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

Optimal. Leaf size=82 \[ -\frac{5 a^3 A \cos ^3(c+d x)}{12 d}-\frac{A \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{4 d}+\frac{5 a^3 A \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 A x \]

[Out]

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

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Rubi [A]  time = 0.105893, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {2736, 2678, 2669, 2635, 8} \[ -\frac{5 a^3 A \cos ^3(c+d x)}{12 d}-\frac{A \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{4 d}+\frac{5 a^3 A \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 A x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 2678

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^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 (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=(a A) \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d}+\frac{1}{4} \left (5 a^2 A\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{5 a^3 A \cos ^3(c+d x)}{12 d}-\frac{A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d}+\frac{1}{4} \left (5 a^3 A\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{5 a^3 A \cos ^3(c+d x)}{12 d}+\frac{5 a^3 A \cos (c+d x) \sin (c+d x)}{8 d}-\frac{A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d}+\frac{1}{8} \left (5 a^3 A\right ) \int 1 \, dx\\ &=\frac{5}{8} a^3 A x-\frac{5 a^3 A \cos ^3(c+d x)}{12 d}+\frac{5 a^3 A \cos (c+d x) \sin (c+d x)}{8 d}-\frac{A \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.357282, size = 54, normalized size = 0.66 \[ \frac{a^3 A (24 \sin (2 (c+d x))-3 \sin (4 (c+d x))-48 \cos (c+d x)-16 \cos (3 (c+d x))+60 d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*A*(60*d*x - 48*Cos[c + d*x] - 16*Cos[3*(c + d*x)] + 24*Sin[2*(c + d*x)] - 3*Sin[4*(c + d*x)]))/(96*d)

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Maple [A]  time = 0.026, size = 89, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( -{a}^{3}A \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{2\,{a}^{3}A \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}}-2\,{a}^{3}A\cos \left ( dx+c \right ) +{a}^{3}A \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.960309, size = 116, normalized size = 1.41 \begin{align*} -\frac{64 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a^{3} + 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 )} A a^{3} - 96 \,{\left (d x + c\right )} A a^{3} + 192 \, A a^{3} \cos \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(64*(cos(d*x + c)^3 - 3*cos(d*x + c))*A*a^3 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*
A*a^3 - 96*(d*x + c)*A*a^3 + 192*A*a^3*cos(d*x + c))/d

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Fricas [A]  time = 1.94899, size = 155, normalized size = 1.89 \begin{align*} -\frac{16 \, A a^{3} \cos \left (d x + c\right )^{3} - 15 \, A a^{3} d x + 3 \,{\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} - 5 \, A a^{3} \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((a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/24*(16*A*a^3*cos(d*x + c)^3 - 15*A*a^3*d*x + 3*(2*A*a^3*cos(d*x + c)^3 - 5*A*a^3*cos(d*x + c))*sin(d*x + c)
)/d

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Sympy [A]  time = 2.45875, size = 196, normalized size = 2.39 \begin{align*} \begin{cases} - \frac{3 A a^{3} x \sin ^{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 )}}{4} - \frac{3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + A a^{3} x + \frac{5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{2 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{3 A a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{4 A a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 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} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [A]  time = 1.12431, size = 104, normalized size = 1.27 \begin{align*} \frac{5}{8} \, A a^{3} x - \frac{A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac{A a^{3} \cos \left (d x + c\right )}{2 \, 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 )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*A*a^3*x - 1/6*A*a^3*cos(3*d*x + 3*c)/d - 1/2*A*a^3*cos(d*x + c)/d - 1/32*A*a^3*sin(4*d*x + 4*c)/d + 1/4*A*
a^3*sin(2*d*x + 2*c)/d