3.14 \(\int \frac{\sin ^2(x)}{(1+\cos (x))^2} \, dx\)

Optimal. Leaf size=14 \[ \frac{2 \sin (x)}{\cos (x)+1}-x \]

[Out]

-x + (2*Sin[x])/(1 + Cos[x])

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Rubi [A]  time = 0.0311965, antiderivative size = 14, 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 = {2680, 8} \[ \frac{2 \sin (x)}{\cos (x)+1}-x \]

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 [A]  time = 0.0055461, size = 18, normalized size = 1.29 \[ 2 \tan \left (\frac{x}{2}\right )-2 \tan ^{-1}\left (\tan \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(1 + Cos[x])^2,x]

[Out]

-2*ArcTan[Tan[x/2]] + 2*Tan[x/2]

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Maple [A]  time = 0.043, size = 11, normalized size = 0.8 \begin{align*} 2\,\tan \left ( x/2 \right ) -x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.72983, size = 31, normalized size = 2.21 \begin{align*} \frac{2 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - 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*sin(x)/(cos(x) + 1) - 2*arctan(sin(x)/(cos(x) + 1))

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Fricas [A]  time = 1.58084, size = 55, normalized size = 3.93 \begin{align*} -\frac{x \cos \left (x\right ) + x - 2 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} \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*cos(x) + x - 2*sin(x))/(cos(x) + 1)

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Sympy [A]  time = 0.677412, size = 7, normalized size = 0.5 \begin{align*} - x + 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 = 1.14553, size = 14, normalized size = 1. \begin{align*} -x + 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)