\(\int \sin (c+d x) (a \sin ^2(c+d x)+b \sin ^3(c+d x)) \, dx\) [934]

Optimal result

Integrand size = 28, antiderivative size = 77 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {3 b x}{8}-\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 b \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \cos (c+d x) \sin ^3(c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4478, 2827, 2713, 2715, 8} \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {a \cos ^3(c+d x)}{3 d}-\frac {a \cos (c+d x)}{d}-\frac {b \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {3 b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 b x}{8} \]

[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 8

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

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2827

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 4478

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]

Rubi steps \begin{align*} \text {integral}& = \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 \text {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] (verified)

Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {3 b (c+d x)}{8 d}-\frac {3 a \cos (c+d x)}{4 d}+\frac {a \cos (3 (c+d x))}{12 d}-\frac {b \sin (2 (c+d x))}{4 d}+\frac {b \sin (4 (c+d x))}{32 d} \]

[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)

Maple [A] (verified)

Time = 1.99 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {b \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}}{d}\)	\(60\)
default	\(\frac {b \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}}{d}\)	\(60\)
parallelrisch	\(\frac {36 d x b -72 \cos \left (d x +c \right ) a +3 b \sin \left (4 d x +4 c \right )+8 a \cos \left (3 d x +3 c \right )-24 b \sin \left (2 d x +2 c \right )-64 a}{96 d}\)	\(60\)
parts	\(-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}+\frac {b \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(62\)
risch	\(\frac {3 x b}{8}-\frac {3 a \cos \left (d x +c \right )}{4 d}+\frac {b \sin \left (4 d x +4 c \right )}{32 d}+\frac {a \cos \left (3 d x +3 c \right )}{12 d}-\frac {b \sin \left (2 d x +2 c \right )}{4 d}\)	\(63\)
norman	\(\frac {\frac {3 x b}{8}-\frac {4 a}{3 d}-\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {11 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {11 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {3 x b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {9 x b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {3 x b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {3 x b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}-\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {16 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(188\)

[In]

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

[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))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {8 \, a \cos \left (d x + c\right )^{3} + 9 \, b d x - 24 \, a \cos \left (d x + c\right ) + 3 \, {\left (2 \, b \cos \left (d x + c\right )^{3} - 5 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

[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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (70) = 140\).

Time = 0.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.95 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\begin {cases} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {2 a \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 b x \cos ^{4}{\left (c + d x \right )}}{8} - \frac {5 b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {3 b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a \sin ^{2}{\left (c \right )} + b \sin ^{3}{\left (c \right )}\right ) \sin {\left (c \right )} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.74 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {32 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b}{96 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {3}{8} \, b x + \frac {a \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {3 \, a \cos \left (d x + c\right )}{4 \, d} + \frac {b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {b \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 29.71 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.44 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {3\,b\,x}{8}-\frac {-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {11\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {11\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {4\,a}{3}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

[In]

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

[Out]

(3*b*x)/8 - ((4*a)/3 + (3*b*tan(c/2 + (d*x)/2))/4 + (16*a*tan(c/2 + (d*x)/2)^2)/3 + 4*a*tan(c/2 + (d*x)/2)^4 +
(11*b*tan(c/2 + (d*x)/2)^3)/4 - (11*b*tan(c/2 + (d*x)/2)^5)/4 - (3*b*tan(c/2 + (d*x)/2)^7)/4)/(d*(tan(c/2 + (
d*x)/2)^2 + 1)^4)