\(\int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx\) [217]

Optimal result

Integrand size = 27, antiderivative size = 91 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^5(c+d x)}{5 d}+\frac {2 a^4 \sin ^6(c+d x)}{3 d}+\frac {6 a^4 \sin ^7(c+d x)}{7 d}+\frac {a^4 \sin ^8(c+d x)}{2 d}+\frac {a^4 \sin ^9(c+d x)}{9 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2912, 12, 45} \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^9(c+d x)}{9 d}+\frac {a^4 \sin ^8(c+d x)}{2 d}+\frac {6 a^4 \sin ^7(c+d x)}{7 d}+\frac {2 a^4 \sin ^6(c+d x)}{3 d}+\frac {a^4 \sin ^5(c+d x)}{5 d} \]

[In]

Int[Cos[c + d*x]*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*Sin[c + d*x]^5)/(5*d) + (2*a^4*Sin[c + d*x]^6)/(3*d) + (6*a^4*Sin[c + d*x]^7)/(7*d) + (a^4*Sin[c + d*x]^8
)/(2*d) + (a^4*Sin[c + d*x]^9)/(9*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4 (a+x)^4}{a^4} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int x^4 (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (a^4 x^4+4 a^3 x^5+6 a^2 x^6+4 a x^7+x^8\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.10 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 (4095-42840 \cos (2 (c+d x))+18900 \cos (4 (c+d x))-4200 \cos (6 (c+d x))+315 \cos (8 (c+d x))+52290 \sin (c+d x)-30660 \sin (3 (c+d x))+9828 \sin (5 (c+d x))-1395 \sin (7 (c+d x))+35 \sin (9 (c+d x)))}{80640 d} \]

[In]

Integrate[Cos[c + d*x]*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^4,x]

[Out]

(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)

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\frac {a^{4} \left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {a^{4} \left (\sin ^{8}\left (d x +c \right )\right )}{2}+\frac {6 a^{4} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {2 a^{4} \left (\sin ^{6}\left (d x +c \right )\right )}{3}+\frac {a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}}{d}\)	\(71\)
default	\(\frac {\frac {a^{4} \left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {a^{4} \left (\sin ^{8}\left (d x +c \right )\right )}{2}+\frac {6 a^{4} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {2 a^{4} \left (\sin ^{6}\left (d x +c \right )\right )}{3}+\frac {a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}}{d}\)	\(71\)
parallelrisch	\(-\frac {a^{4} \left (\sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-5 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (1220 \cos \left (2 d x +2 c \right )-2625 \sin \left (d x +c \right )+315 \sin \left (3 d x +3 c \right )-35 \cos \left (4 d x +4 c \right )-1689\right ) \left (\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+5 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20160 d}\)	\(116\)
risch	\(\frac {83 a^{4} \sin \left (d x +c \right )}{128 d}+\frac {a^{4} \sin \left (9 d x +9 c \right )}{2304 d}+\frac {a^{4} \cos \left (8 d x +8 c \right )}{256 d}-\frac {31 a^{4} \sin \left (7 d x +7 c \right )}{1792 d}-\frac {5 a^{4} \cos \left (6 d x +6 c \right )}{96 d}+\frac {39 a^{4} \sin \left (5 d x +5 c \right )}{320 d}+\frac {15 a^{4} \cos \left (4 d x +4 c \right )}{64 d}-\frac {73 a^{4} \sin \left (3 d x +3 c \right )}{192 d}-\frac {17 a^{4} \cos \left (2 d x +2 c \right )}{32 d}\)	\(152\)
norman	\(\frac {\frac {32 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {4736 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {99136 a^{4} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 d}+\frac {4736 a^{4} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {32 a^{4} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {128 a^{4} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {128 a^{4} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {256 a^{4} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {256 a^{4} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}\)	\(189\)

[In]

int(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.36 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {315 \, a^{4} \cos \left (d x + c\right )^{8} - 1680 \, a^{4} \cos \left (d x + c\right )^{6} + 3150 \, a^{4} \cos \left (d x + c\right )^{4} - 2520 \, a^{4} \cos \left (d x + c\right )^{2} + 2 \, {\left (35 \, a^{4} \cos \left (d x + c\right )^{8} - 410 \, a^{4} \cos \left (d x + c\right )^{6} + 1083 \, a^{4} \cos \left (d x + c\right )^{4} - 1076 \, a^{4} \cos \left (d x + c\right )^{2} + 368 \, a^{4}\right )} \sin \left (d x + c\right )}{630 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\begin {cases} \frac {a^{4} \sin ^{9}{\left (c + d x \right )}}{9 d} + \frac {a^{4} \sin ^{8}{\left (c + d x \right )}}{2 d} + \frac {6 a^{4} \sin ^{7}{\left (c + d x \right )}}{7 d} + \frac {2 a^{4} \sin ^{6}{\left (c + d x \right )}}{3 d} + \frac {a^{4} \sin ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{4} \sin ^{4}{\left (c \right )} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)**4*(a+a*sin(d*x+c))**4,x)

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {70 \, a^{4} \sin \left (d x + c\right )^{9} + 315 \, a^{4} \sin \left (d x + c\right )^{8} + 540 \, a^{4} \sin \left (d x + c\right )^{7} + 420 \, a^{4} \sin \left (d x + c\right )^{6} + 126 \, a^{4} \sin \left (d x + c\right )^{5}}{630 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {70 \, a^{4} \sin \left (d x + c\right )^{9} + 315 \, a^{4} \sin \left (d x + c\right )^{8} + 540 \, a^{4} \sin \left (d x + c\right )^{7} + 420 \, a^{4} \sin \left (d x + c\right )^{6} + 126 \, a^{4} \sin \left (d x + c\right )^{5}}{630 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.47 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\frac {a^4\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {a^4\,{\sin \left (c+d\,x\right )}^8}{2}+\frac {6\,a^4\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {2\,a^4\,{\sin \left (c+d\,x\right )}^6}{3}+\frac {a^4\,{\sin \left (c+d\,x\right )}^5}{5}}{d} \]

[In]

int(cos(c + d*x)*sin(c + d*x)^4*(a + a*sin(c + d*x))^4,x)

[Out]

((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