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

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

[Out]

x/(2*(1 + x^2)) + ArcTan[x]/2

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Rubi [A]  time = 0.002733, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {199, 203} \[ \frac{x}{2 \left (x^2+1\right )}+\frac{1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)^(-2),x]

[Out]

x/(2*(1 + x^2)) + ArcTan[x]/2

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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{1}{\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 = 0.0048138, size = 16, normalized size = 0.84 \[ \frac{1}{2} \left (\frac{x}{x^2+1}+\tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^2)^(-2),x]

[Out]

(x/(1 + x^2) + ArcTan[x])/2

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Maple [A]  time = 0.001, 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(1/(x^2+1)^2,x)

[Out]

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

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Maxima [A]  time = 1.40375, 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(1/(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 = 1.76961, 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(1/(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 = 0.094896, 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(1/(x**2+1)**2,x)

[Out]

x/(2*x**2 + 2) + atan(x)/2

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Giac [A]  time = 1.05088, 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(1/(x^2+1)^2,x, algorithm="giac")

[Out]

1/2*x/(x^2 + 1) + 1/2*arctan(x)