3.30 \(\int \sec (e+f x) (1-2 \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=17 \[ -\frac {\tan (e+f x) \sec (e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {4043} \[ -\frac {\tan (e+f x) \sec (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(1 - 2*Sec[e + f*x]^2),x]

[Out]

-((Sec[e + f*x]*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 (e+f x) \left (1-2 \sec ^2(e+f x)\right ) \, dx &=-\frac {\sec (e+f x) \tan (e+f x)}{f}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \[ -\frac {\tan (e+f x) \sec (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(1 - 2*Sec[e + f*x]^2),x]

[Out]

-((Sec[e + f*x]*Tan[e + f*x])/f)

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fricas [A]  time = 0.43, size = 19, normalized size = 1.12 \[ -\frac {\sin \left (f x + e\right )}{f \cos \left (f x + e\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(1-2*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

-sin(f*x + e)/(f*cos(f*x + e)^2)

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giac [A]  time = 0.25, size = 26, normalized size = 1.53 \[ -\frac {1}{f {\left (\frac {1}{\sin \left (f x + e\right )} - \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(1-2*sec(f*x+e)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.94, size = 18, normalized size = 1.06 \[ -\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(1-2*sec(f*x+e)^2),x)

[Out]

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

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maxima [A]  time = 0.37, size = 22, normalized size = 1.29 \[ \frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{2} - 1\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(1-2*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

sin(f*x + e)/((sin(f*x + e)^2 - 1)*f)

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mupad [B]  time = 0.06, size = 22, normalized size = 1.29 \[ \frac {\sin \left (e+f\,x\right )}{f\,\left ({\sin \left (e+f\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2/cos(e + f*x)^2 - 1)/cos(e + f*x),x)

[Out]

sin(e + f*x)/(f*(sin(e + f*x)^2 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \sec {\left (e + f x \right )}\right )\, dx - \int 2 \sec ^{3}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(1-2*sec(f*x+e)**2),x)

[Out]

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

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