3.28 \(\int \sec ^3(e+f x) (3-4 \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=19 \[ -\frac{\tan (e+f x) \sec ^3(e+f x)}{f} \]

[Out]

-((Sec[e + f*x]^3*Tan[e + f*x])/f)

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Rubi [A]  time = 0.0232353, antiderivative size = 19, normalized size of antiderivative = 1., 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} \[ -\frac{\tan (e+f x) \sec ^3(e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sec[e + f*x]^3*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 ^3(e+f x) \left (3-4 \sec ^2(e+f x)\right ) \, dx &=-\frac{\sec ^3(e+f x) \tan (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 0.0313312, size = 19, normalized size = 1. \[ -\frac{\tan (e+f x) \sec ^3(e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sec[e + f*x]^3*Tan[e + f*x])/f)

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Maple [B]  time = 0.028, size = 47, normalized size = 2.5 \begin{align*}{\frac{1}{f} \left ({\frac{3\,\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2}}+4\, \left ( -1/4\, \left ( \sec \left ( fx+e \right ) \right ) ^{3}-3/8\,\sec \left ( fx+e \right ) \right ) \tan \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.921748, size = 45, normalized size = 2.37 \begin{align*} -\frac{\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.455161, size = 46, normalized size = 2.42 \begin{align*} -\frac{\sin \left (f x + e\right )}{f \cos \left (f x + e\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sin(f*x + e)/(f*cos(f*x + e)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - 3 \sec ^{3}{\left (e + f x \right )}\, dx - \int 4 \sec ^{5}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-3*sec(e + f*x)**3, x) - Integral(4*sec(e + f*x)**5, x)

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Giac [A]  time = 1.19714, size = 34, normalized size = 1.79 \begin{align*} -\frac{\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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