\(\int \frac {1}{1-\sin (x)} \, dx\) [97]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 8, antiderivative size = 11 \[ \int \frac {1}{1-\sin (x)} \, dx=\frac {\cos (x)}{1-\sin (x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2727} \[ \int \frac {1}{1-\sin (x)} \, dx=\frac {\cos (x)}{1-\sin (x)} \]

[In]

Int[(1 - Sin[x])^(-1),x]

[Out]

Cos[x]/(1 - Sin[x])

Rule 2727

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*} \text {integral}& = \frac {\cos (x)}{1-\sin (x)} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(25\) vs. \(2(11)=22\).

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 2.27 \[ \int \frac {1}{1-\sin (x)} \, dx=\frac {2 \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )} \]

[In]

Integrate[(1 - Sin[x])^(-1),x]

[Out]

(2*Sin[x/2])/(Cos[x/2] - Sin[x/2])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00

method result size
default \(-\frac {2}{\tan \left (\frac {x}{2}\right )-1}\) \(11\)
norman \(-\frac {2}{\tan \left (\frac {x}{2}\right )-1}\) \(11\)
parallelrisch \(-\frac {2}{\tan \left (\frac {x}{2}\right )-1}\) \(11\)
risch \(\frac {2}{{\mathrm e}^{i x}-i}\) \(13\)

[In]

int(1/(1-sin(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.55 \[ \int \frac {1}{1-\sin (x)} \, dx=\frac {\cos \left (x\right ) + \sin \left (x\right ) + 1}{\cos \left (x\right ) - \sin \left (x\right ) + 1} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int \frac {1}{1-\sin (x)} \, dx=- \frac {2}{\tan {\left (\frac {x}{2} \right )} - 1} \]

[In]

integrate(1/(1-sin(x)),x)

[Out]

-2/(tan(x/2) - 1)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.36 \[ \int \frac {1}{1-\sin (x)} \, dx=-\frac {2}{\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1} \]

[In]

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

[Out]

-2/(sin(x)/(cos(x) + 1) - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {1}{1-\sin (x)} \, dx=-\frac {2}{\tan \left (\frac {1}{2} \, x\right ) - 1} \]

[In]

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

[Out]

-2/(tan(1/2*x) - 1)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {1}{1-\sin (x)} \, dx=-\frac {2}{\mathrm {tan}\left (\frac {x}{2}\right )-1} \]

[In]

int(-1/(sin(x) - 1),x)

[Out]

-2/(tan(x/2) - 1)