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

Optimal. Leaf size=15 \[ \frac{\cos (x)}{1-\sin (x)}-x \]

[Out]

-x + Cos[x]/(1 - Sin[x])

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Rubi [A]  time = 0.0604004, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4391, 2735, 2648} \[ \frac{\cos (x)}{1-\sin (x)}-x \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(Sec[x] - Tan[x]),x]

[Out]

-x + Cos[x]/(1 - Sin[x])

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0360226, size = 29, normalized size = 1.93 \[ \frac{2 \sin \left (\frac{x}{2}\right )}{\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )}-x \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(Sec[x] - Tan[x]),x]

[Out]

-x + (2*Sin[x/2])/(Cos[x/2] - Sin[x/2])

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Maple [A]  time = 0.067, size = 15, normalized size = 1. \begin{align*} -2\, \left ( \tan \left ( x/2 \right ) -1 \right ) ^{-1}-x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(tan(1/2*x)-1)-x

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Maxima [A]  time = 1.49618, size = 38, normalized size = 2.53 \begin{align*} -\frac{2}{\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1} - 2 \, \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(sin(x)/(cos(x) + 1) - 1) - 2*arctan(sin(x)/(cos(x) + 1))

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Fricas [A]  time = 0.468083, size = 89, normalized size = 5.93 \begin{align*} -\frac{{\left (x - 1\right )} \cos \left (x\right ) -{\left (x + 1\right )} \sin \left (x\right ) + x - 1}{\cos \left (x\right ) - \sin \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((x - 1)*cos(x) - (x + 1)*sin(x) + x - 1)/(cos(x) - sin(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (x \right )}}{- \tan{\left (x \right )} + \sec{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)/(-tan(x) + sec(x)), x)

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Giac [A]  time = 1.14186, size = 19, normalized size = 1.27 \begin{align*} -x - \frac{2}{\tan \left (\frac{1}{2} \, x\right ) - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2/(tan(1/2*x) - 1)