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(f*x+e)^3*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} \[ -\frac {\tan (e+f x) \sec ^3(e+f x)}{f} \]

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*}

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Mathematica [A]  time = 0.04, size = 19, normalized size = 1.00 \[ -\frac {\tan (e+f x) \sec ^3(e+f x)}{f} \]

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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fricas [A]  time = 0.41, size = 19, normalized size = 1.00 \[ -\frac {\sin \left (f x + e\right )}{f \cos \left (f x + e\right )^{4}} \]

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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giac [A]  time = 0.28, size = 25, normalized size = 1.32 \[ -\frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{2} f} \]

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)

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maple [B]  time = 1.29, size = 47, normalized size = 2.47 \[ \frac {\frac {3 \sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+4 \left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )}{f} \]

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 = 0.32, size = 33, normalized size = 1.74 \[ -\frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1\right )} f} \]

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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mupad [B]  time = 2.37, size = 23, normalized size = 1.21 \[ -\frac {\sin \left (e+f\,x\right )}{f\,{\left ({\sin \left (e+f\,x\right )}^2-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- 3 \sec ^{3}{\left (e + f x \right )}\right )\, dx - \int 4 \sec ^{5}{\left (e + f x \right )}\, dx \]

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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