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

Optimal. Leaf size=34 \[ -\frac {1}{4 \left (x^2+1\right )}-\frac {x \tan ^{-1}(x)}{2 \left (x^2+1\right )}+\frac {1}{4} \tan ^{-1}(x)^2 \]

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Rubi [A]  time = 0.05, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4934, 4884} \[ -\frac {1}{4 \left (x^2+1\right )}-\frac {x \tan ^{-1}(x)}{2 \left (x^2+1\right )}+\frac {1}{4} \tan ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcTan[x])/(1 + x^2)^2,x]

[Out]

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

Rule 4884

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

Rule 4934

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q
+ 1))/(4*c^3*d*(q + 1)^2), x] + (-Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x],
x] + Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]))/(2*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && E
qQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 28, normalized size = 0.82 \[ \frac {\left (x^2+1\right ) \tan ^{-1}(x)^2-2 x \tan ^{-1}(x)-1}{4 \left (x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 0.73, size = 26, normalized size = 0.76 \[ \frac {{\left (x^{2} + 1\right )} \arctan \relax (x)^{2} - 2 \, x \arctan \relax (x) - 1}{4 \, {\left (x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \arctan \relax (x)}{{\left (x^{2} + 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.39, size = 29, normalized size = 0.85

method	result	size
default	\(-\frac {1}{4 \left (x^{2}+1\right )}-\frac {x \arctan \relax (x )}{2 \left (x^{2}+1\right )}+\frac {\arctan \relax (x )^{2}}{4}\)	\(29\)
risch	\(-\frac {\ln \left (i x +1\right )^{2}}{16}+\frac {\left (x^{2} \ln \left (-i x +1\right )+\ln \left (-i x +1\right )+2 i x \right ) \ln \left (i x +1\right )}{8 x^{2}+8}-\frac {x^{2} \ln \left (-i x +1\right )^{2}+\ln \left (-i x +1\right )^{2}+4 i \ln \left (-i x +1\right ) x +4}{16 \left (x +i\right ) \left (x -i\right )}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 40, normalized size = 1.18 \[ -\frac {1}{2} \, {\left (\frac {x}{x^{2} + 1} - \arctan \relax (x)\right )} \arctan \relax (x) - \frac {{\left (x^{2} + 1\right )} \arctan \relax (x)^{2} + 1}{4 \, {\left (x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 23, normalized size = 0.68 \[ \frac {{\mathrm {atan}\relax (x)}^2}{4}-\frac {\frac {x\,\mathrm {atan}\relax (x)}{2}+\frac {1}{4}}{x^2+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*atan(x))/(x^2 + 1)^2,x)

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RecursionError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RecursionError

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