3.1.13 \(\int \frac {\sin (x)}{(1-\cos (x))^2} \, dx\) [13]

Optimal. Leaf size=10 \[ -\frac {1}{1-\cos (x)} \]

[Out]

-1/(1-cos(x))

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Rubi [A]
time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2746, 32} \begin {gather*} -\frac {1}{1-\cos (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(1 - Cos[x])^2,x]

[Out]

-(1 - Cos[x])^(-1)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2746

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 (x)}{(1-\cos (x))^2} \, dx &=\text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,-\cos (x)\right )\\ &=-\frac {1}{1-\cos (x)}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 12, normalized size = 1.20 \begin {gather*} -\frac {1}{2} \csc ^2\left (\frac {x}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(1 - Cos[x])^2,x]

[Out]

-1/2*Csc[x/2]^2

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Maple [A]
time = 0.05, size = 11, normalized size = 1.10

method	result	size
derivativedivides	\(-\frac {1}{1-\cos \left (x \right )}\)	\(11\)
default	\(-\frac {1}{1-\cos \left (x \right )}\)	\(11\)
risch	\(\frac {2 \,{\mathrm e}^{i x}}{\left ({\mathrm e}^{i x}-1\right )^{2}}\)	\(17\)
norman	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}-\frac {\tan \left (\frac {x}{2}\right )}{2}}{\left (\tan ^{2}\left (\frac {x}{2}\right )+1\right ) \tan \left (\frac {x}{2}\right )^{3}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(1-cos(x))^2,x,method=_RETURNVERBOSE)

[Out]

-1/(1-cos(x))

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Maxima [A]
time = 0.29, size = 6, normalized size = 0.60 \begin {gather*} \frac {1}{\cos \left (x\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(cos(x) - 1)

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Fricas [A]
time = 0.38, size = 6, normalized size = 0.60 \begin {gather*} \frac {1}{\cos \left (x\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(cos(x) - 1)

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Sympy [A]
time = 0.16, size = 5, normalized size = 0.50 \begin {gather*} \frac {1}{\cos {\left (x \right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(cos(x) - 1)

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Giac [A]
time = 0.45, size = 6, normalized size = 0.60 \begin {gather*} \frac {1}{\cos \left (x\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(cos(x) - 1)

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Mupad [B]
time = 0.25, size = 6, normalized size = 0.60 \begin {gather*} \frac {1}{\cos \left (x\right )-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(cos(x) - 1)^2,x)

[Out]

1/(cos(x) - 1)

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