3.934 \(\int \sin (c+d x) (a \sin ^2(c+d x)+b \sin ^3(c+d x)) \, dx\)

Optimal. Leaf size=77 \[ \frac{a \cos ^3(c+d x)}{3 d}-\frac{a \cos (c+d x)}{d}-\frac{b \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac{3 b \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 b x}{8} \]

[Out]

(3*b*x)/8 - (a*Cos[c + d*x])/d + (a*Cos[c + d*x]^3)/(3*d) - (3*b*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (b*Cos[c +
d*x]*Sin[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.12597, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {4393, 2748, 2633, 2635, 8} \[ \frac{a \cos ^3(c+d x)}{3 d}-\frac{a \cos (c+d x)}{d}-\frac{b \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac{3 b \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 b x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*x)/8 - (a*Cos[c + d*x])/d + (a*Cos[c + d*x]^3)/(3*d) - (3*b*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (b*Cos[c +
d*x]*Sin[c + d*x]^3)/(4*d)

Rule 4393

Int[(u_)*((a_)*(F_)[(c_.) + (d_.)*(x_)]^(p_.) + (b_.)*(F_)[(c_.) + (d_.)*(x_)]^(q_.))^(n_.), x_Symbol] :> Int[
ActivateTrig[u*F[c + d*x]^(n*p)*(a + b*F[c + d*x]^(q - p))^n], x] /; FreeQ[{a, b, c, d, p, q}, x] && InertTrig
Q[F] && IntegerQ[n] && PosQ[q - p]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx &=\int \sin ^3(c+d x) (a+b \sin (c+d x)) \, dx\\ &=a \int \sin ^3(c+d x) \, dx+b \int \sin ^4(c+d x) \, dx\\ &=-\frac{b \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac{1}{4} (3 b) \int \sin ^2(c+d x) \, dx-\frac{a \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a \cos (c+d x)}{d}+\frac{a \cos ^3(c+d x)}{3 d}-\frac{3 b \cos (c+d x) \sin (c+d x)}{8 d}-\frac{b \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac{1}{8} (3 b) \int 1 \, dx\\ &=\frac{3 b x}{8}-\frac{a \cos (c+d x)}{d}+\frac{a \cos ^3(c+d x)}{3 d}-\frac{3 b \cos (c+d x) \sin (c+d x)}{8 d}-\frac{b \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.159127, size = 76, normalized size = 0.99 \[ -\frac{3 a \cos (c+d x)}{4 d}+\frac{a \cos (3 (c+d x))}{12 d}+\frac{3 b (c+d x)}{8 d}-\frac{b \sin (2 (c+d x))}{4 d}+\frac{b \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(c + d*x))/(8*d) - (3*a*Cos[c + d*x])/(4*d) + (a*Cos[3*(c + d*x)])/(12*d) - (b*Sin[2*(c + d*x)])/(4*d) +
(b*Sin[4*(c + d*x)])/(32*d)

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Maple [A]  time = 0.012, size = 60, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( b \left ( -{\frac{\cos \left ( dx+c \right ) }{4} \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\sin \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) -{\frac{a \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.968368, size = 77, normalized size = 1. \begin{align*} \frac{32 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a + 3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(32*(cos(d*x + c)^3 - 3*cos(d*x + c))*a + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*b)/d

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Fricas [A]  time = 2.09194, size = 157, normalized size = 2.04 \begin{align*} \frac{8 \, a \cos \left (d x + c\right )^{3} + 9 \, b d x - 24 \, a \cos \left (d x + c\right ) + 3 \,{\left (2 \, b \cos \left (d x + c\right )^{3} - 5 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*a*cos(d*x + c)^3 + 9*b*d*x - 24*a*cos(d*x + c) + 3*(2*b*cos(d*x + c)^3 - 5*b*cos(d*x + c))*sin(d*x + c
))/d

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Sympy [A]  time = 1.0998, size = 150, normalized size = 1.95 \begin{align*} \begin{cases} - \frac{a \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{2 a \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac{3 b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 b x \cos ^{4}{\left (c + d x \right )}}{8} - \frac{5 b \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{3 b \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (a \sin ^{2}{\left (c \right )} + b \sin ^{3}{\left (c \right )}\right ) \sin{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*sin(c + d*x)**2*cos(c + d*x)/d - 2*a*cos(c + d*x)**3/(3*d) + 3*b*x*sin(c + d*x)**4/8 + 3*b*x*sin
(c + d*x)**2*cos(c + d*x)**2/4 + 3*b*x*cos(c + d*x)**4/8 - 5*b*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 3*b*sin(c
+ d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(a*sin(c)**2 + b*sin(c)**3)*sin(c), True))

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Giac [A]  time = 1.07828, size = 84, normalized size = 1.09 \begin{align*} \frac{3}{8} \, b x + \frac{a \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac{3 \, a \cos \left (d x + c\right )}{4 \, d} + \frac{b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{b \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*b*x + 1/12*a*cos(3*d*x + 3*c)/d - 3/4*a*cos(d*x + c)/d + 1/32*b*sin(4*d*x + 4*c)/d - 1/4*b*sin(2*d*x + 2*c
)/d