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 = 0.03, antiderivative size = 14, 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 = {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 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 [A]  time = 0.01, 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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fricas [A]  time = 1.10, size = 18, normalized size = 1.29 \[ -\frac {x \cos \relax (x) + x - 2 \, \sin \relax (x)}{\cos \relax (x) + 1} \]

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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giac [A]  time = 0.46, size = 10, normalized size = 0.71 \[ -x + 2 \, \tan \left (\frac {1}{2} \, x\right ) \]

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 = 0.04, size = 11, normalized size = 0.79 \[ 2 \tan \left (\frac {x}{2}\right )-x \]

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 = 0.91, size = 23, normalized size = 1.64 \[ \frac {2 \, \sin \relax (x)}{\cos \relax (x) + 1} - 2 \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.30, size = 10, normalized size = 0.71 \[ 2\,\mathrm {tan}\left (\frac {x}{2}\right )-x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(cos(x) + 1)^2,x)

[Out]

2*tan(x/2) - x

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sympy [A]  time = 0.46, size = 7, normalized size = 0.50 \[ - x + 2 \tan {\left (\frac {x}{2} \right )} \]

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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