\(\int (1-2 \sin ^2(e+f x)) \, dx\) [7]

Optimal result

Integrand size = 12, antiderivative size = 16 \[ \int \left (1-2 \sin ^2(e+f x)\right ) \, dx=\frac {\cos (e+f x) \sin (e+f x)}{f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2715, 8} \[ \int \left (1-2 \sin ^2(e+f x)\right ) \, dx=\frac {\sin (e+f x) \cos (e+f x)}{f} \]

[In]

Int[1 - 2*Sin[e + f*x]^2,x]

[Out]

(Cos[e + f*x]*Sin[e + f*x])/f

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

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] && Integ
erQ[2*n]

Rubi steps \begin{align*} \text {integral}& = x-2 \int \sin ^2(e+f x) \, dx \\ & = x+\frac {\cos (e+f x) \sin (e+f x)}{f}-\int 1 \, dx \\ & = \frac {\cos (e+f x) \sin (e+f x)}{f} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(33\) vs. \(2(16)=32\).

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 2.06 \[ \int \left (1-2 \sin ^2(e+f x)\right ) \, dx=\frac {\cos (2 f x) \sin (2 e)}{2 f}+\frac {\cos (2 e) \sin (2 f x)}{2 f} \]

[In]

Integrate[1 - 2*Sin[e + f*x]^2,x]

[Out]

(Cos[2*f*x]*Sin[2*e])/(2*f) + (Cos[2*e]*Sin[2*f*x])/(2*f)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {\sin \left (2 f x +2 e \right )}{2 f}\)	\(15\)
derivativedivides	\(\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{f}\)	\(17\)
parallelrisch	\(-\frac {2 f x -\sin \left (2 f x +2 e \right )}{2 f}+x\)	\(24\)
default	\(x -\frac {2 \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(30\)
parts	\(x -\frac {2 \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(30\)
norman	\(\frac {\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {2 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}\)	\(48\)

[In]

int(1-2*sin(f*x+e)^2,x,method=_RETURNVERBOSE)

[Out]

1/2/f*sin(2*f*x+2*e)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \left (1-2 \sin ^2(e+f x)\right ) \, dx=\frac {\cos \left (f x + e\right ) \sin \left (f x + e\right )}{f} \]

[In]

integrate(1-2*sin(f*x+e)^2,x, algorithm="fricas")

[Out]

cos(f*x + e)*sin(f*x + e)/f

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 3.06 \[ \int \left (1-2 \sin ^2(e+f x)\right ) \, dx=x - 2 \left (\begin {cases} \frac {x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {\sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \sin ^{2}{\left (e \right )} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate(1-2*sin(f*x+e)**2,x)

[Out]

x - 2*Piecewise((x*sin(e + f*x)**2/2 + x*cos(e + f*x)**2/2 - sin(e + f*x)*cos(e + f*x)/(2*f), Ne(f, 0)), (x*si
n(e)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.62 \[ \int \left (1-2 \sin ^2(e+f x)\right ) \, dx=x - \frac {2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )}{2 \, f} \]

[In]

integrate(1-2*sin(f*x+e)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \left (1-2 \sin ^2(e+f x)\right ) \, dx=\frac {\sin \left (2 \, f x + 2 \, e\right )}{2 \, f} \]

[In]

integrate(1-2*sin(f*x+e)^2,x, algorithm="giac")

[Out]

1/2*sin(2*f*x + 2*e)/f

Mupad [B] (verification not implemented)

Time = 13.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \left (1-2 \sin ^2(e+f x)\right ) \, dx=\frac {\sin \left (2\,e+2\,f\,x\right )}{2\,f} \]

[In]

int(1 - 2*sin(e + f*x)^2,x)

[Out]

sin(2*e + 2*f*x)/(2*f)