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

Optimal. Leaf size=89 \[ -\frac{(4 a+3 c) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 a+3 c)+\frac{b \cos ^3(c+d x)}{3 d}-\frac{b \cos (c+d x)}{d}-\frac{c \sin ^3(c+d x) \cos (c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.108528, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4237, 3023, 2748, 2635, 8, 2633} \[ -\frac{(4 a+3 c) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 a+3 c)+\frac{b \cos ^3(c+d x)}{3 d}-\frac{b \cos (c+d x)}{d}-\frac{c \sin ^3(c+d x) \cos (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4237

Int[(u_)*((A_.)*sin[(a_.) + (b_.)*(x_)]^(n_.) + (B_.)*sin[(a_.) + (b_.)*(x_)]^(n1_) + (C_.)*sin[(a_.) + (b_.)*
(x_)]^(n2_)), x_Symbol] :> Int[ActivateTrig[u]*Sin[a + b*x]^n*(A + B*Sin[a + b*x] + C*Sin[a + b*x]^2), x] /; F
reeQ[{a, b, A, B, C, n}, x] && EqQ[n1, n + 1] && EqQ[n2, n + 2]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 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]

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]

Rubi steps

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

Mathematica [A]  time = 0.154986, size = 105, normalized size = 1.18 \[ \frac{a (c+d x)}{2 d}-\frac{a \sin (2 (c+d x))}{4 d}-\frac{3 b \cos (c+d x)}{4 d}+\frac{b \cos (3 (c+d x))}{12 d}+\frac{3 c (c+d x)}{8 d}-\frac{c \sin (2 (c+d x))}{4 d}+\frac{c \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 84, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( c \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{b \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}}+a \left ( -{\frac{\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.970018, size = 107, normalized size = 1.2 \begin{align*} \frac{24 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a + 32 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} b + 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 )} c}{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)+b*sin(d*x+c)^2+c*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

1/96*(24*(2*d*x + 2*c - sin(2*d*x + 2*c))*a + 32*(cos(d*x + c)^3 - 3*cos(d*x + c))*b + 3*(12*d*x + 12*c + sin(
4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*c)/d

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Fricas [A]  time = 2.09546, size = 181, normalized size = 2.03 \begin{align*} \frac{8 \, b \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, a + 3 \, c\right )} d x - 24 \, b \cos \left (d x + c\right ) + 3 \,{\left (2 \, c \cos \left (d x + c\right )^{3} -{\left (4 \, a + 5 \, c\right )} \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)+b*sin(d*x+c)^2+c*sin(d*x+c)^3),x, algorithm="fricas")

[Out]

1/24*(8*b*cos(d*x + c)^3 + 3*(4*a + 3*c)*d*x - 24*b*cos(d*x + c) + 3*(2*c*cos(d*x + c)^3 - (4*a + 5*c)*cos(d*x
+ c))*sin(d*x + c))/d

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Sympy [A]  time = 1.20449, size = 201, normalized size = 2.26 \begin{align*} \begin{cases} \frac{a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{a x \cos ^{2}{\left (c + d x \right )}}{2} - \frac{a \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} - \frac{b \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{2 b \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac{3 c x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 c x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 c x \cos ^{4}{\left (c + d x \right )}}{8} - \frac{5 c \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{3 c \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + b \sin ^{2}{\left (c \right )} + c \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)+b*sin(d*x+c)**2+c*sin(d*x+c)**3),x)

[Out]

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

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Giac [A]  time = 1.08788, size = 95, normalized size = 1.07 \begin{align*} \frac{1}{8} \,{\left (4 \, a + 3 \, c\right )} x + \frac{b \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac{3 \, b \cos \left (d x + c\right )}{4 \, d} + \frac{c \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{{\left (a + c\right )} \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)+b*sin(d*x+c)^2+c*sin(d*x+c)^3),x, algorithm="giac")

[Out]

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