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

Optimal. Leaf size=19 \[ -\frac {\tan (e+f x) \sec ^4(e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {4043} \[ -\frac {\tan (e+f x) \sec ^4(e+f x)}{f} \]

Antiderivative was successfully verified.

[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 4043

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*} \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}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 19, normalized size = 1.00 \[ -\frac {\tan (e+f x) \sec ^4(e+f x)}{f} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.43, size = 19, normalized size = 1.00 \[ -\frac {\sin \left (f x + e\right )}{f \cos \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.24, size = 33, normalized size = 1.74 \[ -\frac {\tan \left (f x + e\right )^{5} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [B]  time = 1.17, size = 56, normalized size = 2.95 \[ \frac {-4 \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+5 \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-4*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e)+5*(-8/15-1/5*sec(f*x+e)^4-4/15*sec(f*x+e)^2)*tan(f*x+e))

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maxima [A]  time = 0.33, size = 30, normalized size = 1.58 \[ -\frac {\tan \left (f x + e\right )^{5} + 2 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 2.39, size = 19, normalized size = 1.00 \[ -\frac {\sin \left (e+f\,x\right )}{f\,{\cos \left (e+f\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- 4 \sec ^{4}{\left (e + f x \right )}\right )\, dx - \int 5 \sec ^{6}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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