3.3 \(\int \sin (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx\)

Optimal. Leaf size=77 \[ -\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \sin ^3(e+f x) \cos (e+f x)}{4 f}-\frac{a^2 c \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a^2 c x \]

[Out]

(a^2*c*x)/8 - (a^2*c*Cos[e + f*x]^3)/(3*f) - (a^2*c*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a^2*c*Cos[e + f*x]*Sin
[e + f*x]^3)/(4*f)

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Rubi [A]  time = 0.102453, antiderivative size = 77, 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} \[ -\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \sin ^3(e+f x) \cos (e+f x)}{4 f}-\frac{a^2 c \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a^2 c x \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x]),x]

[Out]

(a^2*c*x)/8 - (a^2*c*Cos[e + f*x]^3)/(3*f) - (a^2*c*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a^2*c*Cos[e + f*x]*Sin
[e + f*x]^3)/(4*f)

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 (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx &=\int \left (a^2 c \sin (e+f x)+a^2 c \sin ^2(e+f x)-a^2 c \sin ^3(e+f x)-a^2 c \sin ^4(e+f x)\right ) \, dx\\ &=\left (a^2 c\right ) \int \sin (e+f x) \, dx+\left (a^2 c\right ) \int \sin ^2(e+f x) \, dx-\left (a^2 c\right ) \int \sin ^3(e+f x) \, dx-\left (a^2 c\right ) \int \sin ^4(e+f x) \, dx\\ &=-\frac{a^2 c \cos (e+f x)}{f}-\frac{a^2 c \cos (e+f x) \sin (e+f x)}{2 f}+\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f}+\frac{1}{2} \left (a^2 c\right ) \int 1 \, dx-\frac{1}{4} \left (3 a^2 c\right ) \int \sin ^2(e+f x) \, dx+\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac{1}{2} a^2 c x-\frac{a^2 c \cos ^3(e+f x)}{3 f}-\frac{a^2 c \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f}-\frac{1}{8} \left (3 a^2 c\right ) \int 1 \, dx\\ &=\frac{1}{8} a^2 c x-\frac{a^2 c \cos ^3(e+f x)}{3 f}-\frac{a^2 c \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.110709, size = 47, normalized size = 0.61 \[ \frac{a^2 c (-3 \sin (4 (e+f x))-24 \cos (e+f x)-8 \cos (3 (e+f x))+12 e+12 f x)}{96 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x]),x]

[Out]

(a^2*c*(12*e + 12*f*x - 24*Cos[e + f*x] - 8*Cos[3*(e + f*x)] - 3*Sin[4*(e + f*x)]))/(96*f)

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Maple [A]  time = 0.021, size = 106, normalized size = 1.4 \begin{align*}{\frac{1}{f} \left ( -{a}^{2}c \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) +{\frac{{a}^{2}c \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+{a}^{2}c \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -{a}^{2}c\cos \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x)

[Out]

1/f*(-a^2*c*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)+1/3*a^2*c*(2+sin(f*x+e)^2)*cos(f*x+e
)+a^2*c*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-a^2*c*cos(f*x+e))

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Maxima [A]  time = 0.989108, size = 135, normalized size = 1.75 \begin{align*} -\frac{32 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} c + 3 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c - 24 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c + 96 \, a^{2} c \cos \left (f x + e\right )}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

-1/96*(32*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^2*c + 3*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*
a^2*c - 24*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^2*c + 96*a^2*c*cos(f*x + e))/f

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Fricas [A]  time = 1.99197, size = 150, normalized size = 1.95 \begin{align*} -\frac{8 \, a^{2} c \cos \left (f x + e\right )^{3} - 3 \, a^{2} c f x + 3 \,{\left (2 \, a^{2} c \cos \left (f x + e\right )^{3} - a^{2} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

-1/24*(8*a^2*c*cos(f*x + e)^3 - 3*a^2*c*f*x + 3*(2*a^2*c*cos(f*x + e)^3 - a^2*c*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 1.27841, size = 245, normalized size = 3.18 \begin{align*} \begin{cases} - \frac{3 a^{2} c x \sin ^{4}{\left (e + f x \right )}}{8} - \frac{3 a^{2} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac{3 a^{2} c x \cos ^{4}{\left (e + f x \right )}}{8} + \frac{a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + \frac{5 a^{2} c \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} + \frac{a^{2} c \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{3 a^{2} c \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{a^{2} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} + \frac{2 a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{a^{2} c \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (a \sin{\left (e \right )} + a\right )^{2} \left (- c \sin{\left (e \right )} + c\right ) \sin{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(a+a*sin(f*x+e))**2*(c-c*sin(f*x+e)),x)

[Out]

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

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Giac [A]  time = 1.16844, size = 84, normalized size = 1.09 \begin{align*} \frac{1}{8} \, a^{2} c x - \frac{a^{2} c \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac{a^{2} c \cos \left (f x + e\right )}{4 \, f} - \frac{a^{2} c \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

1/8*a^2*c*x - 1/12*a^2*c*cos(3*f*x + 3*e)/f - 1/4*a^2*c*cos(f*x + e)/f - 1/32*a^2*c*sin(4*f*x + 4*e)/f