\(\int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\) [512]

Optimal result

Integrand size = 29, antiderivative size = 109 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {4 (a+a \sin (c+d x))^5}{5 a^3 d}-\frac {2 (a+a \sin (c+d x))^6}{a^4 d}+\frac {13 (a+a \sin (c+d x))^7}{7 a^5 d}-\frac {3 (a+a \sin (c+d x))^8}{4 a^6 d}+\frac {(a+a \sin (c+d x))^9}{9 a^7 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 90} \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {(a \sin (c+d x)+a)^9}{9 a^7 d}-\frac {3 (a \sin (c+d x)+a)^8}{4 a^6 d}+\frac {13 (a \sin (c+d x)+a)^7}{7 a^5 d}-\frac {2 (a \sin (c+d x)+a)^6}{a^4 d}+\frac {4 (a \sin (c+d x)+a)^5}{5 a^3 d} \]

[In]

Int[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]

[Out]

(4*(a + a*Sin[c + d*x])^5)/(5*a^3*d) - (2*(a + a*Sin[c + d*x])^6)/(a^4*d) + (13*(a + a*Sin[c + d*x])^7)/(7*a^5
*d) - (3*(a + a*Sin[c + d*x])^8)/(4*a^6*d) + (a + a*Sin[c + d*x])^9/(9*a^7*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^2 (a+x)^4}{a^2} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int (a-x)^2 x^2 (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = \frac {\text {Subst}\left (\int \left (4 a^4 (a+x)^4-12 a^3 (a+x)^5+13 a^2 (a+x)^6-6 a (a+x)^7+(a+x)^8\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 (7560 \cos (2 (c+d x))+1260 \cos (4 (c+d x))-840 \cos (6 (c+d x))-315 \cos (8 (c+d x))-16380 \sin (c+d x)+1680 \sin (3 (c+d x))+2016 \sin (5 (c+d x))+270 \sin (7 (c+d x))-70 \sin (9 (c+d x)))}{161280 d} \]

[In]

Integrate[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]

[Out]

-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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {2 \left (\sin ^{6}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(79\)
default	\(\frac {a^{2} \left (\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {2 \left (\sin ^{6}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(79\)
parallelrisch	\(\frac {a^{2} \left (-\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (2406 \cos \left (2 d x +2 c \right )+315 \sin \left (5 d x +5 c \right )-70 \cos \left (6 d x +6 c \right )+3150 \sin \left (d x +c \right )+1785 \sin \left (3 d x +3 c \right )+60 \cos \left (4 d x +4 c \right )+4324\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40320 d}\)	\(118\)
risch	\(\frac {13 a^{2} \sin \left (d x +c \right )}{128 d}+\frac {a^{2} \sin \left (9 d x +9 c \right )}{2304 d}+\frac {a^{2} \cos \left (8 d x +8 c \right )}{512 d}-\frac {3 a^{2} \sin \left (7 d x +7 c \right )}{1792 d}+\frac {a^{2} \cos \left (6 d x +6 c \right )}{192 d}-\frac {a^{2} \sin \left (5 d x +5 c \right )}{80 d}-\frac {a^{2} \cos \left (4 d x +4 c \right )}{128 d}-\frac {a^{2} \sin \left (3 d x +3 c \right )}{96 d}-\frac {3 a^{2} \cos \left (2 d x +2 c \right )}{64 d}\)	\(152\)
norman	\(\frac {\frac {8 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {48 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {136 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {11104 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 d}-\frac {136 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {48 a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {8 a^{2} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}\)	\(265\)

[In]

int(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {315 \, a^{2} \cos \left (d x + c\right )^{8} - 420 \, a^{2} \cos \left (d x + c\right )^{6} + 4 \, {\left (35 \, a^{2} \cos \left (d x + c\right )^{8} - 95 \, a^{2} \cos \left (d x + c\right )^{6} + 12 \, a^{2} \cos \left (d x + c\right )^{4} + 16 \, a^{2} \cos \left (d x + c\right )^{2} + 32 \, a^{2}\right )} \sin \left (d x + c\right )}{1260 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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 + 1
2*a^2*cos(d*x + c)^4 + 16*a^2*cos(d*x + c)^2 + 32*a^2)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.74 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {8 a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {4 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {8 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {4 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \cos ^{8}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**2*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((8*a**2*sin(c + d*x)**9/(315*d) + 4*a**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 8*a**2*sin(c + d*x
)**7/(105*d) + a**2*sin(c + d*x)**5*cos(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)**3*cos(c + d*x)**4/(3*d) - a**2*sin(c + d*x)**2*cos(c + d*x)**6/(3*d) - a**2*cos(c + d*x)**8/(
12*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**2*cos(c)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {140 \, a^{2} \sin \left (d x + c\right )^{9} + 315 \, a^{2} \sin \left (d x + c\right )^{8} - 180 \, a^{2} \sin \left (d x + c\right )^{7} - 840 \, a^{2} \sin \left (d x + c\right )^{6} - 252 \, a^{2} \sin \left (d x + c\right )^{5} + 630 \, a^{2} \sin \left (d x + c\right )^{4} + 420 \, a^{2} \sin \left (d x + c\right )^{3}}{1260 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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 - 25
2*a^2*sin(d*x + c)^5 + 630*a^2*sin(d*x + c)^4 + 420*a^2*sin(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.39 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \cos \left (8 \, d x + 8 \, c\right )}{512 \, d} + \frac {a^{2} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a^{2} \cos \left (4 \, d x + 4 \, c\right )}{128 \, d} - \frac {3 \, a^{2} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {a^{2} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {3 \, a^{2} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {a^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a^{2} \sin \left (3 \, d x + 3 \, c\right )}{96 \, d} + \frac {13 \, a^{2} \sin \left (d x + c\right )}{128 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/512*a^2*cos(8*d*x + 8*c)/d + 1/192*a^2*cos(6*d*x + 6*c)/d - 1/128*a^2*cos(4*d*x + 4*c)/d - 3/64*a^2*cos(2*d*
x + 2*c)/d + 1/2304*a^2*sin(9*d*x + 9*c)/d - 3/1792*a^2*sin(7*d*x + 7*c)/d - 1/80*a^2*sin(5*d*x + 5*c)/d - 1/9
6*a^2*sin(3*d*x + 3*c)/d + 13/128*a^2*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 9.47 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {a^2\,{\sin \left (c+d\,x\right )}^8}{4}-\frac {a^2\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^6}{3}-\frac {a^2\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a^2\,{\sin \left (c+d\,x\right )}^4}{2}+\frac {a^2\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]

[In]

int(cos(c + d*x)^5*sin(c + d*x)^2*(a + a*sin(c + d*x))^2,x)

[Out]

((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*si
n(c + d*x)^7)/7 + (a^2*sin(c + d*x)^8)/4 + (a^2*sin(c + d*x)^9)/9)/d