3.5 \(\int \sin ^2(e+f x) (3-4 \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=18 \[ \frac{\sin ^3(e+f x) \cos (e+f x)}{f} \]

[Out]

(Cos[e + f*x]*Sin[e + f*x]^3)/f

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Rubi [A]  time = 0.0214714, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {3011} \[ \frac{\sin ^3(e+f x) \cos (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2*(3 - 4*Sin[e + f*x]^2),x]

[Out]

(Cos[e + f*x]*Sin[e + f*x]^3)/f

Rule 3011

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*Cos[e
+ f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] /; FreeQ[{b, e, f, A, C, m}, x] && EqQ[A*(m + 2) + C*(m +
1), 0]

Rubi steps

\begin{align*} \int \sin ^2(e+f x) \left (3-4 \sin ^2(e+f x)\right ) \, dx &=\frac{\cos (e+f x) \sin ^3(e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 0.0656382, size = 31, normalized size = 1.72 \[ \frac{2 \sin (2 (e+f x))-\sin (4 (e+f x))+4 e}{8 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^2*(3 - 4*Sin[e + f*x]^2),x]

[Out]

(4*e + 2*Sin[2*(e + f*x)] - Sin[4*(e + f*x)])/(8*f)

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Maple [B]  time = 0.026, size = 44, normalized size = 2.4 \begin{align*}{\frac{1}{f} \left ( \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) \cos \left ( fx+e \right ) -{\frac{3\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^2*(3-4*sin(f*x+e)^2),x)

[Out]

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

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Maxima [A]  time = 0.958813, size = 46, normalized size = 2.56 \begin{align*} \frac{\tan \left (f x + e\right )^{3}}{{\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(3-4*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

tan(f*x + e)^3/((tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)*f)

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Fricas [A]  time = 1.62775, size = 66, normalized size = 3.67 \begin{align*} -\frac{{\left (\cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(3-4*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

-(cos(f*x + e)^3 - cos(f*x + e))*sin(f*x + e)/f

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Sympy [A]  time = 1.75751, size = 148, normalized size = 8.22 \begin{align*} \begin{cases} - \frac{3 x \sin ^{4}{\left (e + f x \right )}}{2} - 3 x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )} + \frac{3 x \sin ^{2}{\left (e + f x \right )}}{2} - \frac{3 x \cos ^{4}{\left (e + f x \right )}}{2} + \frac{3 x \cos ^{2}{\left (e + f x \right )}}{2} + \frac{5 \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} + \frac{3 \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{2 f} - \frac{3 \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} & \text{for}\: f \neq 0 \\x \left (3 - 4 \sin ^{2}{\left (e \right )}\right ) \sin ^{2}{\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)**2*(3-4*sin(f*x+e)**2),x)

[Out]

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

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Giac [A]  time = 1.1625, size = 42, normalized size = 2.33 \begin{align*} -\frac{\sin \left (4 \, f x + 4 \, e\right )}{8 \, f} + \frac{\sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(3-4*sin(f*x+e)^2),x, algorithm="giac")

[Out]

-1/8*sin(4*f*x + 4*e)/f + 1/4*sin(2*f*x + 2*e)/f