Integrand size = 29, antiderivative size = 109 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {4 (a+a \sin (c+d x))^5}{5 a^3 d}-\frac {2 (a+a \sin (c+d x))^6}{a^4 d}+\frac {13 (a+a \sin (c+d x))^7}{7 a^5 d}-\frac {3 (a+a \sin (c+d x))^8}{4 a^6 d}+\frac {(a+a \sin (c+d x))^9}{9 a^7 d} \] Output:
4/5*(a+a*sin(d*x+c))^5/a^3/d-2*(a+a*sin(d*x+c))^6/a^4/d+13/7*(a+a*sin(d*x+ c))^7/a^5/d-3/4*(a+a*sin(d*x+c))^8/a^6/d+1/9*(a+a*sin(d*x+c))^9/a^7/d
Time = 0.62 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 (7560 \cos (2 (c+d x))+1260 \cos (4 (c+d x))-840 \cos (6 (c+d x))-315 \cos (8 (c+d x))-16380 \sin (c+d x)+1680 \sin (3 (c+d x))+2016 \sin (5 (c+d x))+270 \sin (7 (c+d x))-70 \sin (9 (c+d x)))}{161280 d} \] Input:
Integrate[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
-1/161280*(a^2*(7560*Cos[2*(c + d*x)] + 1260*Cos[4*(c + d*x)] - 840*Cos[6* (c + d*x)] - 315*Cos[8*(c + d*x)] - 16380*Sin[c + d*x] + 1680*Sin[3*(c + d *x)] + 2016*Sin[5*(c + d*x)] + 270*Sin[7*(c + d*x)] - 70*Sin[9*(c + d*x)]) )/d
Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88, 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.
\(\displaystyle \int \sin ^2(c+d x) \cos ^5(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^5 (a \sin (c+d x)+a)^2dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^2 \sin ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left ((\sin (c+d x) a+a)^8-6 a (\sin (c+d x) a+a)^7+13 a^2 (\sin (c+d x) a+a)^6-12 a^3 (\sin (c+d x) a+a)^5+4 a^4 (\sin (c+d x) a+a)^4\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {4}{5} a^4 (a \sin (c+d x)+a)^5-2 a^3 (a \sin (c+d x)+a)^6+\frac {13}{7} a^2 (a \sin (c+d x)+a)^7+\frac {1}{9} (a \sin (c+d x)+a)^9-\frac {3}{4} a (a \sin (c+d x)+a)^8}{a^7 d}\) |
Input:
Int[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
((4*a^4*(a + a*Sin[c + d*x])^5)/5 - 2*a^3*(a + a*Sin[c + d*x])^6 + (13*a^2 *(a + a*Sin[c + d*x])^7)/7 - (3*a*(a + a*Sin[c + d*x])^8)/4 + (a + a*Sin[c + d*x])^9/9)/(a^7*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 = 0.72
\[\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{9}}{9}+\frac {\sin \left (d x +c \right )^{8}}{4}-\frac {\sin \left (d x +c \right )^{7}}{7}-\frac {2 \sin \left (d x +c \right )^{6}}{3}-\frac {\sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4}}{2}+\frac {\sin \left (d x +c \right )^{3}}{3}\right )}{d}\]
Input:
int(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
Output:
a^2/d*(1/9*sin(d*x+c)^9+1/4*sin(d*x+c)^8-1/7*sin(d*x+c)^7-2/3*sin(d*x+c)^6 -1/5*sin(d*x+c)^5+1/2*sin(d*x+c)^4+1/3*sin(d*x+c)^3)
Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {315 \, a^{2} \cos \left (d x + c\right )^{8} - 420 \, a^{2} \cos \left (d x + c\right )^{6} + 4 \, {\left (35 \, a^{2} \cos \left (d x + c\right )^{8} - 95 \, a^{2} \cos \left (d x + c\right )^{6} + 12 \, a^{2} \cos \left (d x + c\right )^{4} + 16 \, a^{2} \cos \left (d x + c\right )^{2} + 32 \, a^{2}\right )} \sin \left (d x + c\right )}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/1260*(315*a^2*cos(d*x + c)^8 - 420*a^2*cos(d*x + c)^6 + 4*(35*a^2*cos(d* x + c)^8 - 95*a^2*cos(d*x + c)^6 + 12*a^2*cos(d*x + c)^4 + 16*a^2*cos(d*x + c)^2 + 32*a^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (97) = 194\).
Time = 0.95 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.96 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {8 a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {a^{2} \sin ^{8}{\left (c + d x \right )}}{12 d} + \frac {4 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {8 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {a^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {4 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**2*(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((8*a**2*sin(c + d*x)**9/(315*d) + a**2*sin(c + d*x)**8/(12*d) + 4*a**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 8*a**2*sin(c + d*x)**7/(10 5*d) + a**2*sin(c + d*x)**6*cos(c + d*x)**2/(3*d) + a**2*sin(c + d*x)**5*c os(c + d*x)**4/(5*d) + 4*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + a** 2*sin(c + d*x)**4*cos(c + d*x)**4/(2*d) + a**2*sin(c + d*x)**3*cos(c + d*x )**4/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**2*cos(c)**5, True))
Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {140 \, a^{2} \sin \left (d x + c\right )^{9} + 315 \, a^{2} \sin \left (d x + c\right )^{8} - 180 \, a^{2} \sin \left (d x + c\right )^{7} - 840 \, a^{2} \sin \left (d x + c\right )^{6} - 252 \, a^{2} \sin \left (d x + c\right )^{5} + 630 \, a^{2} \sin \left (d x + c\right )^{4} + 420 \, a^{2} \sin \left (d x + c\right )^{3}}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/1260*(140*a^2*sin(d*x + c)^9 + 315*a^2*sin(d*x + c)^8 - 180*a^2*sin(d*x + c)^7 - 840*a^2*sin(d*x + c)^6 - 252*a^2*sin(d*x + c)^5 + 630*a^2*sin(d*x + c)^4 + 420*a^2*sin(d*x + c)^3)/d
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {140 \, a^{2} \sin \left (d x + c\right )^{9} + 315 \, a^{2} \sin \left (d x + c\right )^{8} - 180 \, a^{2} \sin \left (d x + c\right )^{7} - 840 \, a^{2} \sin \left (d x + c\right )^{6} - 252 \, a^{2} \sin \left (d x + c\right )^{5} + 630 \, a^{2} \sin \left (d x + c\right )^{4} + 420 \, a^{2} \sin \left (d x + c\right )^{3}}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/1260*(140*a^2*sin(d*x + c)^9 + 315*a^2*sin(d*x + c)^8 - 180*a^2*sin(d*x + c)^7 - 840*a^2*sin(d*x + c)^6 - 252*a^2*sin(d*x + c)^5 + 630*a^2*sin(d*x + c)^4 + 420*a^2*sin(d*x + c)^3)/d
Time = 20.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {a^2\,{\sin \left (c+d\,x\right )}^8}{4}-\frac {a^2\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^6}{3}-\frac {a^2\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a^2\,{\sin \left (c+d\,x\right )}^4}{2}+\frac {a^2\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \] Input:
int(cos(c + d*x)^5*sin(c + d*x)^2*(a + a*sin(c + d*x))^2,x)
Output:
((a^2*sin(c + d*x)^3)/3 + (a^2*sin(c + d*x)^4)/2 - (a^2*sin(c + d*x)^5)/5 - (2*a^2*sin(c + d*x)^6)/3 - (a^2*sin(c + d*x)^7)/7 + (a^2*sin(c + d*x)^8) /4 + (a^2*sin(c + d*x)^9)/9)/d
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.70 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\sin \left (d x +c \right )^{3} a^{2} \left (140 \sin \left (d x +c \right )^{6}+315 \sin \left (d x +c \right )^{5}-180 \sin \left (d x +c \right )^{4}-840 \sin \left (d x +c \right )^{3}-252 \sin \left (d x +c \right )^{2}+630 \sin \left (d x +c \right )+420\right )}{1260 d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
Output:
(sin(c + d*x)**3*a**2*(140*sin(c + d*x)**6 + 315*sin(c + d*x)**5 - 180*sin (c + d*x)**4 - 840*sin(c + d*x)**3 - 252*sin(c + d*x)**2 + 630*sin(c + d*x ) + 420))/(1260*d)