\(\int \cos ^2(e+f x) (-2+\sec ^2(e+f x)) \, dx\) [33]

Optimal result

Integrand size = 19, antiderivative size = 17 \[ \int \cos ^2(e+f x) \left (-2+\sec ^2(e+f x)\right ) \, dx=-\frac {\cos (e+f x) \sin (e+f x)}{f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {4128} \[ \int \cos ^2(e+f x) \left (-2+\sec ^2(e+f x)\right ) \, dx=-\frac {\sin (e+f x) \cos (e+f x)}{f} \]

[In]

Int[Cos[e + f*x]^2*(-2 + Sec[e + f*x]^2),x]

[Out]

-((Cos[e + f*x]*Sin[e + f*x])/f)

Rule 4128

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x) \sin (e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.94 \[ \int \cos ^2(e+f x) \left (-2+\sec ^2(e+f x)\right ) \, dx=-\frac {\cos (2 f x) \sin (2 e)}{2 f}-\frac {\cos (2 e) \sin (2 f x)}{2 f} \]

[In]

Integrate[Cos[e + f*x]^2*(-2 + Sec[e + f*x]^2),x]

[Out]

-1/2*(Cos[2*f*x]*Sin[2*e])/f - (Cos[2*e]*Sin[2*f*x])/(2*f)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {\sin \left (2 f x +2 e \right )}{2 f}\)	\(15\)
parallelrisch	\(-\frac {\sin \left (2 f x +2 e \right )}{2 f}\)	\(15\)
derivativedivides	\(-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{f}\)	\(18\)
default	\(-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{f}\)	\(18\)
norman	\(\frac {\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}+\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}\)	\(79\)

[In]

int(cos(f*x+e)^2*(-2+sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \cos ^2(e+f x) \left (-2+\sec ^2(e+f x)\right ) \, dx=-\frac {\cos \left (f x + e\right ) \sin \left (f x + e\right )}{f} \]

[In]

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

[Out]

-cos(f*x + e)*sin(f*x + e)/f

Sympy [A] (verification not implemented)

Time = 3.63 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.88 \[ \int \cos ^2(e+f x) \left (-2+\sec ^2(e+f x)\right ) \, dx=x - 2 \left (\begin {cases} \frac {x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {x \cos ^{2}{\left (e + f x \right )}}{2} + \frac {\sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \cos ^{2}{\left (e \right )} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate(cos(f*x+e)**2*(-2+sec(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*co
s(e)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \cos ^2(e+f x) \left (-2+\sec ^2(e+f x)\right ) \, dx=-\frac {\tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )} f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \cos ^2(e+f x) \left (-2+\sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (2 \, f x + 2 \, e\right )}{2 \, f} \]

[In]

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

[Out]

-1/2*sin(2*f*x + 2*e)/f

Mupad [B] (verification not implemented)

Time = 15.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \cos ^2(e+f x) \left (-2+\sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (2\,e+2\,f\,x\right )}{2\,f} \]

[In]

int(cos(e + f*x)^2*(1/cos(e + f*x)^2 - 2),x)

[Out]

-sin(2*e + 2*f*x)/(2*f)