\(\int \frac {x^2 \arctan (x)}{1+x^2} \, dx\) [671]

Optimal result

Integrand size = 13, antiderivative size = 23 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \arctan (x)-\frac {\arctan (x)^2}{2}-\frac {1}{2} \log \left (1+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5036, 4930, 266, 5004} \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=-\frac {1}{2} \arctan (x)^2+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right ) \]

[In]

Int[(x^2*ArcTan[x])/(1 + x^2),x]

[Out]

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

Rule 266

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 4930

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

Rule 5004

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 5036

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]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \arctan (x)-\frac {\arctan (x)^2}{2}-\frac {1}{2} \log \left (1+x^2\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(x \arctan \left (x \right )-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(20\)
parallelrisch	\(x \arctan \left (x \right )-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(20\)
parts	\(x \arctan \left (x \right )-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(20\)
risch	\(\frac {\ln \left (i x +1\right )^{2}}{8}+\frac {i \left (-x +\frac {i \ln \left (-i x +1\right )}{2}\right ) \ln \left (i x +1\right )}{2}+\frac {\ln \left (-i x +1\right )^{2}}{8}+\frac {i x \ln \left (-i x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(67\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \arctan \left (x\right ) - \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \operatorname {atan}{\left (x \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx={\left (x - \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \arctan \left (x\right ) - \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=-\frac {{\mathrm {atan}\left (x\right )}^2}{2}+x\,\mathrm {atan}\left (x\right )-\frac {\ln \left (x^2+1\right )}{2} \]

[In]

int((x^2*atan(x))/(x^2 + 1),x)

[Out]

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