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

Optimal. Leaf size=52 \[ \frac{a c^2 \cos ^3(e+f x)}{3 f}+\frac{a c^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac{1}{2} a c^2 x \]

[Out]

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

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Rubi [A]  time = 0.0649879, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2736, 2669, 2635, 8} \[ \frac{a c^2 \cos ^3(e+f x)}{3 f}+\frac{a c^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac{1}{2} a c^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.287442, size = 42, normalized size = 0.81 \[ \frac{a c^2 (3 \sin (2 (e+f x))+3 \cos (e+f x)+\cos (3 (e+f x))+6 f x)}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^2*(6*f*x + 3*Cos[e + f*x] + Cos[3*(e + f*x)] + 3*Sin[2*(e + f*x)]))/(12*f)

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Maple [A]  time = 0.015, size = 77, normalized size = 1.5 \begin{align*}{\frac{1}{f} \left ( -{\frac{a{c}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-a{c}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) +a{c}^{2}\cos \left ( fx+e \right ) +a{c}^{2} \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.12239, size = 104, normalized size = 2. \begin{align*} \frac{4 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a c^{2} - 3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a c^{2} + 12 \,{\left (f x + e\right )} a c^{2} + 12 \, a c^{2} \cos \left (f x + e\right )}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.33194, size = 111, normalized size = 2.13 \begin{align*} \frac{2 \, a c^{2} \cos \left (f x + e\right )^{3} + 3 \, a c^{2} f x + 3 \, a c^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*c^2*cos(f*x + e)^3 + 3*a*c^2*f*x + 3*a*c^2*cos(f*x + e)*sin(f*x + e))/f

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Sympy [A]  time = 0.699654, size = 133, normalized size = 2.56 \begin{align*} \begin{cases} - \frac{a c^{2} x \sin ^{2}{\left (e + f x \right )}}{2} - \frac{a c^{2} x \cos ^{2}{\left (e + f x \right )}}{2} + a c^{2} x - \frac{a c^{2} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{a c^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 a c^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{a c^{2} \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (a \sin{\left (e \right )} + a\right ) \left (- c \sin{\left (e \right )} + c\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.91836, size = 84, normalized size = 1.62 \begin{align*} \frac{1}{2} \, a c^{2} x + \frac{a c^{2} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} + \frac{a c^{2} \cos \left (f x + e\right )}{4 \, f} + \frac{a c^{2} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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