\(\int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx\) [495]

Optimal result

Integrand size = 27, antiderivative size = 97 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^6(c+d x)}{6 d}-\frac {2 a \sin ^7(c+d x)}{7 d}-\frac {a \sin ^8(c+d x)}{4 d}+\frac {a \sin ^9(c+d x)}{9 d}+\frac {a \sin ^{10}(c+d x)}{10 d} \]

[Out]

1/5*a*sin(d*x+c)^5/d+1/6*a*sin(d*x+c)^6/d-2/7*a*sin(d*x+c)^7/d-1/4*a*sin(d*x+c)^8/d+1/9*a*sin(d*x+c)^9/d+1/10*
a*sin(d*x+c)^10/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 97, 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 = {2915, 12, 90} \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^{10}(c+d x)}{10 d}+\frac {a \sin ^9(c+d x)}{9 d}-\frac {a \sin ^8(c+d x)}{4 d}-\frac {2 a \sin ^7(c+d x)}{7 d}+\frac {a \sin ^6(c+d x)}{6 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[In]

Int[Cos[c + d*x]^5*Sin[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

(a*Sin[c + d*x]^5)/(5*d) + (a*Sin[c + d*x]^6)/(6*d) - (2*a*Sin[c + d*x]^7)/(7*d) - (a*Sin[c + d*x]^8)/(4*d) +
(a*Sin[c + d*x]^9)/(9*d) + (a*Sin[c + d*x]^10)/(10*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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(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]))

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[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] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2 x^4 (a+x)^3}{a^4} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int (a-x)^2 x^4 (a+x)^3 \, dx,x,a \sin (c+d x)\right )}{a^9 d} \\ & = \frac {\text {Subst}\left (\int \left (a^5 x^4+a^4 x^5-2 a^3 x^6-2 a^2 x^7+a x^8+x^9\right ) \, dx,x,a \sin (c+d x)\right )}{a^9 d} \\ & = \frac {a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^6(c+d x)}{6 d}-\frac {2 a \sin ^7(c+d x)}{7 d}-\frac {a \sin ^8(c+d x)}{4 d}+\frac {a \sin ^9(c+d x)}{9 d}+\frac {a \sin ^{10}(c+d x)}{10 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a (3150 \cos (2 (c+d x))-525 \cos (6 (c+d x))+63 \cos (10 (c+d x))-7560 \sin (c+d x)+1680 \sin (3 (c+d x))+1008 \sin (5 (c+d x))-180 \sin (7 (c+d x))-140 \sin (9 (c+d x)))}{322560 d} \]

[In]

Integrate[Cos[c + d*x]^5*Sin[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

-1/322560*(a*(3150*Cos[2*(c + d*x)] - 525*Cos[6*(c + d*x)] + 63*Cos[10*(c + d*x)] - 7560*Sin[c + d*x] + 1680*S
in[3*(c + d*x)] + 1008*Sin[5*(c + d*x)] - 180*Sin[7*(c + d*x)] - 140*Sin[9*(c + d*x)]))/d

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}-\frac {2 \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}\right )}{d}\)	\(67\)
default	\(\frac {a \left (\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}-\frac {2 \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}\right )}{d}\)	\(67\)
risch	\(\frac {3 a \sin \left (d x +c \right )}{128 d}-\frac {a \cos \left (10 d x +10 c \right )}{5120 d}+\frac {a \sin \left (9 d x +9 c \right )}{2304 d}+\frac {a \sin \left (7 d x +7 c \right )}{1792 d}+\frac {5 a \cos \left (6 d x +6 c \right )}{3072 d}-\frac {a \sin \left (5 d x +5 c \right )}{320 d}-\frac {a \sin \left (3 d x +3 c \right )}{192 d}-\frac {5 a \cos \left (2 d x +2 c \right )}{512 d}\)	\(119\)
parallelrisch	\(\frac {a \left (\sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-5 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (996+880 \cos \left (2 d x +2 c \right )+63 \sin \left (5 d x +5 c \right )+420 \sin \left (d x +c \right )+315 \sin \left (3 d x +3 c \right )+140 \cos \left (4 d x +4 c \right )\right ) \left (\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+5 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{80640 d}\)	\(125\)

[In]

int(cos(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

a/d*(1/10*sin(d*x+c)^10+1/9*sin(d*x+c)^9-1/4*sin(d*x+c)^8-2/7*sin(d*x+c)^7+1/6*sin(d*x+c)^6+1/5*sin(d*x+c)^5)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {126 \, a \cos \left (d x + c\right )^{10} - 315 \, a \cos \left (d x + c\right )^{8} + 210 \, a \cos \left (d x + c\right )^{6} - 4 \, {\left (35 \, a \cos \left (d x + c\right )^{8} - 50 \, a \cos \left (d x + c\right )^{6} + 3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{1260 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/1260*(126*a*cos(d*x + c)^10 - 315*a*cos(d*x + c)^8 + 210*a*cos(d*x + c)^6 - 4*(35*a*cos(d*x + c)^8 - 50*a*c
os(d*x + c)^6 + 3*a*cos(d*x + c)^4 + 4*a*cos(d*x + c)^2 + 8*a)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 1.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.40 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {4 a \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {a \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} - \frac {a \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{12 d} - \frac {a \cos ^{10}{\left (c + d x \right )}}{60 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{4}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**4*(a+a*sin(d*x+c)),x)

[Out]

Piecewise((8*a*sin(c + d*x)**9/(315*d) + 4*a*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + a*sin(c + d*x)**5*cos(c
+ d*x)**4/(5*d) - a*sin(c + d*x)**4*cos(c + d*x)**6/(6*d) - a*sin(c + d*x)**2*cos(c + d*x)**8/(12*d) - a*cos(c
+ d*x)**10/(60*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**4*cos(c)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {126 \, a \sin \left (d x + c\right )^{10} + 140 \, a \sin \left (d x + c\right )^{9} - 315 \, a \sin \left (d x + c\right )^{8} - 360 \, a \sin \left (d x + c\right )^{7} + 210 \, a \sin \left (d x + c\right )^{6} + 252 \, a \sin \left (d x + c\right )^{5}}{1260 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/1260*(126*a*sin(d*x + c)^10 + 140*a*sin(d*x + c)^9 - 315*a*sin(d*x + c)^8 - 360*a*sin(d*x + c)^7 + 210*a*sin
(d*x + c)^6 + 252*a*sin(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {5 \, a \cos \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac {5 \, a \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} + \frac {a \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {a \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {a \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {a \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {3 \, a \sin \left (d x + c\right )}{128 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/5120*a*cos(10*d*x + 10*c)/d + 5/3072*a*cos(6*d*x + 6*c)/d - 5/512*a*cos(2*d*x + 2*c)/d + 1/2304*a*sin(9*d*x
+ 9*c)/d + 1/1792*a*sin(7*d*x + 7*c)/d - 1/320*a*sin(5*d*x + 5*c)/d - 1/192*a*sin(3*d*x + 3*c)/d + 3/128*a*si
n(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 9.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {a\,{\sin \left (c+d\,x\right )}^{10}}{10}+\frac {a\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a\,{\sin \left (c+d\,x\right )}^8}{4}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}}{d} \]

[In]

int(cos(c + d*x)^5*sin(c + d*x)^4*(a + a*sin(c + d*x)),x)

[Out]

((a*sin(c + d*x)^5)/5 + (a*sin(c + d*x)^6)/6 - (2*a*sin(c + d*x)^7)/7 - (a*sin(c + d*x)^8)/4 + (a*sin(c + d*x)
^9)/9 + (a*sin(c + d*x)^10)/10)/d