3.456 \(\int \frac{x^2}{(1+x^2)^2} \, dx\)

Optimal. Leaf size=19 \[ \frac{1}{2} \tan ^{-1}(x)-\frac{x}{2 \left (x^2+1\right )} \]

[Out]

-x/(2*(1 + x^2)) + ArcTan[x]/2

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Rubi [A]  time = 0.0037943, antiderivative size = 19, 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 = {288, 203} \[ \frac{1}{2} \tan ^{-1}(x)-\frac{x}{2 \left (x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(1 + x^2)^2,x]

[Out]

-x/(2*(1 + x^2)) + ArcTan[x]/2

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^2}{\left (1+x^2\right )^2} \, dx &=-\frac{x}{2 \left (1+x^2\right )}+\frac{1}{2} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{x}{2 \left (1+x^2\right )}+\frac{1}{2} \tan ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0078803, size = 19, normalized size = 1. \[ \frac{1}{2} \tan ^{-1}(x)-\frac{x}{2 \left (x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(1 + x^2)^2,x]

[Out]

-x/(2*(1 + x^2)) + ArcTan[x]/2

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Maple [A]  time = 0.004, size = 16, normalized size = 0.8 \begin{align*} -{\frac{x}{2\,{x}^{2}+2}}+{\frac{\arctan \left ( x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(x^2+1)+1/2*arctan(x)

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Maxima [A]  time = 1.5673, size = 20, normalized size = 1.05 \begin{align*} -\frac{x}{2 \,{\left (x^{2} + 1\right )}} + \frac{1}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(x^2 + 1) + 1/2*arctan(x)

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Fricas [A]  time = 0.948795, size = 55, normalized size = 2.89 \begin{align*} \frac{{\left (x^{2} + 1\right )} \arctan \left (x\right ) - x}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^2+1)^2,x, algorithm="fricas")

[Out]

1/2*((x^2 + 1)*arctan(x) - x)/(x^2 + 1)

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Sympy [A]  time = 0.093209, size = 12, normalized size = 0.63 \begin{align*} - \frac{x}{2 x^{2} + 2} + \frac{\operatorname{atan}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(x**2+1)**2,x)

[Out]

-x/(2*x**2 + 2) + atan(x)/2

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Giac [A]  time = 1.22345, size = 20, normalized size = 1.05 \begin{align*} -\frac{x}{2 \,{\left (x^{2} + 1\right )}} + \frac{1}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^2+1)^2,x, algorithm="giac")

[Out]

-1/2*x/(x^2 + 1) + 1/2*arctan(x)