\(\int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\) [197]

Optimal result

Integrand size = 27, antiderivative size = 55 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin ^4(c+d x)}{2 d}+\frac {a^2 \sin ^5(c+d x)}{5 d} \]

[Out]

1/3*a^2*sin(d*x+c)^3/d+1/2*a^2*sin(d*x+c)^4/d+1/5*a^2*sin(d*x+c)^5/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 55, 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} \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^5(c+d x)}{5 d}+\frac {a^2 \sin ^4(c+d x)}{2 d}+\frac {a^2 \sin ^3(c+d x)}{3 d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (15 \cos (4 (c+d x))+104 \sin ^3(c+d x)-12 \cos (2 (c+d x)) \left (5+2 \sin ^3(c+d x)\right )\right )}{240 d} \]

[In]

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

[Out]

(a^2*(15*Cos[4*(c + d*x)] + 104*Sin[c + d*x]^3 - 12*Cos[2*(c + d*x)]*(5 + 2*Sin[c + d*x]^3)))/(240*d)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(45\)
default	\(\frac {\frac {a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(45\)
parallelrisch	\(\frac {a^{2} \left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (-13+3 \cos \left (2 d x +2 c \right )-15 \sin \left (d x +c \right )\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d}\)	\(72\)
risch	\(\frac {3 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {a^{2} \sin \left (5 d x +5 c \right )}{80 d}+\frac {a^{2} \cos \left (4 d x +4 c \right )}{16 d}-\frac {7 a^{2} \sin \left (3 d x +3 c \right )}{48 d}-\frac {a^{2} \cos \left (2 d x +2 c \right )}{4 d}\)	\(84\)
norman	\(\frac {\frac {8 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {176 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {8 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(113\)

[In]

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

[Out]

1/d*(1/5*a^2*sin(d*x+c)^5+1/2*a^2*sin(d*x+c)^4+1/3*a^2*sin(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.31 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {15 \, a^{2} \cos \left (d x + c\right )^{4} - 30 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{4} - 11 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

1/30*(15*a^2*cos(d*x + c)^4 - 30*a^2*cos(d*x + c)^2 + 2*(3*a^2*cos(d*x + c)^4 - 11*a^2*cos(d*x + c)^2 + 8*a^2)
*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {a^{2} \sin ^{4}{\left (c + d x \right )}}{2 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{2}{\left (c \right )} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \sin \left (d x + c\right )^{5} + 15 \, a^{2} \sin \left (d x + c\right )^{4} + 10 \, a^{2} \sin \left (d x + c\right )^{3}}{30 \, d} \]

[In]

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

[Out]

1/30*(6*a^2*sin(d*x + c)^5 + 15*a^2*sin(d*x + c)^4 + 10*a^2*sin(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \sin \left (d x + c\right )^{5} + 15 \, a^{2} \sin \left (d x + c\right )^{4} + 10 \, a^{2} \sin \left (d x + c\right )^{3}}{30 \, d} \]

[In]

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

[Out]

1/30*(6*a^2*sin(d*x + c)^5 + 15*a^2*sin(d*x + c)^4 + 10*a^2*sin(d*x + c)^3)/d

Mupad [B] (verification not implemented)

Time = 9.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\sin \left (c+d\,x\right )}^3\,\left (6\,{\sin \left (c+d\,x\right )}^2+15\,\sin \left (c+d\,x\right )+10\right )}{30\,d} \]

[In]

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

[Out]

(a^2*sin(c + d*x)^3*(15*sin(c + d*x) + 6*sin(c + d*x)^2 + 10))/(30*d)