\(\int \sin ^2(e+f x) (3-4 \sin ^2(e+f x)) \, dx\) [5]

Optimal result

Integrand size = 21, antiderivative size = 18 \[ \int \sin ^2(e+f x) \left (3-4 \sin ^2(e+f x)\right ) \, dx=\frac {\cos (e+f x) \sin ^3(e+f x)}{f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3090} \[ \int \sin ^2(e+f x) \left (3-4 \sin ^2(e+f x)\right ) \, dx=\frac {\sin ^3(e+f x) \cos (e+f x)}{f} \]

[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 3090

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*} \text {integral}& = \frac {\cos (e+f x) \sin ^3(e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \sin ^2(e+f x) \left (3-4 \sin ^2(e+f x)\right ) \, dx=\frac {4 e+2 \sin (2 (e+f x))-\sin (4 (e+f x))}{8 f} \]

[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)

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61

method	result	size
parallelrisch	\(\frac {-\sin \left (4 f x +4 e \right )+2 \sin \left (2 f x +2 e \right )}{8 f}\)	\(29\)
risch	\(-\frac {\sin \left (4 f x +4 e \right )}{8 f}+\frac {\sin \left (2 f x +2 e \right )}{4 f}\)	\(30\)
derivativedivides	\(\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 \cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}}{f}\)	\(44\)
default	\(\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 \cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}}{f}\)	\(44\)
norman	\(\frac {\frac {8 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {8 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{4}}\)	\(50\)
parts	\(\frac {-\frac {3 \cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {3 f x}{2}+\frac {3 e}{2}}{f}-\frac {4 \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\)	\(67\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56 \[ \int \sin ^2(e+f x) \left (3-4 \sin ^2(e+f x)\right ) \, dx=-\frac {{\left (\cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{f} \]

[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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (15) = 30\).

Time = 0.16 (sec) , antiderivative size = 148, normalized size of antiderivative = 8.22 \[ \int \sin ^2(e+f x) \left (3-4 \sin ^2(e+f x)\right ) \, dx=\begin {cases} - \frac {3 x \sin ^{4}{\left (e + f x \right )}}{2} - 3 x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )} + \frac {3 x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {3 x \cos ^{4}{\left (e + f x \right )}}{2} + \frac {3 x \cos ^{2}{\left (e + f x \right )}}{2} + \frac {5 \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {3 \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{2 f} - \frac {3 \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (3 - 4 \sin ^{2}{\left (e \right )}\right ) \sin ^{2}{\left (e \right )} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.89 \[ \int \sin ^2(e+f x) \left (3-4 \sin ^2(e+f x)\right ) \, dx=\frac {\tan \left (f x + e\right )^{3}}{{\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1\right )} f} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \sin ^2(e+f x) \left (3-4 \sin ^2(e+f x)\right ) \, dx=-\frac {\sin \left (4 \, f x + 4 \, e\right )}{8 \, f} + \frac {\sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

[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

Mupad [B] (verification not implemented)

Time = 13.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \sin ^2(e+f x) \left (3-4 \sin ^2(e+f x)\right ) \, dx=\frac {\cos \left (e+f\,x\right )\,{\sin \left (e+f\,x\right )}^3}{f} \]

[In]

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

[Out]

(cos(e + f*x)*sin(e + f*x)^3)/f