\(\int \cos ^4(e+f x) (-4+3 \sec ^2(e+f x)) \, dx\) [35]

Optimal result

Integrand size = 21, antiderivative size = 19 \[ \int \cos ^4(e+f x) \left (-4+3 \sec ^2(e+f x)\right ) \, dx=-\frac {\cos ^3(e+f x) \sin (e+f x)}{f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, 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 = {4128} \[ \int \cos ^4(e+f x) \left (-4+3 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin (e+f x) \cos ^3(e+f x)}{f} \]

[In]

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

[Out]

-((Cos[e + f*x]^3*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 ^3(e+f x) \sin (e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \cos ^4(e+f x) \left (-4+3 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin (2 (e+f x))}{4 f}-\frac {\sin (4 (e+f x))}{8 f} \]

[In]

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

[Out]

-1/4*Sin[2*(e + f*x)]/f - Sin[4*(e + f*x)]/(8*f)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53

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	\(\frac {-\left (\cos \left (f x +e \right )^{3}+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )+\frac {3 \cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}}{f}\)	\(45\)
default	\(\frac {-\left (\cos \left (f x +e \right )^{3}+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )+\frac {3 \cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}}{f}\)	\(45\)
norman	\(\frac {\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}+\frac {12 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}-\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}+\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}\)	\(111\)

[In]

int(cos(f*x+e)^4*(-4+3*sec(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 = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \cos ^4(e+f x) \left (-4+3 \sec ^2(e+f x)\right ) \, dx=-\frac {\cos \left (f x + e\right )^{3} \sin \left (f x + e\right )}{f} \]

[In]

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

[Out]

-cos(f*x + e)^3*sin(f*x + e)/f

Sympy [F]

\[ \int \cos ^4(e+f x) \left (-4+3 \sec ^2(e+f x)\right ) \, dx=\int \left (3 \sec ^{2}{\left (e + f x \right )} - 4\right ) \cos ^{4}{\left (e + f x \right )}\, dx \]

[In]

integrate(cos(f*x+e)**4*(-4+3*sec(f*x+e)**2),x)

[Out]

Integral((3*sec(e + f*x)**2 - 4)*cos(e + f*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \cos ^4(e+f x) \left (-4+3 \sec ^2(e+f x)\right ) \, dx=-\frac {\tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1\right )} f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \cos ^4(e+f x) \left (-4+3 \sec ^2(e+f x)\right ) \, dx=-\frac {\tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.43 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \cos ^4(e+f x) \left (-4+3 \sec ^2(e+f x)\right ) \, dx=-\frac {{\cos \left (e+f\,x\right )}^3\,\sin \left (e+f\,x\right )}{f} \]

[In]

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

[Out]

-(cos(e + f*x)^3*sin(e + f*x))/f