\(\int x \arctan (x) \, dx\) [92]

Optimal result

Integrand size = 4, antiderivative size = 21 \[ \int x \arctan (x) \, dx=-\frac {x}{2}+\frac {\arctan (x)}{2}+\frac {1}{2} x^2 \arctan (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {4946, 327, 209} \[ \int x \arctan (x) \, dx=\frac {1}{2} x^2 \arctan (x)+\frac {\arctan (x)}{2}-\frac {x}{2} \]

[In]

Int[x*ArcTan[x],x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 327

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

Rule 4946

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int x \arctan (x) \, dx=\frac {1}{2} \left (-x+\left (1+x^2\right ) \arctan (x)\right ) \]

[In]

Integrate[x*ArcTan[x],x]

[Out]

(-x + (1 + x^2)*ArcTan[x])/2

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {x}{2}+\frac {\arctan \left (x \right )}{2}+\frac {x^{2} \arctan \left (x \right )}{2}\)	\(16\)
meijerg	\(-\frac {x}{2}+\frac {\left (3 x^{2}+3\right ) \arctan \left (x \right )}{6}\)	\(16\)
parallelrisch	\(-\frac {x}{2}+\frac {\arctan \left (x \right )}{2}+\frac {x^{2} \arctan \left (x \right )}{2}\)	\(16\)
parts	\(-\frac {x}{2}+\frac {\arctan \left (x \right )}{2}+\frac {x^{2} \arctan \left (x \right )}{2}\)	\(16\)
risch	\(-\frac {i x^{2} \ln \left (i x +1\right )}{4}+\frac {i x^{2} \ln \left (-i x +1\right )}{4}-\frac {x}{2}+\frac {\arctan \left (x \right )}{2}\)	\(35\)

[In]

int(x*arctan(x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int x \arctan (x) \, dx=\frac {1}{2} \, {\left (x^{2} + 1\right )} \arctan \left (x\right ) - \frac {1}{2} \, x \]

[In]

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

[Out]

1/2*(x^2 + 1)*arctan(x) - 1/2*x

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int x \arctan (x) \, dx=\frac {x^{2} \operatorname {atan}{\left (x \right )}}{2} - \frac {x}{2} + \frac {\operatorname {atan}{\left (x \right )}}{2} \]

[In]

integrate(x*atan(x),x)

[Out]

x**2*atan(x)/2 - x/2 + atan(x)/2

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int x \arctan (x) \, dx=\frac {1}{2} \, x^{2} \arctan \left (x\right ) - \frac {1}{2} \, x + \frac {1}{2} \, \arctan \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int x \arctan (x) \, dx=\frac {1}{2} \, x^{2} \arctan \left (x\right ) - \frac {1}{2} \, x + \frac {1}{2} \, \arctan \left (x\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int x \arctan (x) \, dx=\mathrm {atan}\left (x\right )\,\left (\frac {x^2}{2}+\frac {1}{2}\right )-\frac {x}{2} \]

[In]

int(x*atan(x),x)

[Out]

atan(x)*(x^2/2 + 1/2) - x/2