\(\int \sec ^4(e+f x) (4-5 \sec ^2(e+f x)) \, dx\) [27]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.02 (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} \[ \int \sec ^4(e+f x) \left (4-5 \sec ^2(e+f x)\right ) \, dx=-\frac {\tan (e+f x) \sec ^4(e+f x)}{f} \]

[In]

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

[Out]

-((Sec[e + f*x]^4*Tan[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 {\sec ^4(e+f x) \tan (e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \sec ^4(e+f x) \left (4-5 \sec ^2(e+f x)\right ) \, dx=-\frac {\sec ^4(e+f x) \tan (e+f x)}{f} \]

[In]

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

[Out]

-((Sec[e + f*x]^4*Tan[e + f*x])/f)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16

method	result	size
risch	\(\frac {16 i \left ({\mathrm e}^{6 i \left (f x +e \right )}-{\mathrm e}^{4 i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\)	\(41\)
parallelrisch	\(-\frac {16 \sin \left (f x +e \right )}{f \left (\cos \left (5 f x +5 e \right )+5 \cos \left (3 f x +3 e \right )+10 \cos \left (f x +e \right )\right )}\)	\(43\)
derivativedivides	\(\frac {-4 \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )+5 \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\)	\(56\)
default	\(\frac {-4 \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )+5 \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\)	\(56\)
parts	\(-\frac {4 \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )}{f}+\frac {5 \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\)	\(58\)
norman	\(\frac {\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}+\frac {12 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}+\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}+\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{5}}\)	\(96\)

[In]

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

[Out]

16*I/f/(exp(2*I*(f*x+e))+1)^5*(exp(6*I*(f*x+e))-exp(4*I*(f*x+e)))

Fricas [A] (verification not implemented)

none

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

[In]

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

[Out]

-sin(f*x + e)/(f*cos(f*x + e)^5)

Sympy [F]

\[ \int \sec ^4(e+f x) \left (4-5 \sec ^2(e+f x)\right ) \, dx=- \int \left (- 4 \sec ^{4}{\left (e + f x \right )}\right )\, dx - \int 5 \sec ^{6}{\left (e + f x \right )}\, dx \]

[In]

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

[Out]

-Integral(-4*sec(e + f*x)**4, x) - Integral(5*sec(e + f*x)**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \sec ^4(e+f x) \left (4-5 \sec ^2(e+f x)\right ) \, dx=-\frac {\tan \left (f x + e\right )^{5} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )}{f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \sec ^4(e+f x) \left (4-5 \sec ^2(e+f x)\right ) \, dx=-\frac {\tan \left (f x + e\right )^{5} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \sec ^4(e+f x) \left (4-5 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (e+f\,x\right )}{f\,{\cos \left (e+f\,x\right )}^5} \]

[In]

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

[Out]

-sin(e + f*x)/(f*cos(e + f*x)^5)