3.195 \(\int \frac{1}{\sec (x)-\tan (x)} \, dx\)

Optimal. Leaf size=9 \[ -\log (1-\sin (x)) \]

[Out]

-Log[1 - Sin[x]]

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Rubi [A]  time = 0.0270939, antiderivative size = 9, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3159, 2667, 31} \[ -\log (1-\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x] - Tan[x])^(-1),x]

[Out]

-Log[1 - Sin[x]]

Rule 3159

Int[((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Int[Cos[d + e*x
]/(b + a*Cos[d + e*x] + c*Sin[d + e*x]), x] /; FreeQ[{a, b, c, d, e}, x]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{\sec (x)-\tan (x)} \, dx &=\int \frac{\cos (x)}{1-\sin (x)} \, dx\\ &=-\operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,-\sin (x)\right )\\ &=-\log (1-\sin (x))\\ \end{align*}

Mathematica [A]  time = 0.0195591, size = 18, normalized size = 2. \[ -2 \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x] - Tan[x])^(-1),x]

[Out]

-2*Log[Cos[x/2] - Sin[x/2]]

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Maple [A]  time = 0.067, size = 8, normalized size = 0.9 \begin{align*} -\ln \left ( \sin \left ( x \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sec(x)-tan(x)),x)

[Out]

-ln(sin(x)-1)

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Maxima [B]  time = 1.09662, size = 39, normalized size = 4.33 \begin{align*} -2 \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right ) + \log \left (\frac{\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)-tan(x)),x, algorithm="maxima")

[Out]

-2*log(sin(x)/(cos(x) + 1) - 1) + log(sin(x)^2/(cos(x) + 1)^2 + 1)

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Fricas [A]  time = 0.482665, size = 26, normalized size = 2.89 \begin{align*} -\log \left (-\sin \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)-tan(x)),x, algorithm="fricas")

[Out]

-log(-sin(x) + 1)

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Sympy [B]  time = 0.233267, size = 17, normalized size = 1.89 \begin{align*} - \log{\left (- \tan{\left (x \right )} + \sec{\left (x \right )} \right )} + \frac{\log{\left (\tan ^{2}{\left (x \right )} + 1 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)-tan(x)),x)

[Out]

-log(-tan(x) + sec(x)) + log(tan(x)**2 + 1)/2

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Giac [B]  time = 1.11513, size = 27, normalized size = 3. \begin{align*} \log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right ) - 2 \, \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)-tan(x)),x, algorithm="giac")

[Out]

log(tan(1/2*x)^2 + 1) - 2*log(abs(tan(1/2*x) - 1))