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

Optimal. Leaf size=16 \[ -x-\frac {2 \sin (x)}{1-\cos (x)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2680, 8} \[ -x-\frac {2 \sin (x)}{1-\cos (x)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(1 - Cos[x])^2,x]

[Out]

-x - (2*Sin[x])/(1 - Cos[x])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rubi steps

\begin {align*} \int \frac {\sin ^2(x)}{(1-\cos (x))^2} \, dx &=-\frac {2 \sin (x)}{1-\cos (x)}-\int 1 \, dx\\ &=-x-\frac {2 \sin (x)}{1-\cos (x)}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 26, normalized size = 1.62 \[ -2 \cot \left (\frac {x}{2}\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\tan ^2\left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(1 - Cos[x])^2,x]

[Out]

-2*Cot[x/2]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[x/2]^2]

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fricas [A]  time = 1.41, size = 16, normalized size = 1.00 \[ -\frac {x \sin \relax (x) + 2 \, \cos \relax (x) + 2}{\sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*sin(x) + 2*cos(x) + 2)/sin(x)

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giac [A]  time = 0.37, size = 12, normalized size = 0.75 \[ -x - \frac {2}{\tan \left (\frac {1}{2} \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2/tan(1/2*x)

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maple [A]  time = 0.07, size = 13, normalized size = 0.81 \[ -\frac {2}{\tan \left (\frac {x}{2}\right )}-x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/tan(1/2*x)-x

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maxima [A]  time = 1.14, size = 23, normalized size = 1.44 \[ -\frac {2 \, {\left (\cos \relax (x) + 1\right )}}{\sin \relax (x)} - 2 \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(cos(x) + 1)/sin(x) - 2*arctan(sin(x)/(cos(x) + 1))

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mupad [B]  time = 0.31, size = 10, normalized size = 0.62 \[ -x-2\,\mathrm {cot}\left (\frac {x}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(cos(x) - 1)^2,x)

[Out]

- x - 2*cot(x/2)

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sympy [A]  time = 0.88, size = 8, normalized size = 0.50 \[ - x - \frac {2}{\tan {\left (\frac {x}{2} \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2/tan(x/2)

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