Integrand size = 27, antiderivative size = 91 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^5(c+d x)}{5 d}+\frac {2 a^4 \sin ^6(c+d x)}{3 d}+\frac {6 a^4 \sin ^7(c+d x)}{7 d}+\frac {a^4 \sin ^8(c+d x)}{2 d}+\frac {a^4 \sin ^9(c+d x)}{9 d} \] Output:
1/5*a^4*sin(d*x+c)^5/d+2/3*a^4*sin(d*x+c)^6/d+6/7*a^4*sin(d*x+c)^7/d+1/2*a ^4*sin(d*x+c)^8/d+1/9*a^4*sin(d*x+c)^9/d
Time = 0.53 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.10 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 (4095-42840 \cos (2 (c+d x))+18900 \cos (4 (c+d x))-4200 \cos (6 (c+d x))+315 \cos (8 (c+d x))+52290 \sin (c+d x)-30660 \sin (3 (c+d x))+9828 \sin (5 (c+d x))-1395 \sin (7 (c+d x))+35 \sin (9 (c+d x)))}{80640 d} \] Input:
Integrate[Cos[c + d*x]*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^4,x]
Output:
(a^4*(4095 - 42840*Cos[2*(c + d*x)] + 18900*Cos[4*(c + d*x)] - 4200*Cos[6* (c + d*x)] + 315*Cos[8*(c + d*x)] + 52290*Sin[c + d*x] - 30660*Sin[3*(c + d*x)] + 9828*Sin[5*(c + d*x)] - 1395*Sin[7*(c + d*x)] + 35*Sin[9*(c + d*x) ]))/(80640*d)
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 49, 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.
\(\displaystyle \int \sin ^4(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^4 \cos (c+d x) (a \sin (c+d x)+a)^4dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \sin ^4(c+d x) (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^4 \sin ^4(c+d x) (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\sin ^8(c+d x) a^8+4 \sin ^7(c+d x) a^8+6 \sin ^6(c+d x) a^8+4 \sin ^5(c+d x) a^8+\sin ^4(c+d x) a^8\right )d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{9} a^9 \sin ^9(c+d x)+\frac {1}{2} a^9 \sin ^8(c+d x)+\frac {6}{7} a^9 \sin ^7(c+d x)+\frac {2}{3} a^9 \sin ^6(c+d x)+\frac {1}{5} a^9 \sin ^5(c+d x)}{a^5 d}\) |
Input:
Int[Cos[c + d*x]*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^4,x]
Output:
((a^9*Sin[c + d*x]^5)/5 + (2*a^9*Sin[c + d*x]^6)/3 + (6*a^9*Sin[c + d*x]^7 )/7 + (a^9*Sin[c + d*x]^8)/2 + (a^9*Sin[c + d*x]^9)/9)/(a^5*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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78
\[\frac {\frac {a^{4} \sin \left (d x +c \right )^{9}}{9}+\frac {a^{4} \sin \left (d x +c \right )^{8}}{2}+\frac {6 a^{4} \sin \left (d x +c \right )^{7}}{7}+\frac {2 a^{4} \sin \left (d x +c \right )^{6}}{3}+\frac {a^{4} \sin \left (d x +c \right )^{5}}{5}}{d}\]
Input:
int(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^4,x)
Output:
1/d*(1/9*a^4*sin(d*x+c)^9+1/2*a^4*sin(d*x+c)^8+6/7*a^4*sin(d*x+c)^7+2/3*a^ 4*sin(d*x+c)^6+1/5*a^4*sin(d*x+c)^5)
Time = 0.08 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.36 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {315 \, a^{4} \cos \left (d x + c\right )^{8} - 1680 \, a^{4} \cos \left (d x + c\right )^{6} + 3150 \, a^{4} \cos \left (d x + c\right )^{4} - 2520 \, a^{4} \cos \left (d x + c\right )^{2} + 2 \, {\left (35 \, a^{4} \cos \left (d x + c\right )^{8} - 410 \, a^{4} \cos \left (d x + c\right )^{6} + 1083 \, a^{4} \cos \left (d x + c\right )^{4} - 1076 \, a^{4} \cos \left (d x + c\right )^{2} + 368 \, a^{4}\right )} \sin \left (d x + c\right )}{630 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^4,x, algorithm="fricas" )
Output:
1/630*(315*a^4*cos(d*x + c)^8 - 1680*a^4*cos(d*x + c)^6 + 3150*a^4*cos(d*x + c)^4 - 2520*a^4*cos(d*x + c)^2 + 2*(35*a^4*cos(d*x + c)^8 - 410*a^4*cos (d*x + c)^6 + 1083*a^4*cos(d*x + c)^4 - 1076*a^4*cos(d*x + c)^2 + 368*a^4) *sin(d*x + c))/d
Time = 0.97 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\begin {cases} \frac {a^{4} \sin ^{9}{\left (c + d x \right )}}{9 d} + \frac {a^{4} \sin ^{8}{\left (c + d x \right )}}{2 d} + \frac {6 a^{4} \sin ^{7}{\left (c + d x \right )}}{7 d} + \frac {2 a^{4} \sin ^{6}{\left (c + d x \right )}}{3 d} + \frac {a^{4} \sin ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{4} \sin ^{4}{\left (c \right )} \cos {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**4*(a+a*sin(d*x+c))**4,x)
Output:
Piecewise((a**4*sin(c + d*x)**9/(9*d) + a**4*sin(c + d*x)**8/(2*d) + 6*a** 4*sin(c + d*x)**7/(7*d) + 2*a**4*sin(c + d*x)**6/(3*d) + a**4*sin(c + d*x) **5/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**4*sin(c)**4*cos(c), True))
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {70 \, a^{4} \sin \left (d x + c\right )^{9} + 315 \, a^{4} \sin \left (d x + c\right )^{8} + 540 \, a^{4} \sin \left (d x + c\right )^{7} + 420 \, a^{4} \sin \left (d x + c\right )^{6} + 126 \, a^{4} \sin \left (d x + c\right )^{5}}{630 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^4,x, algorithm="maxima" )
Output:
1/630*(70*a^4*sin(d*x + c)^9 + 315*a^4*sin(d*x + c)^8 + 540*a^4*sin(d*x + c)^7 + 420*a^4*sin(d*x + c)^6 + 126*a^4*sin(d*x + c)^5)/d
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {70 \, a^{4} \sin \left (d x + c\right )^{9} + 315 \, a^{4} \sin \left (d x + c\right )^{8} + 540 \, a^{4} \sin \left (d x + c\right )^{7} + 420 \, a^{4} \sin \left (d x + c\right )^{6} + 126 \, a^{4} \sin \left (d x + c\right )^{5}}{630 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
1/630*(70*a^4*sin(d*x + c)^9 + 315*a^4*sin(d*x + c)^8 + 540*a^4*sin(d*x + c)^7 + 420*a^4*sin(d*x + c)^6 + 126*a^4*sin(d*x + c)^5)/d
Time = 18.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\frac {a^4\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {a^4\,{\sin \left (c+d\,x\right )}^8}{2}+\frac {6\,a^4\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {2\,a^4\,{\sin \left (c+d\,x\right )}^6}{3}+\frac {a^4\,{\sin \left (c+d\,x\right )}^5}{5}}{d} \] Input:
int(cos(c + d*x)*sin(c + d*x)^4*(a + a*sin(c + d*x))^4,x)
Output:
((a^4*sin(c + d*x)^5)/5 + (2*a^4*sin(c + d*x)^6)/3 + (6*a^4*sin(c + d*x)^7 )/7 + (a^4*sin(c + d*x)^8)/2 + (a^4*sin(c + d*x)^9)/9)/d
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\sin \left (d x +c \right )^{5} a^{4} \left (70 \sin \left (d x +c \right )^{4}+315 \sin \left (d x +c \right )^{3}+540 \sin \left (d x +c \right )^{2}+420 \sin \left (d x +c \right )+126\right )}{630 d} \] Input:
int(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^4,x)
Output:
(sin(c + d*x)**5*a**4*(70*sin(c + d*x)**4 + 315*sin(c + d*x)**3 + 540*sin( c + d*x)**2 + 420*sin(c + d*x) + 126))/(630*d)