3.19 \(\int \frac{\sin (x)}{(1-\cos (x))^3} \, dx\)

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

[Out]

-1/(2*(1 - Cos[x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0188377, antiderivative size = 12, normalized size of antiderivative = 1., 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 = {2667, 32} \[ -\frac{1}{2 (1-\cos (x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(1 - Cos[x])^3,x]

[Out]

-1/(2*(1 - Cos[x])^2)

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])

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]

Rubi steps

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

Mathematica [A]  time = 0.0101411, size = 12, normalized size = 1. \[ -\frac{1}{8} \csc ^4\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(1 - Cos[x])^3,x]

[Out]

-Csc[x/2]^4/8

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 11, normalized size = 0.9 \begin{align*} -{\frac{1}{2\, \left ( 1-\cos \left ( x \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.4021, size = 11, normalized size = 0.92 \begin{align*} -\frac{1}{2 \,{\left (\cos \left (x\right ) - 1\right )}^{2}} \end{align*}

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

________________________________________________________________________________________

Fricas [A]  time = 1.5605, size = 43, normalized size = 3.58 \begin{align*} -\frac{1}{2 \,{\left (\cos \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right )}} \end{align*}

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)

________________________________________________________________________________________

Sympy [A]  time = 0.816951, size = 15, normalized size = 1.25 \begin{align*} - \frac{1}{2 \cos ^{2}{\left (x \right )} - 4 \cos{\left (x \right )} + 2} \end{align*}

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)

________________________________________________________________________________________

Giac [A]  time = 1.1406, size = 11, normalized size = 0.92 \begin{align*} -\frac{1}{2 \,{\left (\cos \left (x\right ) - 1\right )}^{2}} \end{align*}

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