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

Optimal. Leaf size=288 \[ \frac{a^2 \cos ^3(c+d x)}{3 d}-\frac{a^2 \cos (c+d x)}{d}-\frac{\left (2 a c+b^2\right ) \cos ^5(c+d x)}{5 d}+\frac{2 \left (2 a c+b^2\right ) \cos ^3(c+d x)}{3 d}-\frac{\left (2 a c+b^2\right ) \cos (c+d x)}{d}-\frac{a b \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac{3 a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a b x}{4}-\frac{b c \sin ^5(c+d x) \cos (c+d x)}{3 d}-\frac{5 b c \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac{5 b c \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5 b c x}{8}+\frac{c^2 \cos ^7(c+d x)}{7 d}-\frac{3 c^2 \cos ^5(c+d x)}{5 d}+\frac{c^2 \cos ^3(c+d x)}{d}-\frac{c^2 \cos (c+d x)}{d} \]

[Out]

(3*a*b*x)/4 + (5*b*c*x)/8 - (a^2*Cos[c + d*x])/d - (c^2*Cos[c + d*x])/d - ((b^2 + 2*a*c)*Cos[c + d*x])/d + (a^
2*Cos[c + d*x]^3)/(3*d) + (c^2*Cos[c + d*x]^3)/d + (2*(b^2 + 2*a*c)*Cos[c + d*x]^3)/(3*d) - (3*c^2*Cos[c + d*x
]^5)/(5*d) - ((b^2 + 2*a*c)*Cos[c + d*x]^5)/(5*d) + (c^2*Cos[c + d*x]^7)/(7*d) - (3*a*b*Cos[c + d*x]*Sin[c + d
*x])/(4*d) - (5*b*c*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*b*Cos[c + d*x]*Sin[c + d*x]^3)/(2*d) - (5*b*c*Cos[c
+ d*x]*Sin[c + d*x]^3)/(12*d) - (b*c*Cos[c + d*x]*Sin[c + d*x]^5)/(3*d)

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Rubi [A]  time = 0.400114, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {4394, 3256, 2633, 2635, 8} \[ \frac{a^2 \cos ^3(c+d x)}{3 d}-\frac{a^2 \cos (c+d x)}{d}-\frac{\left (2 a c+b^2\right ) \cos ^5(c+d x)}{5 d}+\frac{2 \left (2 a c+b^2\right ) \cos ^3(c+d x)}{3 d}-\frac{\left (2 a c+b^2\right ) \cos (c+d x)}{d}-\frac{a b \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac{3 a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a b x}{4}-\frac{b c \sin ^5(c+d x) \cos (c+d x)}{3 d}-\frac{5 b c \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac{5 b c \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5 b c x}{8}+\frac{c^2 \cos ^7(c+d x)}{7 d}-\frac{3 c^2 \cos ^5(c+d x)}{5 d}+\frac{c^2 \cos ^3(c+d x)}{d}-\frac{c^2 \cos (c+d x)}{d} \]

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)^2,x]

[Out]

(3*a*b*x)/4 + (5*b*c*x)/8 - (a^2*Cos[c + d*x])/d - (c^2*Cos[c + d*x])/d - ((b^2 + 2*a*c)*Cos[c + d*x])/d + (a^
2*Cos[c + d*x]^3)/(3*d) + (c^2*Cos[c + d*x]^3)/d + (2*(b^2 + 2*a*c)*Cos[c + d*x]^3)/(3*d) - (3*c^2*Cos[c + d*x
]^5)/(5*d) - ((b^2 + 2*a*c)*Cos[c + d*x]^5)/(5*d) + (c^2*Cos[c + d*x]^7)/(7*d) - (3*a*b*Cos[c + d*x]*Sin[c + d
*x])/(4*d) - (5*b*c*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*b*Cos[c + d*x]*Sin[c + d*x]^3)/(2*d) - (5*b*c*Cos[c
+ d*x]*Sin[c + d*x]^3)/(12*d) - (b*c*Cos[c + d*x]*Sin[c + d*x]^5)/(3*d)

Rule 4394

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

Rule 3256

Int[sin[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]
^(n2_.))^(p_), x_Symbol] :> Int[ExpandTrig[sin[d + e*x]^m*(a + b*sin[d + e*x]^n + c*sin[d + e*x]^(2*n))^p, x],
x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IntegersQ[m, n, p]

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 (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right )^2 \, dx &=\int \sin ^3(c+d x) \left (a+b \sin (c+d x)+c \sin ^2(c+d x)\right )^2 \, dx\\ &=\int \left (a^2 \sin ^3(c+d x)+2 a b \sin ^4(c+d x)+\left (b^2+2 a c\right ) \sin ^5(c+d x)+2 b c \sin ^6(c+d x)+c^2 \sin ^7(c+d x)\right ) \, dx\\ &=a^2 \int \sin ^3(c+d x) \, dx+(2 a b) \int \sin ^4(c+d x) \, dx+(2 b c) \int \sin ^6(c+d x) \, dx+c^2 \int \sin ^7(c+d x) \, dx+\left (b^2+2 a c\right ) \int \sin ^5(c+d x) \, dx\\ &=-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{2 d}-\frac{b c \cos (c+d x) \sin ^5(c+d x)}{3 d}+\frac{1}{2} (3 a b) \int \sin ^2(c+d x) \, dx+\frac{1}{3} (5 b c) \int \sin ^4(c+d x) \, dx-\frac{a^2 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac{c^2 \operatorname{Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (b^2+2 a c\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^2 \cos (c+d x)}{d}-\frac{c^2 \cos (c+d x)}{d}-\frac{\left (b^2+2 a c\right ) \cos (c+d x)}{d}+\frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{c^2 \cos ^3(c+d x)}{d}+\frac{2 \left (b^2+2 a c\right ) \cos ^3(c+d x)}{3 d}-\frac{3 c^2 \cos ^5(c+d x)}{5 d}-\frac{\left (b^2+2 a c\right ) \cos ^5(c+d x)}{5 d}+\frac{c^2 \cos ^7(c+d x)}{7 d}-\frac{3 a b \cos (c+d x) \sin (c+d x)}{4 d}-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{2 d}-\frac{5 b c \cos (c+d x) \sin ^3(c+d x)}{12 d}-\frac{b c \cos (c+d x) \sin ^5(c+d x)}{3 d}+\frac{1}{4} (3 a b) \int 1 \, dx+\frac{1}{4} (5 b c) \int \sin ^2(c+d x) \, dx\\ &=\frac{3 a b x}{4}-\frac{a^2 \cos (c+d x)}{d}-\frac{c^2 \cos (c+d x)}{d}-\frac{\left (b^2+2 a c\right ) \cos (c+d x)}{d}+\frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{c^2 \cos ^3(c+d x)}{d}+\frac{2 \left (b^2+2 a c\right ) \cos ^3(c+d x)}{3 d}-\frac{3 c^2 \cos ^5(c+d x)}{5 d}-\frac{\left (b^2+2 a c\right ) \cos ^5(c+d x)}{5 d}+\frac{c^2 \cos ^7(c+d x)}{7 d}-\frac{3 a b \cos (c+d x) \sin (c+d x)}{4 d}-\frac{5 b c \cos (c+d x) \sin (c+d x)}{8 d}-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{2 d}-\frac{5 b c \cos (c+d x) \sin ^3(c+d x)}{12 d}-\frac{b c \cos (c+d x) \sin ^5(c+d x)}{3 d}+\frac{1}{8} (5 b c) \int 1 \, dx\\ &=\frac{3 a b x}{4}+\frac{5 b c x}{8}-\frac{a^2 \cos (c+d x)}{d}-\frac{c^2 \cos (c+d x)}{d}-\frac{\left (b^2+2 a c\right ) \cos (c+d x)}{d}+\frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{c^2 \cos ^3(c+d x)}{d}+\frac{2 \left (b^2+2 a c\right ) \cos ^3(c+d x)}{3 d}-\frac{3 c^2 \cos ^5(c+d x)}{5 d}-\frac{\left (b^2+2 a c\right ) \cos ^5(c+d x)}{5 d}+\frac{c^2 \cos ^7(c+d x)}{7 d}-\frac{3 a b \cos (c+d x) \sin (c+d x)}{4 d}-\frac{5 b c \cos (c+d x) \sin (c+d x)}{8 d}-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{2 d}-\frac{5 b c \cos (c+d x) \sin ^3(c+d x)}{12 d}-\frac{b c \cos (c+d x) \sin ^5(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.479222, size = 167, normalized size = 0.58 \[ \frac{-105 \left (48 a^2+80 a c+40 b^2+35 c^2\right ) \cos (c+d x)+35 \left (16 a^2+40 a c+20 b^2+21 c^2\right ) \cos (3 (c+d x))-21 \left (c (8 a+7 c)+4 b^2\right ) \cos (5 (c+d x))+840 b (6 a+5 c) (c+d x)-210 b (16 a+15 c) \sin (2 (c+d x))+210 b (2 a+3 c) \sin (4 (c+d x))-70 b c \sin (6 (c+d x))+15 c^2 \cos (7 (c+d x))}{6720 d} \]

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)^2,x]

[Out]

(840*b*(6*a + 5*c)*(c + d*x) - 105*(48*a^2 + 40*b^2 + 80*a*c + 35*c^2)*Cos[c + d*x] + 35*(16*a^2 + 20*b^2 + 40
*a*c + 21*c^2)*Cos[3*(c + d*x)] - 21*(4*b^2 + c*(8*a + 7*c))*Cos[5*(c + d*x)] + 15*c^2*Cos[7*(c + d*x)] - 210*
b*(16*a + 15*c)*Sin[2*(c + d*x)] + 210*b*(2*a + 3*c)*Sin[4*(c + d*x)] - 70*b*c*Sin[6*(c + d*x)])/(6720*d)

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Maple [A]  time = 0.023, size = 213, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{\frac{{c}^{2}\cos \left ( dx+c \right ) }{7} \left ({\frac{16}{5}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) }+2\,cb \left ( -1/6\, \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+5/4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) \cos \left ( dx+c \right ) +{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) -{\frac{2\,ac\cos \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }-{\frac{{b}^{2}\cos \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+2\,ab \left ( -1/4\, \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+3/2\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) -{\frac{{a}^{2} \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)+b*sin(d*x+c)^2+c*sin(d*x+c)^3)^2,x)

[Out]

1/d*(-1/7*c^2*(16/5+sin(d*x+c)^6+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c)+2*c*b*(-1/6*(sin(d*x+c)^5+5/4*s
in(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/16*d*x+5/16*c)-2/5*a*c*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c
)-1/5*b^2*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)+2*a*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*(2+sin(d*x+c)^2)*cos(d*x+c))

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Maxima [A]  time = 0.977532, size = 294, normalized size = 1.02 \begin{align*} \frac{1120 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} + 210 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 224 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} b^{2} - 448 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} a c + 35 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b c + 96 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 21 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3} - 35 \, \cos \left (d x + c\right )\right )} c^{2}}{3360 \, 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)^2,x, algorithm="maxima")

[Out]

1/3360*(1120*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^2 + 210*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c
))*a*b - 224*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*b^2 - 448*(3*cos(d*x + c)^5 - 10*cos(d*x
+ c)^3 + 15*cos(d*x + c))*a*c + 35*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x
+ 2*c))*b*c + 96*(5*cos(d*x + c)^7 - 21*cos(d*x + c)^5 + 35*cos(d*x + c)^3 - 35*cos(d*x + c))*c^2)/d

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Fricas [A]  time = 2.29698, size = 420, normalized size = 1.46 \begin{align*} \frac{120 \, c^{2} \cos \left (d x + c\right )^{7} - 168 \,{\left (b^{2} + 2 \, a c + 3 \, c^{2}\right )} \cos \left (d x + c\right )^{5} + 280 \,{\left (a^{2} + 2 \, b^{2} + 4 \, a c + 3 \, c^{2}\right )} \cos \left (d x + c\right )^{3} + 105 \,{\left (6 \, a b + 5 \, b c\right )} d x - 840 \,{\left (a^{2} + b^{2} + 2 \, a c + c^{2}\right )} \cos \left (d x + c\right ) - 35 \,{\left (8 \, b c \cos \left (d x + c\right )^{5} - 2 \,{\left (6 \, a b + 13 \, b c\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (10 \, a b + 11 \, b c\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, 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)^2,x, algorithm="fricas")

[Out]

1/840*(120*c^2*cos(d*x + c)^7 - 168*(b^2 + 2*a*c + 3*c^2)*cos(d*x + c)^5 + 280*(a^2 + 2*b^2 + 4*a*c + 3*c^2)*c
os(d*x + c)^3 + 105*(6*a*b + 5*b*c)*d*x - 840*(a^2 + b^2 + 2*a*c + c^2)*cos(d*x + c) - 35*(8*b*c*cos(d*x + c)^
5 - 2*(6*a*b + 13*b*c)*cos(d*x + c)^3 + 3*(10*a*b + 11*b*c)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 8.98825, size = 541, normalized size = 1.88 \begin{align*} \begin{cases} - \frac{a^{2} \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{2 a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac{3 a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac{3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 a b x \cos ^{4}{\left (c + d x \right )}}{4} - \frac{5 a b \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} - \frac{3 a b \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac{2 a c \sin ^{4}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{8 a c \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{16 a c \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac{b^{2} \sin ^{4}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{4 b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{8 b^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac{5 b c x \sin ^{6}{\left (c + d x \right )}}{8} + \frac{15 b c x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac{15 b c x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac{5 b c x \cos ^{6}{\left (c + d x \right )}}{8} - \frac{11 b c \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{5 b c \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{5 b c \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac{c^{2} \sin ^{6}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{2 c^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac{8 c^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{16 c^{2} \cos ^{7}{\left (c + d x \right )}}{35 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 )^{2} \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)**2,x)

[Out]

Piecewise((-a**2*sin(c + d*x)**2*cos(c + d*x)/d - 2*a**2*cos(c + d*x)**3/(3*d) + 3*a*b*x*sin(c + d*x)**4/4 + 3
*a*b*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + 3*a*b*x*cos(c + d*x)**4/4 - 5*a*b*sin(c + d*x)**3*cos(c + d*x)/(4*d
) - 3*a*b*sin(c + d*x)*cos(c + d*x)**3/(4*d) - 2*a*c*sin(c + d*x)**4*cos(c + d*x)/d - 8*a*c*sin(c + d*x)**2*co
s(c + d*x)**3/(3*d) - 16*a*c*cos(c + d*x)**5/(15*d) - b**2*sin(c + d*x)**4*cos(c + d*x)/d - 4*b**2*sin(c + d*x
)**2*cos(c + d*x)**3/(3*d) - 8*b**2*cos(c + d*x)**5/(15*d) + 5*b*c*x*sin(c + d*x)**6/8 + 15*b*c*x*sin(c + d*x)
**4*cos(c + d*x)**2/8 + 15*b*c*x*sin(c + d*x)**2*cos(c + d*x)**4/8 + 5*b*c*x*cos(c + d*x)**6/8 - 11*b*c*sin(c
+ d*x)**5*cos(c + d*x)/(8*d) - 5*b*c*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) - 5*b*c*sin(c + d*x)*cos(c + d*x)**
5/(8*d) - c**2*sin(c + d*x)**6*cos(c + d*x)/d - 2*c**2*sin(c + d*x)**4*cos(c + d*x)**3/d - 8*c**2*sin(c + d*x)
**2*cos(c + d*x)**5/(5*d) - 16*c**2*cos(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a*sin(c) + b*sin(c)**2 + c*sin(c)**
3)**2*sin(c), True))

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Giac [A]  time = 1.15043, size = 251, normalized size = 0.87 \begin{align*} \frac{1}{8} \,{\left (6 \, a b + 5 \, b c\right )} x + \frac{c^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac{b c \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac{{\left (4 \, b^{2} + 8 \, a c + 7 \, c^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac{{\left (16 \, a^{2} + 20 \, b^{2} + 40 \, a c + 21 \, c^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac{{\left (48 \, a^{2} + 40 \, b^{2} + 80 \, a c + 35 \, c^{2}\right )} \cos \left (d x + c\right )}{64 \, d} + \frac{{\left (2 \, a b + 3 \, b c\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{{\left (16 \, a b + 15 \, b c\right )} \sin \left (2 \, d x + 2 \, c\right )}{32 \, 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)^2,x, algorithm="giac")

[Out]

1/8*(6*a*b + 5*b*c)*x + 1/448*c^2*cos(7*d*x + 7*c)/d - 1/96*b*c*sin(6*d*x + 6*c)/d - 1/320*(4*b^2 + 8*a*c + 7*
c^2)*cos(5*d*x + 5*c)/d + 1/192*(16*a^2 + 20*b^2 + 40*a*c + 21*c^2)*cos(3*d*x + 3*c)/d - 1/64*(48*a^2 + 40*b^2
+ 80*a*c + 35*c^2)*cos(d*x + c)/d + 1/32*(2*a*b + 3*b*c)*sin(4*d*x + 4*c)/d - 1/32*(16*a*b + 15*b*c)*sin(2*d*
x + 2*c)/d