3.94 \(\int x \tan ^{-1}(x)^2 \, dx\)

Optimal. Leaf size=35 \[ \frac{1}{2} \log \left (x^2+1\right )+\frac{1}{2} x^2 \tan ^{-1}(x)^2+\frac{1}{2} \tan ^{-1}(x)^2-x \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0557313, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.833, Rules used = {4852, 4916, 4846, 260, 4884} \[ \frac{1}{2} \log \left (x^2+1\right )+\frac{1}{2} x^2 \tan ^{-1}(x)^2+\frac{1}{2} \tan ^{-1}(x)^2-x \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTan[x]^2,x]

[Out]

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

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4916

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

Rule 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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]

Rubi steps

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

Mathematica [A]  time = 0.0070251, size = 26, normalized size = 0.74 \[ \frac{1}{2} \left (\log \left (x^2+1\right )+\left (x^2+1\right ) \tan ^{-1}(x)^2-2 x \tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTan[x]^2,x]

[Out]

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

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Maple [A]  time = 0.009, size = 30, normalized size = 0.9 \begin{align*} -x\arctan \left ( x \right ) +{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{2}}+{\frac{{x}^{2} \left ( \arctan \left ( x \right ) \right ) ^{2}}{2}}+{\frac{\ln \left ({x}^{2}+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(x)^2,x)

[Out]

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

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Maxima [A]  time = 1.46837, size = 46, normalized size = 1.31 \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (x\right )^{2} -{\left (x - \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac{1}{2} \, \arctan \left (x\right )^{2} + \frac{1}{2} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.523172, size = 81, normalized size = 2.31 \begin{align*} \frac{1}{2} \,{\left (x^{2} + 1\right )} \arctan \left (x\right )^{2} - x \arctan \left (x\right ) + \frac{1}{2} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.351926, size = 29, normalized size = 0.83 \begin{align*} \frac{x^{2} \operatorname{atan}^{2}{\left (x \right )}}{2} - x \operatorname{atan}{\left (x \right )} + \frac{\log{\left (x^{2} + 1 \right )}}{2} + \frac{\operatorname{atan}^{2}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(x)**2,x)

[Out]

x**2*atan(x)**2/2 - x*atan(x) + log(x**2 + 1)/2 + atan(x)**2/2

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Giac [A]  time = 1.07895, size = 39, normalized size = 1.11 \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (x\right )^{2} - x \arctan \left (x\right ) + \frac{1}{2} \, \arctan \left (x\right )^{2} + \frac{1}{2} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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