3.224 \(\int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\)

Optimal. Leaf size=140 \[ -\frac{a^3 A \cos ^7(c+d x)}{7 d}+\frac{3 a^3 A \cos ^5(c+d x)}{5 d}-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \sin ^5(c+d x) \cos (c+d x)}{3 d}-\frac{a^3 A \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac{a^3 A \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a^3 A x \]

[Out]

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

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Rubi [A]  time = 0.185963, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2966, 2633, 2635, 8} \[ -\frac{a^3 A \cos ^7(c+d x)}{7 d}+\frac{3 a^3 A \cos ^5(c+d x)}{5 d}-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \sin ^5(c+d x) \cos (c+d x)}{3 d}-\frac{a^3 A \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac{a^3 A \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a^3 A x \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

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

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

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

Mathematica [A]  time = 0.15218, size = 87, normalized size = 0.62 \[ \frac{a^3 A (-210 \sin (2 (c+d x))-210 \sin (4 (c+d x))+70 \sin (6 (c+d x))-1365 \cos (c+d x)-175 \cos (3 (c+d x))+147 \cos (5 (c+d x))-15 \cos (7 (c+d x))+840 c+840 d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(a^3*A*(840*c + 840*d*x - 1365*Cos[c + d*x] - 175*Cos[3*(c + d*x)] + 147*Cos[5*(c + d*x)] - 15*Cos[7*(c + d*x)
] - 210*Sin[2*(c + d*x)] - 210*Sin[4*(c + d*x)] + 70*Sin[6*(c + d*x)]))/(6720*d)

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Maple [A]  time = 0.031, size = 158, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{3}A\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\,{a}^{3}A \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 ) +2\,{a}^{3}A \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}^{3}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)^3*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.992001, size = 212, normalized size = 1.51 \begin{align*} -\frac{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 )} A a^{3} - 1120 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a^{3} + 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 )} A a^{3} - 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 a^{3}}{3360 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3360*(96*(5*cos(d*x + c)^7 - 21*cos(d*x + c)^5 + 35*cos(d*x + c)^3 - 35*cos(d*x + c))*A*a^3 - 1120*(cos(d*x
+ c)^3 - 3*cos(d*x + c))*A*a^3 + 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))*A*a^3 - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*A*a^3)/d

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Fricas [A]  time = 2.08523, size = 269, normalized size = 1.92 \begin{align*} -\frac{120 \, A a^{3} \cos \left (d x + c\right )^{7} - 504 \, A a^{3} \cos \left (d x + c\right )^{5} + 560 \, A a^{3} \cos \left (d x + c\right )^{3} - 105 \, A a^{3} d x - 35 \,{\left (8 \, A a^{3} \cos \left (d x + c\right )^{5} - 14 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, A a^{3} \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)^3*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/840*(120*A*a^3*cos(d*x + c)^7 - 504*A*a^3*cos(d*x + c)^5 + 560*A*a^3*cos(d*x + c)^3 - 105*A*a^3*d*x - 35*(8
*A*a^3*cos(d*x + c)^5 - 14*A*a^3*cos(d*x + c)^3 + 3*A*a^3*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 15.4764, size = 440, normalized size = 3.14 \begin{align*} \begin{cases} - \frac{5 A a^{3} x \sin ^{6}{\left (c + d x \right )}}{8} - \frac{15 A a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac{3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{4} - \frac{15 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} - \frac{5 A a^{3} x \cos ^{6}{\left (c + d x \right )}}{8} + \frac{3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac{A a^{3} \sin ^{6}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{11 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{2 A a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} + \frac{5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} + \frac{8 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{A a^{3} \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{5 A a^{3} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac{3 A a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac{16 A a^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac{2 A a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (- A \sin{\left (c \right )} + A\right ) \left (a \sin{\left (c \right )} + a\right )^{3} \sin ^{3}{\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)**3*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.13997, size = 177, normalized size = 1.26 \begin{align*} \frac{1}{8} \, A a^{3} x - \frac{A a^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{7 \, A a^{3} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{5 \, A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac{13 \, A a^{3} \cos \left (d x + c\right )}{64 \, d} + \frac{A a^{3} \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac{A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{A a^{3} \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)^3*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*A*a^3*x - 1/448*A*a^3*cos(7*d*x + 7*c)/d + 7/320*A*a^3*cos(5*d*x + 5*c)/d - 5/192*A*a^3*cos(3*d*x + 3*c)/d
- 13/64*A*a^3*cos(d*x + c)/d + 1/96*A*a^3*sin(6*d*x + 6*c)/d - 1/32*A*a^3*sin(4*d*x + 4*c)/d - 1/32*A*a^3*sin
(2*d*x + 2*c)/d