\(\int \cos ^5(e+f x) (-5+4 \sec ^2(e+f x)) \, dx\) [36]

Optimal result

Integrand size = 21, antiderivative size = 19 \[ \int \cos ^5(e+f x) \left (-5+4 \sec ^2(e+f x)\right ) \, dx=-\frac {\cos ^4(e+f x) \sin (e+f x)}{f} \]

[Out]

-cos(f*x+e)^4*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} \[ \int \cos ^5(e+f x) \left (-5+4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin (e+f x) \cos ^4(e+f x)}{f} \]

[In]

Int[Cos[e + f*x]^5*(-5 + 4*Sec[e + f*x]^2),x]

[Out]

-((Cos[e + f*x]^4*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 ^4(e+f x) \sin (e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.00 \[ \int \cos ^5(e+f x) \left (-5+4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin (e+f x)}{f}+\frac {2 \sin ^3(e+f x)}{f}-\frac {\sin ^5(e+f x)}{f} \]

[In]

Integrate[Cos[e + f*x]^5*(-5 + 4*Sec[e + f*x]^2),x]

[Out]

-(Sin[e + f*x]/f) + (2*Sin[e + f*x]^3)/f - Sin[e + f*x]^5/f

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.95

method	result	size
parallelrisch	\(\frac {-2 \sin \left (f x +e \right )-\sin \left (5 f x +5 e \right )-3 \sin \left (3 f x +3 e \right )}{16 f}\)	\(37\)
risch	\(-\frac {\sin \left (f x +e \right )}{8 f}-\frac {\sin \left (5 f x +5 e \right )}{16 f}-\frac {3 \sin \left (3 f x +3 e \right )}{16 f}\)	\(41\)
derivativedivides	\(\frac {-\left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )+\frac {4 \left (2+\cos \left (f x +e \right )^{2}\right ) \sin \left (f x +e \right )}{3}}{f}\)	\(52\)
default	\(\frac {-\left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )+\frac {4 \left (2+\cos \left (f x +e \right )^{2}\right ) \sin \left (f x +e \right )}{3}}{f}\)	\(52\)
norman	\(\frac {\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}+\frac {20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}-\frac {20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}+\frac {10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}\)	\(127\)

[In]

int(cos(f*x+e)^5*(-5+4*sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

1/16*(-2*sin(f*x+e)-sin(5*f*x+5*e)-3*sin(3*f*x+3*e))/f

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \cos ^5(e+f x) \left (-5+4 \sec ^2(e+f x)\right ) \, dx=-\frac {\cos \left (f x + e\right )^{4} \sin \left (f x + e\right )}{f} \]

[In]

integrate(cos(f*x+e)^5*(-5+4*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

-cos(f*x + e)^4*sin(f*x + e)/f

Sympy [F]

\[ \int \cos ^5(e+f x) \left (-5+4 \sec ^2(e+f x)\right ) \, dx=\int \left (4 \sec ^{2}{\left (e + f x \right )} - 5\right ) \cos ^{5}{\left (e + f x \right )}\, dx \]

[In]

integrate(cos(f*x+e)**5*(-5+4*sec(f*x+e)**2),x)

[Out]

Integral((4*sec(e + f*x)**2 - 5)*cos(e + f*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \cos ^5(e+f x) \left (-5+4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (f x + e\right )^{5} - 2 \, \sin \left (f x + e\right )^{3} + \sin \left (f x + e\right )}{f} \]

[In]

integrate(cos(f*x+e)^5*(-5+4*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

-(sin(f*x + e)^5 - 2*sin(f*x + e)^3 + sin(f*x + e))/f

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \cos ^5(e+f x) \left (-5+4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (f x + e\right )^{5} - 2 \, \sin \left (f x + e\right )^{3} + \sin \left (f x + e\right )}{f} \]

[In]

integrate(cos(f*x+e)^5*(-5+4*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-(sin(f*x + e)^5 - 2*sin(f*x + e)^3 + sin(f*x + e))/f

Mupad [B] (verification not implemented)

Time = 15.53 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \cos ^5(e+f x) \left (-5+4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (e+f\,x\right )\,{\left ({\sin \left (e+f\,x\right )}^2-1\right )}^2}{f} \]

[In]

int(cos(e + f*x)^5*(4/cos(e + f*x)^2 - 5),x)

[Out]

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