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 = 0.02, 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 = {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 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 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 (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 = 0.01, size = 12, normalized size = 1.00 \[ -\frac {1}{8} \csc ^4\left (\frac {x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(1 - Cos[x])^3,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.94, size = 14, normalized size = 1.17 \[ -\frac {1}{2 \, {\left (\cos \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.48, size = 8, normalized size = 0.67 \[ -\frac {1}{2 \, {\left (\cos \relax (x) - 1\right )}^{2}} \]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 11, normalized size = 0.92 \[ -\frac {1}{2 \left (1-\cos \relax (x )\right )^{2}} \]

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 = 0.50, size = 8, normalized size = 0.67 \[ -\frac {1}{2 \, {\left (\cos \relax (x) - 1\right )}^{2}} \]

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

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 8, normalized size = 0.67 \[ -\frac {1}{2\,{\left (\cos \relax (x)-1\right )}^2} \]

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)

________________________________________________________________________________________

sympy [A]  time = 0.55, size = 15, normalized size = 1.25 \[ - \frac {1}{2 \cos ^{2}{\relax (x )} - 4 \cos {\relax (x )} + 2} \]

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)

________________________________________________________________________________________