Integrand size = 29, antiderivative size = 127 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^4(c+d x)}{4 d}+\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^6(c+d x)}{6 d}-\frac {4 a^2 \sin ^7(c+d x)}{7 d}-\frac {a^2 \sin ^8(c+d x)}{8 d}+\frac {2 a^2 \sin ^9(c+d x)}{9 d}+\frac {a^2 \sin ^{10}(c+d x)}{10 d} \] Output:
1/4*a^2*sin(d*x+c)^4/d+2/5*a^2*sin(d*x+c)^5/d-1/6*a^2*sin(d*x+c)^6/d-4/7*a ^2*sin(d*x+c)^7/d-1/8*a^2*sin(d*x+c)^8/d+2/9*a^2*sin(d*x+c)^9/d+1/10*a^2*s in(d*x+c)^10/d
Time = 0.78 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 (-2625+10710 \cos (2 (c+d x))+1260 \cos (4 (c+d x))-1365 \cos (6 (c+d x))-315 \cos (8 (c+d x))+63 \cos (10 (c+d x))-15120 \sin (c+d x)+3360 \sin (3 (c+d x))+2016 \sin (5 (c+d x))-360 \sin (7 (c+d x))-280 \sin (9 (c+d x)))}{322560 d} \] Input:
Integrate[Cos[c + d*x]^5*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]
Output:
-1/322560*(a^2*(-2625 + 10710*Cos[2*(c + d*x)] + 1260*Cos[4*(c + d*x)] - 1 365*Cos[6*(c + d*x)] - 315*Cos[8*(c + d*x)] + 63*Cos[10*(c + d*x)] - 15120 *Sin[c + d*x] + 3360*Sin[3*(c + d*x)] + 2016*Sin[5*(c + d*x)] - 360*Sin[7* (c + d*x)] - 280*Sin[9*(c + d*x)]))/d
Time = 0.34 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.89, 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, 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.
\(\displaystyle \int \sin ^3(c+d x) \cos ^5(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^3 \cos (c+d x)^5 (a \sin (c+d x)+a)^2dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^3 \sin ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\sin ^9(c+d x) a^9+2 \sin ^8(c+d x) a^9-\sin ^7(c+d x) a^9-4 \sin ^6(c+d x) a^9-\sin ^5(c+d x) a^9+2 \sin ^4(c+d x) a^9+\sin ^3(c+d x) a^9\right )d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{10} a^{10} \sin ^{10}(c+d x)+\frac {2}{9} a^{10} \sin ^9(c+d x)-\frac {1}{8} a^{10} \sin ^8(c+d x)-\frac {4}{7} a^{10} \sin ^7(c+d x)-\frac {1}{6} a^{10} \sin ^6(c+d x)+\frac {2}{5} a^{10} \sin ^5(c+d x)+\frac {1}{4} a^{10} \sin ^4(c+d x)}{a^8 d}\) |
Input:
Int[Cos[c + d*x]^5*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]
Output:
((a^10*Sin[c + d*x]^4)/4 + (2*a^10*Sin[c + d*x]^5)/5 - (a^10*Sin[c + d*x]^ 6)/6 - (4*a^10*Sin[c + d*x]^7)/7 - (a^10*Sin[c + d*x]^8)/8 + (2*a^10*Sin[c + d*x]^9)/9 + (a^10*Sin[c + d*x]^10)/10)/(a^8*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 = 79, normalized size of antiderivative = 0.62
\[\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{10}}{10}+\frac {2 \sin \left (d x +c \right )^{9}}{9}-\frac {\sin \left (d x +c \right )^{8}}{8}-\frac {4 \sin \left (d x +c \right )^{7}}{7}-\frac {\sin \left (d x +c \right )^{6}}{6}+\frac {2 \sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4}}{4}\right )}{d}\]
Input:
int(cos(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)
Output:
a^2/d*(1/10*sin(d*x+c)^10+2/9*sin(d*x+c)^9-1/8*sin(d*x+c)^8-4/7*sin(d*x+c) ^7-1/6*sin(d*x+c)^6+2/5*sin(d*x+c)^5+1/4*sin(d*x+c)^4)
Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.87 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {252 \, a^{2} \cos \left (d x + c\right )^{10} - 945 \, a^{2} \cos \left (d x + c\right )^{8} + 840 \, a^{2} \cos \left (d x + c\right )^{6} - 16 \, {\left (35 \, a^{2} \cos \left (d x + c\right )^{8} - 50 \, a^{2} \cos \left (d x + c\right )^{6} + 3 \, a^{2} \cos \left (d x + c\right )^{4} + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{2520 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/2520*(252*a^2*cos(d*x + c)^10 - 945*a^2*cos(d*x + c)^8 + 840*a^2*cos(d* x + c)^6 - 16*(35*a^2*cos(d*x + c)^8 - 50*a^2*cos(d*x + c)^6 + 3*a^2*cos(d *x + c)^4 + 4*a^2*cos(d*x + c)^2 + 8*a^2)*sin(d*x + c))/d
Time = 1.31 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.67 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} \sin ^{10}{\left (c + d x \right )}}{60 d} + \frac {16 a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {a^{2} \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{12 d} + \frac {a^{2} \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac {8 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {a^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{6 d} + \frac {a^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac {2 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{3}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**3*(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((a**2*sin(c + d*x)**10/(60*d) + 16*a**2*sin(c + d*x)**9/(315*d) + a**2*sin(c + d*x)**8*cos(c + d*x)**2/(12*d) + a**2*sin(c + d*x)**8/(24*d ) + 8*a**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + a**2*sin(c + d*x)**6*c os(c + d*x)**4/(6*d) + a**2*sin(c + d*x)**6*cos(c + d*x)**2/(6*d) + 2*a**2 *sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + a**2*sin(c + d*x)**4*cos(c + d*x) **4/(4*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**3*cos(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.76 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {252 \, a^{2} \sin \left (d x + c\right )^{10} + 560 \, a^{2} \sin \left (d x + c\right )^{9} - 315 \, a^{2} \sin \left (d x + c\right )^{8} - 1440 \, a^{2} \sin \left (d x + c\right )^{7} - 420 \, a^{2} \sin \left (d x + c\right )^{6} + 1008 \, a^{2} \sin \left (d x + c\right )^{5} + 630 \, a^{2} \sin \left (d x + c\right )^{4}}{2520 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/2520*(252*a^2*sin(d*x + c)^10 + 560*a^2*sin(d*x + c)^9 - 315*a^2*sin(d*x + c)^8 - 1440*a^2*sin(d*x + c)^7 - 420*a^2*sin(d*x + c)^6 + 1008*a^2*sin( d*x + c)^5 + 630*a^2*sin(d*x + c)^4)/d
Time = 0.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.76 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {252 \, a^{2} \sin \left (d x + c\right )^{10} + 560 \, a^{2} \sin \left (d x + c\right )^{9} - 315 \, a^{2} \sin \left (d x + c\right )^{8} - 1440 \, a^{2} \sin \left (d x + c\right )^{7} - 420 \, a^{2} \sin \left (d x + c\right )^{6} + 1008 \, a^{2} \sin \left (d x + c\right )^{5} + 630 \, a^{2} \sin \left (d x + c\right )^{4}}{2520 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/2520*(252*a^2*sin(d*x + c)^10 + 560*a^2*sin(d*x + c)^9 - 315*a^2*sin(d*x + c)^8 - 1440*a^2*sin(d*x + c)^7 - 420*a^2*sin(d*x + c)^6 + 1008*a^2*sin( d*x + c)^5 + 630*a^2*sin(d*x + c)^4)/d
Time = 18.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^{10}}{10}+\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a^2\,{\sin \left (c+d\,x\right )}^8}{8}-\frac {4\,a^2\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a^2\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a^2\,{\sin \left (c+d\,x\right )}^4}{4}}{d} \] Input:
int(cos(c + d*x)^5*sin(c + d*x)^3*(a + a*sin(c + d*x))^2,x)
Output:
((a^2*sin(c + d*x)^4)/4 + (2*a^2*sin(c + d*x)^5)/5 - (a^2*sin(c + d*x)^6)/ 6 - (4*a^2*sin(c + d*x)^7)/7 - (a^2*sin(c + d*x)^8)/8 + (2*a^2*sin(c + d*x )^9)/9 + (a^2*sin(c + d*x)^10)/10)/d
Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.60 \[ \int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\sin \left (d x +c \right )^{4} a^{2} \left (252 \sin \left (d x +c \right )^{6}+560 \sin \left (d x +c \right )^{5}-315 \sin \left (d x +c \right )^{4}-1440 \sin \left (d x +c \right )^{3}-420 \sin \left (d x +c \right )^{2}+1008 \sin \left (d x +c \right )+630\right )}{2520 d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)
Output:
(sin(c + d*x)**4*a**2*(252*sin(c + d*x)**6 + 560*sin(c + d*x)**5 - 315*sin (c + d*x)**4 - 1440*sin(c + d*x)**3 - 420*sin(c + d*x)**2 + 1008*sin(c + d *x) + 630))/(2520*d)