )}{1536 d}-\frac {5 a \cos \left (2 d x +2 c \right )}{512 d}\) | \(134\) |
a/d*(1/11*sin(d*x+c)^11+1/10*sin(d*x+c)^10-2/9*sin(d*x+c)^9-1/4*sin(d*x+c) ^8+1/7*sin(d*x+c)^7+1/6*sin(d*x+c)^6)
Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {1386 \, a \cos \left (d x + c\right )^{10} - 3465 \, a \cos \left (d x + c\right )^{8} + 2310 \, a \cos \left (d x + c\right )^{6} + 20 \, {\left (63 \, a \cos \left (d x + c\right )^{10} - 161 \, a \cos \left (d x + c\right )^{8} + 113 \, a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} - 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\right )} \sin \left (d x + c\right )}{13860 \, d} \]
-1/13860*(1386*a*cos(d*x + c)^10 - 3465*a*cos(d*x + c)^8 + 2310*a*cos(d*x + c)^6 + 20*(63*a*cos(d*x + c)^10 - 161*a*cos(d*x + c)^8 + 113*a*cos(d*x + c)^6 - 3*a*cos(d*x + c)^4 - 4*a*cos(d*x + c)^2 - 8*a)*sin(d*x + c))/d
Time = 1.81 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.40 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{11}{\left (c + d x \right )}}{693 d} + \frac {4 a \sin ^{9}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{63 d} + \frac {a \sin ^{7}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{7 d} - \frac {a \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{12 d} - \frac {a \cos ^{10}{\left (c + d x \right )}}{60 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{5}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((8*a*sin(c + d*x)**11/(693*d) + 4*a*sin(c + d*x)**9*cos(c + d*x) **2/(63*d) + a*sin(c + d*x)**7*cos(c + d*x)**4/(7*d) - a*sin(c + d*x)**4*c os(c + d*x)**6/(6*d) - a*sin(c + d*x)**2*cos(c + d*x)**8/(12*d) - a*cos(c + d*x)**10/(60*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**5*cos(c)**5, True) )
Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {1260 \, a \sin \left (d x + c\right )^{11} + 1386 \, a \sin \left (d x + c\right )^{10} - 3080 \, a \sin \left (d x + c\right )^{9} - 3465 \, a \sin \left (d x + c\right )^{8} + 1980 \, a \sin \left (d x + c\right )^{7} + 2310 \, a \sin \left (d x + c\right )^{6}}{13860 \, d} \]
1/13860*(1260*a*sin(d*x + c)^11 + 1386*a*sin(d*x + c)^10 - 3080*a*sin(d*x + c)^9 - 3465*a*sin(d*x + c)^8 + 1980*a*sin(d*x + c)^7 + 2310*a*sin(d*x + c)^6)/d
Time = 0.44 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.37 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {5 \, a \cos \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac {5 \, a \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac {a \sin \left (11 \, d x + 11 \, c\right )}{11264 \, d} + \frac {a \sin \left (9 \, d x + 9 \, c\right )}{9216 \, d} + \frac {5 \, a \sin \left (7 \, d x + 7 \, c\right )}{7168 \, d} - \frac {a \sin \left (5 \, d x + 5 \, c\right )}{1024 \, d} - \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{1536 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{512 \, d} \]
-1/5120*a*cos(10*d*x + 10*c)/d + 5/3072*a*cos(6*d*x + 6*c)/d - 5/512*a*cos (2*d*x + 2*c)/d - 1/11264*a*sin(11*d*x + 11*c)/d + 1/9216*a*sin(9*d*x + 9* c)/d + 5/7168*a*sin(7*d*x + 7*c)/d - 1/1024*a*sin(5*d*x + 5*c)/d - 5/1536* a*sin(3*d*x + 3*c)/d + 5/512*a*sin(d*x + c)/d
Time = 9.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {a\,{\sin \left (c+d\,x\right )}^{11}}{11}+\frac {a\,{\sin \left (c+d\,x\right )}^{10}}{10}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a\,{\sin \left (c+d\,x\right )}^8}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}}{d} \]