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 = 0.030765, antiderivative size = 16, normalized size of antiderivative = 1., 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 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]

Rule 8

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

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*}

Mathematica [C]  time = 0.0080317, 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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Maple [A]  time = 0.061, size = 13, normalized size = 0.8 \begin{align*} -2\, \left ( \tan \left ( x/2 \right ) \right ) ^{-1}-x \end{align*}

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 = 1.68795, size = 31, normalized size = 1.94 \begin{align*} -\frac{2 \,{\left (\cos \left (x\right ) + 1\right )}}{\sin \left (x\right )} - 2 \, \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right ) \end{align*}

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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Fricas [A]  time = 1.60362, size = 47, normalized size = 2.94 \begin{align*} -\frac{x \sin \left (x\right ) + 2 \, \cos \left (x\right ) + 2}{\sin \left (x\right )} \end{align*}

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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Sympy [A]  time = 1.40056, size = 8, normalized size = 0.5 \begin{align*} - x - \frac{2}{\tan{\left (\frac{x}{2} \right )}} \end{align*}

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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Giac [A]  time = 1.20147, size = 16, normalized size = 1. \begin{align*} -x - \frac{2}{\tan \left (\frac{1}{2} \, x\right )} \end{align*}

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)