Integrand size = 27, antiderivative size = 89 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {2 (a+a \sin (c+d x))^6}{3 a^3 d}+\frac {8 (a+a \sin (c+d x))^7}{7 a^4 d}-\frac {5 (a+a \sin (c+d x))^8}{8 a^5 d}+\frac {(a+a \sin (c+d x))^9}{9 a^6 d} \] Output:
-2/3*(a+a*sin(d*x+c))^6/a^3/d+8/7*(a+a*sin(d*x+c))^7/a^4/d-5/8*(a+a*sin(d* x+c))^8/a^5/d+1/9*(a+a*sin(d*x+c))^9/a^6/d
Time = 0.63 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.12 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 (4662-9576 \cos (2 (c+d x))-2772 \cos (4 (c+d x))+168 \cos (6 (c+d x))+189 \cos (8 (c+d x))+16632 \sin (c+d x)-1344 \sin (3 (c+d x))-2016 \sin (5 (c+d x))-396 \sin (7 (c+d x))+28 \sin (9 (c+d x)))}{64512 d} \] Input:
Integrate[Cos[c + d*x]^5*Sin[c + d*x]*(a + a*Sin[c + d*x])^3,x]
Output:
(a^3*(4662 - 9576*Cos[2*(c + d*x)] - 2772*Cos[4*(c + d*x)] + 168*Cos[6*(c + d*x)] + 189*Cos[8*(c + d*x)] + 16632*Sin[c + d*x] - 1344*Sin[3*(c + d*x) ] - 2016*Sin[5*(c + d*x)] - 396*Sin[7*(c + d*x)] + 28*Sin[9*(c + d*x)]))/( 64512*d)
Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 86, 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.
\(\displaystyle \int \sin (c+d x) \cos ^5(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x) \cos (c+d x)^5 (a \sin (c+d x)+a)^3dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin (c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a \sin (c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left ((\sin (c+d x) a+a)^8-5 a (\sin (c+d x) a+a)^7+8 a^2 (\sin (c+d x) a+a)^6-4 a^3 (\sin (c+d x) a+a)^5\right )d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2}{3} a^3 (a \sin (c+d x)+a)^6+\frac {8}{7} a^2 (a \sin (c+d x)+a)^7+\frac {1}{9} (a \sin (c+d x)+a)^9-\frac {5}{8} a (a \sin (c+d x)+a)^8}{a^6 d}\) |
Input:
Int[Cos[c + d*x]^5*Sin[c + d*x]*(a + a*Sin[c + d*x])^3,x]
Output:
((-2*a^3*(a + a*Sin[c + d*x])^6)/3 + (8*a^2*(a + a*Sin[c + d*x])^7)/7 - (5 *a*(a + a*Sin[c + d*x])^8)/8 + (a + a*Sin[c + d*x])^9/9)/(a^6*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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 87, normalized size of antiderivative = 0.98
\[\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{9}}{9}+\frac {3 \sin \left (d x +c \right )^{8}}{8}+\frac {\sin \left (d x +c \right )^{7}}{7}-\frac {5 \sin \left (d x +c \right )^{6}}{6}-\sin \left (d x +c \right )^{5}+\frac {\sin \left (d x +c \right )^{4}}{4}+\sin \left (d x +c \right )^{3}+\frac {\sin \left (d x +c \right )^{2}}{2}\right )}{d}\]
Input:
int(cos(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c))^3,x)
Output:
a^3/d*(1/9*sin(d*x+c)^9+3/8*sin(d*x+c)^8+1/7*sin(d*x+c)^7-5/6*sin(d*x+c)^6 -sin(d*x+c)^5+1/4*sin(d*x+c)^4+sin(d*x+c)^3+1/2*sin(d*x+c)^2)
Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.10 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {189 \, a^{3} \cos \left (d x + c\right )^{8} - 336 \, a^{3} \cos \left (d x + c\right )^{6} + 8 \, {\left (7 \, a^{3} \cos \left (d x + c\right )^{8} - 37 \, a^{3} \cos \left (d x + c\right )^{6} + 6 \, a^{3} \cos \left (d x + c\right )^{4} + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 16 \, a^{3}\right )} \sin \left (d x + c\right )}{504 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
1/504*(189*a^3*cos(d*x + c)^8 - 336*a^3*cos(d*x + c)^6 + 8*(7*a^3*cos(d*x + c)^8 - 37*a^3*cos(d*x + c)^6 + 6*a^3*cos(d*x + c)^4 + 8*a^3*cos(d*x + c) ^2 + 16*a^3)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (78) = 156\).
Time = 0.98 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.56 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx=\begin {cases} \frac {8 a^{3} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {a^{3} \sin ^{8}{\left (c + d x \right )}}{8 d} + \frac {4 a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {8 a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {a^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac {a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {4 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {3 a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a^{3} \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{3} \sin {\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)*(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((8*a**3*sin(c + d*x)**9/(315*d) + a**3*sin(c + d*x)**8/(8*d) + 4 *a**3*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 8*a**3*sin(c + d*x)**7/(35* d) + a**3*sin(c + d*x)**6*cos(c + d*x)**2/(2*d) + a**3*sin(c + d*x)**5*cos (c + d*x)**4/(5*d) + 4*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 3*a**3 *sin(c + d*x)**4*cos(c + d*x)**4/(4*d) + a**3*sin(c + d*x)**3*cos(c + d*x) **4/d - a**3*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c) *cos(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.24 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {56 \, a^{3} \sin \left (d x + c\right )^{9} + 189 \, a^{3} \sin \left (d x + c\right )^{8} + 72 \, a^{3} \sin \left (d x + c\right )^{7} - 420 \, a^{3} \sin \left (d x + c\right )^{6} - 504 \, a^{3} \sin \left (d x + c\right )^{5} + 126 \, a^{3} \sin \left (d x + c\right )^{4} + 504 \, a^{3} \sin \left (d x + c\right )^{3} + 252 \, a^{3} \sin \left (d x + c\right )^{2}}{504 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
1/504*(56*a^3*sin(d*x + c)^9 + 189*a^3*sin(d*x + c)^8 + 72*a^3*sin(d*x + c )^7 - 420*a^3*sin(d*x + c)^6 - 504*a^3*sin(d*x + c)^5 + 126*a^3*sin(d*x + c)^4 + 504*a^3*sin(d*x + c)^3 + 252*a^3*sin(d*x + c)^2)/d
Time = 0.15 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.24 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {56 \, a^{3} \sin \left (d x + c\right )^{9} + 189 \, a^{3} \sin \left (d x + c\right )^{8} + 72 \, a^{3} \sin \left (d x + c\right )^{7} - 420 \, a^{3} \sin \left (d x + c\right )^{6} - 504 \, a^{3} \sin \left (d x + c\right )^{5} + 126 \, a^{3} \sin \left (d x + c\right )^{4} + 504 \, a^{3} \sin \left (d x + c\right )^{3} + 252 \, a^{3} \sin \left (d x + c\right )^{2}}{504 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
1/504*(56*a^3*sin(d*x + c)^9 + 189*a^3*sin(d*x + c)^8 + 72*a^3*sin(d*x + c )^7 - 420*a^3*sin(d*x + c)^6 - 504*a^3*sin(d*x + c)^5 + 126*a^3*sin(d*x + c)^4 + 504*a^3*sin(d*x + c)^3 + 252*a^3*sin(d*x + c)^2)/d
Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.21 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {a^3\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {5\,a^3\,{\sin \left (c+d\,x\right )}^6}{6}-a^3\,{\sin \left (c+d\,x\right )}^5+\frac {a^3\,{\sin \left (c+d\,x\right )}^4}{4}+a^3\,{\sin \left (c+d\,x\right )}^3+\frac {a^3\,{\sin \left (c+d\,x\right )}^2}{2}}{d} \] Input:
int(cos(c + d*x)^5*sin(c + d*x)*(a + a*sin(c + d*x))^3,x)
Output:
((a^3*sin(c + d*x)^2)/2 + a^3*sin(c + d*x)^3 + (a^3*sin(c + d*x)^4)/4 - a^ 3*sin(c + d*x)^5 - (5*a^3*sin(c + d*x)^6)/6 + (a^3*sin(c + d*x)^7)/7 + (3* a^3*sin(c + d*x)^8)/8 + (a^3*sin(c + d*x)^9)/9)/d
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\sin \left (d x +c \right )^{2} a^{3} \left (56 \sin \left (d x +c \right )^{7}+189 \sin \left (d x +c \right )^{6}+72 \sin \left (d x +c \right )^{5}-420 \sin \left (d x +c \right )^{4}-504 \sin \left (d x +c \right )^{3}+126 \sin \left (d x +c \right )^{2}+504 \sin \left (d x +c \right )+252\right )}{504 d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c))^3,x)
Output:
(sin(c + d*x)**2*a**3*(56*sin(c + d*x)**7 + 189*sin(c + d*x)**6 + 72*sin(c + d*x)**5 - 420*sin(c + d*x)**4 - 504*sin(c + d*x)**3 + 126*sin(c + d*x)* *2 + 504*sin(c + d*x) + 252))/(504*d)