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

Optimal. Leaf size=96 \[ \frac{a^3 A \cos ^5(c+d x)}{5 d}-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac{a^3 A \sin (c+d x) \cos (c+d x)}{4 d}+\frac{1}{4} a^3 A x \]

[Out]

(a^3*A*x)/4 - (2*a^3*A*Cos[c + d*x]^3)/(3*d) + (a^3*A*Cos[c + d*x]^5)/(5*d) - (a^3*A*Cos[c + d*x]*Sin[c + d*x]
)/(4*d) + (a^3*A*Cos[c + d*x]*Sin[c + d*x]^3)/(2*d)

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Rubi [A]  time = 0.116473, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2966, 2638, 2635, 8, 2633} \[ \frac{a^3 A \cos ^5(c+d x)}{5 d}-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac{a^3 A \sin (c+d x) \cos (c+d x)}{4 d}+\frac{1}{4} a^3 A x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*A*x)/4 - (2*a^3*A*Cos[c + d*x]^3)/(3*d) + (a^3*A*Cos[c + d*x]^5)/(5*d) - (a^3*A*Cos[c + d*x]*Sin[c + d*x]
)/(4*d) + (a^3*A*Cos[c + d*x]*Sin[c + d*x]^3)/(2*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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=\int \left (a^3 A \sin (c+d x)+2 a^3 A \sin ^2(c+d x)-2 a^3 A \sin ^4(c+d x)-a^3 A \sin ^5(c+d x)\right ) \, dx\\ &=\left (a^3 A\right ) \int \sin (c+d x) \, dx-\left (a^3 A\right ) \int \sin ^5(c+d x) \, dx+\left (2 a^3 A\right ) \int \sin ^2(c+d x) \, dx-\left (2 a^3 A\right ) \int \sin ^4(c+d x) \, dx\\ &=-\frac{a^3 A \cos (c+d x)}{d}-\frac{a^3 A \cos (c+d x) \sin (c+d x)}{d}+\frac{a^3 A \cos (c+d x) \sin ^3(c+d x)}{2 d}+\left (a^3 A\right ) \int 1 \, dx-\frac{1}{2} \left (3 a^3 A\right ) \int \sin ^2(c+d x) \, dx+\frac{\left (a^3 A\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=a^3 A x-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \cos ^5(c+d x)}{5 d}-\frac{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)}{2 d}-\frac{1}{4} \left (3 a^3 A\right ) \int 1 \, dx\\ &=\frac{1}{4} a^3 A x-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \cos ^5(c+d x)}{5 d}-\frac{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)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.482205, size = 55, normalized size = 0.57 \[ \frac{a^3 A (-90 \cos (c+d x)-25 \cos (3 (c+d x))+3 (-5 \sin (4 (c+d x))+\cos (5 (c+d x))+20 d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*A*(-90*Cos[c + d*x] - 25*Cos[3*(c + d*x)] + 3*(20*d*x + Cos[5*(c + d*x)] - 5*Sin[4*(c + d*x)])))/(240*d)

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Maple [A]  time = 0.026, size = 117, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{3}A\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\,{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 ) +2\,{a}^{3}A \left ( -1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) -{a}^{3}A\cos \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.02804, size = 151, normalized size = 1.57 \begin{align*} \frac{16 \,{\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 a^{3} - 15 \,{\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} + 120 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 240 \, A a^{3} \cos \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(16*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*A*a^3 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c
) - 8*sin(2*d*x + 2*c))*A*a^3 + 120*(2*d*x + 2*c - sin(2*d*x + 2*c))*A*a^3 - 240*A*a^3*cos(d*x + c))/d

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Fricas [A]  time = 2.17571, size = 188, normalized size = 1.96 \begin{align*} \frac{12 \, A a^{3} \cos \left (d x + c\right )^{5} - 40 \, A a^{3} \cos \left (d x + c\right )^{3} + 15 \, A a^{3} d x - 15 \,{\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} - A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*A*a^3*cos(d*x + c)^5 - 40*A*a^3*cos(d*x + c)^3 + 15*A*a^3*d*x - 15*(2*A*a^3*cos(d*x + c)^3 - A*a^3*co
s(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 6.07396, size = 267, normalized size = 2.78 \begin{align*} \begin{cases} - \frac{3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{4} - \frac{3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + A a^{3} x \sin ^{2}{\left (c + d x \right )} - \frac{3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{4} + A a^{3} x \cos ^{2}{\left (c + d x \right )} + \frac{A a^{3} \sin ^{4}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} + \frac{4 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac{3 A a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac{A a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{8 A a^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac{A a^{3} \cos{\left (c + d x \right )}}{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{\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+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)

[Out]

Piecewise((-3*A*a**3*x*sin(c + d*x)**4/4 - 3*A*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + A*a**3*x*sin(c + d*x
)**2 - 3*A*a**3*x*cos(c + d*x)**4/4 + A*a**3*x*cos(c + d*x)**2 + A*a**3*sin(c + d*x)**4*cos(c + d*x)/d + 5*A*a
**3*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 4*A*a**3*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) + 3*A*a**3*sin(c + d*x
)*cos(c + d*x)**3/(4*d) - A*a**3*sin(c + d*x)*cos(c + d*x)/d + 8*A*a**3*cos(c + d*x)**5/(15*d) - A*a**3*cos(c
+ d*x)/d, Ne(d, 0)), (x*(-A*sin(c) + A)*(a*sin(c) + a)**3*sin(c), True))

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Giac [A]  time = 1.11934, size = 104, normalized size = 1.08 \begin{align*} \frac{1}{4} \, A a^{3} x + \frac{A a^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{5 \, A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{3 \, A a^{3} \cos \left (d x + c\right )}{8 \, d} - \frac{A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*A*a^3*x + 1/80*A*a^3*cos(5*d*x + 5*c)/d - 5/48*A*a^3*cos(3*d*x + 3*c)/d - 3/8*A*a^3*cos(d*x + c)/d - 1/16*
A*a^3*sin(4*d*x + 4*c)/d