3.1.19 \(\int \frac {\sin (x)}{(1-\cos (x))^3} \, dx\) [19]

Optimal. Leaf size=12 \[ -\frac {1}{2 (1-\cos (x))^2} \]

[Out]

-1/2/(1-cos(x))^2

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Rubi [A]
time = 0.01, antiderivative size = 12, 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}{2 (1-\cos (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(1 - Cos[x])^3,x]

[Out]

-1/2*1/(1 - Cos[x])^2

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))^3} \, dx &=\text {Subst}\left (\int \frac {1}{(1+x)^3} \, dx,x,-\cos (x)\right )\\ &=-\frac {1}{2 (1-\cos (x))^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} -\frac {1}{8} \csc ^4\left (\frac {x}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(1 - Cos[x])^3,x]

[Out]

-1/8*Csc[x/2]^4

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Maple [A]
time = 0.05, size = 11, normalized size = 0.92

method	result	size
derivativedivides	\(-\frac {1}{2 \left (1-\cos \left (x \right )\right )^{2}}\)	\(11\)
default	\(-\frac {1}{2 \left (1-\cos \left (x \right )\right )^{2}}\)	\(11\)
risch	\(-\frac {2 \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{i x}-1\right )^{4}}\)	\(17\)
norman	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4}-\frac {3 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{8}-\frac {\tan \left (\frac {x}{2}\right )}{8}}{\left (\tan ^{2}\left (\frac {x}{2}\right )+1\right ) \tan \left (\frac {x}{2}\right )^{5}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(1-cos(x))^3,x,method=_RETURNVERBOSE)

[Out]

-1/2/(1-cos(x))^2

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Maxima [A]
time = 0.30, size = 8, normalized size = 0.67 \begin {gather*} -\frac {1}{2 \, {\left (\cos \left (x\right ) - 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(1-cos(x))^3,x, algorithm="maxima")

[Out]

-1/2/(cos(x) - 1)^2

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Fricas [A]
time = 0.35, size = 14, normalized size = 1.17 \begin {gather*} -\frac {1}{2 \, {\left (\cos \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(1-cos(x))^3,x, algorithm="fricas")

[Out]

-1/2/(cos(x)^2 - 2*cos(x) + 1)

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Sympy [A]
time = 0.24, size = 15, normalized size = 1.25 \begin {gather*} - \frac {1}{2 \cos ^{2}{\left (x \right )} - 4 \cos {\left (x \right )} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(1-cos(x))**3,x)

[Out]

-1/(2*cos(x)**2 - 4*cos(x) + 2)

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Giac [A]
time = 0.45, size = 8, normalized size = 0.67 \begin {gather*} -\frac {1}{2 \, {\left (\cos \left (x\right ) - 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(1-cos(x))^3,x, algorithm="giac")

[Out]

-1/2/(cos(x) - 1)^2

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Mupad [B]
time = 0.04, size = 8, normalized size = 0.67 \begin {gather*} -\frac {1}{2\,{\left (\cos \left (x\right )-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-sin(x)/(cos(x) - 1)^3,x)

[Out]

-1/(2*(cos(x) - 1)^2)

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