3.11 \(\int \frac {\sin (2 x)}{1-\cos (2 x)} \, dx\)

Optimal. Leaf size=3 \[ \log (\sin (x)) \]

[Out]

ln(sin(x))

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Rubi [B]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 4.33, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2667, 31} \[ \frac {1}{2} \log (1-\cos (2 x)) \]

Antiderivative was successfully verified.

[In]

Int[Sin[2*x]/(1 - Cos[2*x]),x]

[Out]

Log[1 - Cos[2*x]]/2

Rule 31

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

Rubi steps

\begin {align*} \int \frac {\sin (2 x)}{1-\cos (2 x)} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,-\cos (2 x)\right )\\ &=\frac {1}{2} \log (1-\cos (2 x))\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 3, normalized size = 1.00 \[ \log (\sin (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[2*x]/(1 - Cos[2*x]),x]

[Out]

Log[Sin[x]]

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fricas [B]  time = 0.97, size = 11, normalized size = 3.67 \[ \frac {1}{2} \, \log \left (-\frac {1}{2} \, \cos \left (2 \, x\right ) + \frac {1}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(1-cos(2*x)),x, algorithm="fricas")

[Out]

1/2*log(-1/2*cos(2*x) + 1/2)

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giac [B]  time = 0.41, size = 11, normalized size = 3.67 \[ \frac {1}{2} \, \log \left (-\cos \left (2 \, x\right ) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(1-cos(2*x)),x, algorithm="giac")

[Out]

1/2*log(-cos(2*x) + 1)

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maple [B]  time = 0.02, size = 12, normalized size = 4.00 \[ \frac {\ln \left (1-\cos \left (2 x \right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)/(1-cos(2*x)),x)

[Out]

1/2*ln(1-cos(2*x))

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maxima [B]  time = 0.31, size = 9, normalized size = 3.00 \[ \frac {1}{2} \, \log \left (\cos \left (2 \, x\right ) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(1-cos(2*x)),x, algorithm="maxima")

[Out]

1/2*log(cos(2*x) - 1)

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mupad [B]  time = 0.08, size = 9, normalized size = 3.00 \[ \frac {\ln \left (-{\sin \relax (x)}^2\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-sin(2*x)/(cos(2*x) - 1),x)

[Out]

log(-sin(x)^2)/2

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sympy [B]  time = 0.11, size = 8, normalized size = 2.67 \[ \frac {\log {\left (\cos {\left (2 x \right )} - 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(1-cos(2*x)),x)

[Out]

log(cos(2*x) - 1)/2

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