3.669 \(\int \frac{x \tan ^{-1}(x)}{(1+x^2)^2} \, dx\)

Optimal. Leaf size=32 \[ \frac{x}{4 \left (x^2+1\right )}-\frac{\tan ^{-1}(x)}{2 \left (x^2+1\right )}+\frac{1}{4} \tan ^{-1}(x) \]

[Out]

x/(4*(1 + x^2)) + ArcTan[x]/4 - ArcTan[x]/(2*(1 + x^2))

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Rubi [A]  time = 0.0259001, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4930, 199, 203} \[ \frac{x}{4 \left (x^2+1\right )}-\frac{\tan ^{-1}(x)}{2 \left (x^2+1\right )}+\frac{1}{4} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTan[x])/(1 + x^2)^2,x]

[Out]

x/(4*(1 + x^2)) + ArcTan[x]/4 - ArcTan[x]/(2*(1 + x^2))

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

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{x \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx &=-\frac{\tan ^{-1}(x)}{2 \left (1+x^2\right )}+\frac{1}{2} \int \frac{1}{\left (1+x^2\right )^2} \, dx\\ &=\frac{x}{4 \left (1+x^2\right )}-\frac{\tan ^{-1}(x)}{2 \left (1+x^2\right )}+\frac{1}{4} \int \frac{1}{1+x^2} \, dx\\ &=\frac{x}{4 \left (1+x^2\right )}+\frac{1}{4} \tan ^{-1}(x)-\frac{\tan ^{-1}(x)}{2 \left (1+x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0199502, size = 21, normalized size = 0.66 \[ \frac{\left (x^2-1\right ) \tan ^{-1}(x)+x}{4 \left (x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcTan[x])/(1 + x^2)^2,x]

[Out]

(x + (-1 + x^2)*ArcTan[x])/(4*(1 + x^2))

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Maple [A]  time = 0.004, size = 27, normalized size = 0.8 \begin{align*}{\frac{x}{4\,{x}^{2}+4}}+{\frac{\arctan \left ( x \right ) }{4}}-{\frac{\arctan \left ( x \right ) }{2\,{x}^{2}+2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x/(x^2+1)+1/4*arctan(x)-1/2*arctan(x)/(x^2+1)

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Maxima [A]  time = 1.40762, size = 35, normalized size = 1.09 \begin{align*} \frac{x}{4 \,{\left (x^{2} + 1\right )}} - \frac{\arctan \left (x\right )}{2 \,{\left (x^{2} + 1\right )}} + \frac{1}{4} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x/(x^2 + 1) - 1/2*arctan(x)/(x^2 + 1) + 1/4*arctan(x)

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Fricas [A]  time = 2.36349, size = 55, normalized size = 1.72 \begin{align*} \frac{{\left (x^{2} - 1\right )} \arctan \left (x\right ) + x}{4 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((x^2 - 1)*arctan(x) + x)/(x^2 + 1)

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Sympy [A]  time = 0.687117, size = 31, normalized size = 0.97 \begin{align*} \frac{x^{2} \operatorname{atan}{\left (x \right )}}{4 x^{2} + 4} + \frac{x}{4 x^{2} + 4} - \frac{\operatorname{atan}{\left (x \right )}}{4 x^{2} + 4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*atan(x)/(4*x**2 + 4) + x/(4*x**2 + 4) - atan(x)/(4*x**2 + 4)

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Giac [A]  time = 1.06447, size = 35, normalized size = 1.09 \begin{align*} \frac{x}{4 \,{\left (x^{2} + 1\right )}} - \frac{\arctan \left (x\right )}{2 \,{\left (x^{2} + 1\right )}} + \frac{1}{4} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x/(x^2 + 1) - 1/2*arctan(x)/(x^2 + 1) + 1/4*arctan(x)