3.1.30 \(\int \sec (e+f x) (1-2 \sec ^2(e+f x)) \, dx\) [30]

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {4128} \begin {gather*} -\frac {\tan (e+f x) \sec (e+f x)}{f} \end {gather*}

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 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*} \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 = 0.01, size = 17, normalized size = 1.00 \begin {gather*} -\frac {\sec (e+f x) \tan (e+f x)}{f} \end {gather*}

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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Maple [A]
time = 0.30, size = 18, normalized size = 1.06

method	result	size
derivativedivides	\(-\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{f}\)	\(18\)
default	\(-\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{f}\)	\(18\)
risch	\(\frac {2 i \left ({\mathrm e}^{3 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}\)	\(41\)
norman	\(\frac {-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {2 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 24, normalized size = 1.41 \begin {gather*} \frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{2} - 1\right )} f} \end {gather*}

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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Fricas [A]
time = 3.74, size = 21, normalized size = 1.24 \begin {gather*} -\frac {\sin \left (f x + e\right )}{f \cos \left (f x + e\right )^{2}} \end {gather*}

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

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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Giac [A]
time = 0.45, size = 24, normalized size = 1.41 \begin {gather*} -\frac {1}{f {\left (\frac {1}{\sin \left (f x + e\right )} - \sin \left (f x + e\right )\right )}} \end {gather*}

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

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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