Integrand size = 27, antiderivative size = 97 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^6(c+d x)}{6 d}+\frac {a \sin ^7(c+d x)}{7 d}-\frac {a \sin ^8(c+d x)}{4 d}-\frac {2 a \sin ^9(c+d x)}{9 d}+\frac {a \sin ^{10}(c+d x)}{10 d}+\frac {a \sin ^{11}(c+d x)}{11 d} \] Output:
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 = 0.62 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a (34650 \cos (2 (c+d x))-5775 \cos (6 (c+d x))+693 \cos (10 (c+d x))-34650 \sin (c+d x)+11550 \sin (3 (c+d x))+3465 \sin (5 (c+d x))-2475 \sin (7 (c+d x))-385 \sin (9 (c+d x))+315 \sin (11 (c+d x)))}{3548160 d} \] Input:
Integrate[Cos[c + d*x]^5*Sin[c + d*x]^5*(a + a*Sin[c + d*x]),x]
Output:
-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 = 0.30 (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 |
\(\displaystyle \frac {\frac {1}{11} a^{11} \sin ^{11}(c+d x)+\frac {1}{10} a^{11} \sin ^{10}(c+d x)-\frac {2}{9} a^{11} \sin ^9(c+d x)-\frac {1}{4} a^{11} \sin ^8(c+d x)+\frac {1}{7} a^{11} \sin ^7(c+d x)+\frac {1}{6} a^{11} \sin ^6(c+d x)}{a^{10} d}\) |
Input:
Int[Cos[c + d*x]^5*Sin[c + d*x]^5*(a + a*Sin[c + d*x]),x]
Output:
((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)
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 = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69
\[\frac {a \left (\frac {\sin \left (d x +c \right )^{11}}{11}+\frac {\sin \left (d x +c \right )^{10}}{10}-\frac {2 \sin \left (d x +c \right )^{9}}{9}-\frac {\sin \left (d x +c \right )^{8}}{4}+\frac {\sin \left (d x +c \right )^{7}}{7}+\frac {\sin \left (d x +c \right )^{6}}{6}\right )}{d}\]
Input:
int(cos(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c)),x)
Output:
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 = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {1386 \, a \cos \left (d x + c\right )^{10} - 3465 \, a \cos \left (d x + c\right )^{8} + 2310 \, a \cos \left (d x + c\right )^{6} + 20 \, {\left (63 \, a \cos \left (d x + c\right )^{10} - 161 \, a \cos \left (d x + c\right )^{8} + 113 \, a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} - 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\right )} \sin \left (d x + c\right )}{13860 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-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 = 1.90 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.40 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{11}{\left (c + d x \right )}}{693 d} + \frac {a \sin ^{10}{\left (c + d x \right )}}{60 d} + \frac {4 a \sin ^{9}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{63 d} + \frac {a \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{12 d} + \frac {a \sin ^{7}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{7 d} + \frac {a \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{5}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**5*(a+a*sin(d*x+c)),x)
Output:
Piecewise((8*a*sin(c + d*x)**11/(693*d) + a*sin(c + d*x)**10/(60*d) + 4*a* sin(c + d*x)**9*cos(c + d*x)**2/(63*d) + a*sin(c + d*x)**8*cos(c + d*x)**2 /(12*d) + a*sin(c + d*x)**7*cos(c + d*x)**4/(7*d) + a*sin(c + d*x)**6*cos( c + d*x)**4/(6*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**5*cos(c)**5, True) )
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {1260 \, a \sin \left (d x + c\right )^{11} + 1386 \, a \sin \left (d x + c\right )^{10} - 3080 \, a \sin \left (d x + c\right )^{9} - 3465 \, a \sin \left (d x + c\right )^{8} + 1980 \, a \sin \left (d x + c\right )^{7} + 2310 \, a \sin \left (d x + c\right )^{6}}{13860 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
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 = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {1260 \, a \sin \left (d x + c\right )^{11} + 1386 \, a \sin \left (d x + c\right )^{10} - 3080 \, a \sin \left (d x + c\right )^{9} - 3465 \, a \sin \left (d x + c\right )^{8} + 1980 \, a \sin \left (d x + c\right )^{7} + 2310 \, a \sin \left (d x + c\right )^{6}}{13860 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
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 = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {a\,{\sin \left (c+d\,x\right )}^{11}}{11}+\frac {a\,{\sin \left (c+d\,x\right )}^{10}}{10}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a\,{\sin \left (c+d\,x\right )}^8}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}}{d} \] Input:
int(cos(c + d*x)^5*sin(c + d*x)^5*(a + a*sin(c + d*x)),x)
Output:
((a*sin(c + d*x)^6)/6 + (a*sin(c + d*x)^7)/7 - (a*sin(c + d*x)^8)/4 - (2*a *sin(c + d*x)^9)/9 + (a*sin(c + d*x)^10)/10 + (a*sin(c + d*x)^11)/11)/d
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.66 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right )^{6} a \left (1260 \sin \left (d x +c \right )^{5}+1386 \sin \left (d x +c \right )^{4}-3080 \sin \left (d x +c \right )^{3}-3465 \sin \left (d x +c \right )^{2}+1980 \sin \left (d x +c \right )+2310\right )}{13860 d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)**6*a*(1260*sin(c + d*x)**5 + 1386*sin(c + d*x)**4 - 3080*sin (c + d*x)**3 - 3465*sin(c + d*x)**2 + 1980*sin(c + d*x) + 2310))/(13860*d)