\(\int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\) [207]

Optimal result

Integrand size = 27, antiderivative size = 73 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^4(c+d x)}{4 d}+\frac {3 a^3 \sin ^5(c+d x)}{5 d}+\frac {a^3 \sin ^6(c+d x)}{6 d} \]

[Out]

1/3*a^3*sin(d*x+c)^3/d+3/4*a^3*sin(d*x+c)^4/d+3/5*a^3*sin(d*x+c)^5/d+1/6*a^3*sin(d*x+c)^6/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 73, 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} \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^6(c+d x)}{6 d}+\frac {3 a^3 \sin ^5(c+d x)}{5 d}+\frac {3 a^3 \sin ^4(c+d x)}{4 d}+\frac {a^3 \sin ^3(c+d x)}{3 d} \]

[In]

Int[Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 (-45+870 \cos (2 (c+d x))-240 \cos (4 (c+d x))+10 \cos (6 (c+d x))-1200 \sin (c+d x)+520 \sin (3 (c+d x))-72 \sin (5 (c+d x)))}{1920 d} \]

[In]

Integrate[Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

-1/1920*(a^3*(-45 + 870*Cos[2*(c + d*x)] - 240*Cos[4*(c + d*x)] + 10*Cos[6*(c + d*x)] - 1200*Sin[c + d*x] + 52
0*Sin[3*(c + d*x)] - 72*Sin[5*(c + d*x)]))/d

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\frac {a^{3} \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {3 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {3 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(58\)
default	\(\frac {\frac {a^{3} \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {3 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {3 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(58\)
parallelrisch	\(-\frac {a^{3} \left (-\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (-76+36 \cos \left (2 d x +2 c \right )+5 \sin \left (3 d x +3 c \right )-105 \sin \left (d x +c \right )\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}\)	\(85\)
risch	\(\frac {5 a^{3} \sin \left (d x +c \right )}{8 d}-\frac {a^{3} \cos \left (6 d x +6 c \right )}{192 d}+\frac {3 a^{3} \sin \left (5 d x +5 c \right )}{80 d}+\frac {a^{3} \cos \left (4 d x +4 c \right )}{8 d}-\frac {13 a^{3} \sin \left (3 d x +3 c \right )}{48 d}-\frac {29 a^{3} \cos \left (2 d x +2 c \right )}{64 d}\)	\(101\)
norman	\(\frac {\frac {8 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {136 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {136 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {8 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {12 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {12 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {104 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(151\)

[In]

int(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/6*a^3*sin(d*x+c)^6+3/5*a^3*sin(d*x+c)^5+3/4*a^3*sin(d*x+c)^4+1/3*a^3*sin(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.16 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {10 \, a^{3} \cos \left (d x + c\right )^{6} - 75 \, a^{3} \cos \left (d x + c\right )^{4} + 120 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (9 \, a^{3} \cos \left (d x + c\right )^{4} - 23 \, a^{3} \cos \left (d x + c\right )^{2} + 14 \, a^{3}\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(10*a^3*cos(d*x + c)^6 - 75*a^3*cos(d*x + c)^4 + 120*a^3*cos(d*x + c)^2 - 4*(9*a^3*cos(d*x + c)^4 - 23*a
^3*cos(d*x + c)^2 + 14*a^3)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.12 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\begin {cases} \frac {a^{3} \sin ^{6}{\left (c + d x \right )}}{6 d} + \frac {3 a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{3} \sin ^{2}{\left (c \right )} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((a**3*sin(c + d*x)**6/(6*d) + 3*a**3*sin(c + d*x)**5/(5*d) + 3*a**3*sin(c + d*x)**4/(4*d) + a**3*sin
(c + d*x)**3/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)**2*cos(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {10 \, a^{3} \sin \left (d x + c\right )^{6} + 36 \, a^{3} \sin \left (d x + c\right )^{5} + 45 \, a^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{3} \sin \left (d x + c\right )^{3}}{60 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(10*a^3*sin(d*x + c)^6 + 36*a^3*sin(d*x + c)^5 + 45*a^3*sin(d*x + c)^4 + 20*a^3*sin(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {10 \, a^{3} \sin \left (d x + c\right )^{6} + 36 \, a^{3} \sin \left (d x + c\right )^{5} + 45 \, a^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{3} \sin \left (d x + c\right )^{3}}{60 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(10*a^3*sin(d*x + c)^6 + 36*a^3*sin(d*x + c)^5 + 45*a^3*sin(d*x + c)^4 + 20*a^3*sin(d*x + c)^3)/d

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a^3\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]

[In]

int(cos(c + d*x)*sin(c + d*x)^2*(a + a*sin(c + d*x))^3,x)

[Out]

((a^3*sin(c + d*x)^3)/3 + (3*a^3*sin(c + d*x)^4)/4 + (3*a^3*sin(c + d*x)^5)/5 + (a^3*sin(c + d*x)^6)/6)/d