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

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

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.02, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.10, size = 19, normalized size = 0.59 \[ \frac {{\left (x^{2} - 1\right )} \arctan \relax (x) + x}{4 \, {\left (x^{2} + 1\right )}} \]

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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giac [A]  time = 1.06, size = 26, normalized size = 0.81 \[ \frac {x}{4 \, {\left (x^{2} + 1\right )}} - \frac {\arctan \relax (x)}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{4} \, \arctan \relax (x) \]

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)

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maple [A]  time = 0.36, size = 27, normalized size = 0.84

method	result	size
default	\(\frac {x}{4 x^{2}+4}+\frac {\arctan \relax (x )}{4}-\frac {\arctan \relax (x )}{2 \left (x^{2}+1\right )}\)	\(27\)
risch	\(\frac {i \ln \left (i x +1\right )}{4 x^{2}+4}-\frac {i \left (2 \ln \left (-i x +1\right )+\ln \left (x -i\right ) x^{2}+\ln \left (x -i\right )-\ln \left (x +i\right ) x^{2}-\ln \left (x +i\right )+2 i x \right )}{8 \left (x +i\right ) \left (x -i\right )}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.96, size = 26, normalized size = 0.81 \[ \frac {x}{4 \, {\left (x^{2} + 1\right )}} - \frac {\arctan \relax (x)}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{4} \, \arctan \relax (x) \]

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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mupad [B]  time = 0.08, size = 21, normalized size = 0.66 \[ \frac {\mathrm {atan}\relax (x)}{4}+\frac {\frac {x}{4}-\frac {\mathrm {atan}\relax (x)}{2}}{x^2+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*atan(x))/(x^2 + 1)^2,x)

[Out]

atan(x)/4 + (x/4 - atan(x)/2)/(x^2 + 1)

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sympy [A]  time = 0.64, size = 31, normalized size = 0.97 \[ \frac {x^{2} \operatorname {atan}{\relax (x )}}{4 x^{2} + 4} + \frac {x}{4 x^{2} + 4} - \frac {\operatorname {atan}{\relax (x )}}{4 x^{2} + 4} \]

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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