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

Optimal. Leaf size=14 \[ -\sin (x)-\log (1-\sin (x)) \]

[Out]

-Log[1 - Sin[x]] - Sin[x]

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Rubi [A]  time = 0.0819025, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4391, 2833, 43} \[ -\sin (x)-\log (1-\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(Sec[x] - Tan[x]),x]

[Out]

-Log[1 - Sin[x]] - 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 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0220718, size = 23, normalized size = 1.64 \[ -\sin (x)-2 \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(Sec[x] - Tan[x]),x]

[Out]

-2*Log[Cos[x/2] - Sin[x/2]] - Sin[x]

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Maple [A]  time = 0.083, size = 13, normalized size = 0.9 \begin{align*} -\sin \left ( x \right ) -\ln \left ( \sin \left ( x \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sin(x)-ln(sin(x)-1)

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Maxima [B]  time = 1.66421, size = 73, normalized size = 5.21 \begin{align*} -\frac{2 \, \sin \left (x\right )}{{\left (\frac{\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}{\left (\cos \left (x\right ) + 1\right )}} - 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(sin(x)/(sec(x)-tan(x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.48163, size = 38, normalized size = 2.71 \begin{align*} -\log \left (-\sin \left (x\right ) + 1\right ) - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-sin(x) + 1) - sin(x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\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(sin(x)/(sec(x)-tan(x)),x)

[Out]

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

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Giac [A]  time = 1.10918, size = 19, normalized size = 1.36 \begin{align*} -\log \left (-\sin \left (x\right ) + 1\right ) - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-sin(x) + 1) - sin(x)