\(\int \frac {x \arctan (x)}{(1+x^2)^3} \, dx\) [670]

Optimal result

Integrand size = 11, antiderivative size = 44 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^3} \, dx=\frac {x}{16 \left (1+x^2\right )^2}+\frac {3 x}{32 \left (1+x^2\right )}+\frac {3 \arctan (x)}{32}-\frac {\arctan (x)}{4 \left (1+x^2\right )^2} \]

[Out]

1/16*x/(x^2+1)^2+3/32*x/(x^2+1)+3/32*arctan(x)-1/4*arctan(x)/(x^2+1)^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5050, 205, 209} \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^3} \, dx=-\frac {\arctan (x)}{4 \left (x^2+1\right )^2}+\frac {3 \arctan (x)}{32}+\frac {3 x}{32 \left (x^2+1\right )}+\frac {x}{16 \left (x^2+1\right )^2} \]

[In]

Int[(x*ArcTan[x])/(1 + x^2)^3,x]

[Out]

x/(16*(1 + x^2)^2) + (3*x)/(32*(1 + x^2)) + (3*ArcTan[x])/32 - ArcTan[x]/(4*(1 + x^2)^2)

Rule 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 5050

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*} \text {integral}& = -\frac {\arctan (x)}{4 \left (1+x^2\right )^2}+\frac {1}{4} \int \frac {1}{\left (1+x^2\right )^3} \, dx \\ & = \frac {x}{16 \left (1+x^2\right )^2}-\frac {\arctan (x)}{4 \left (1+x^2\right )^2}+\frac {3}{16} \int \frac {1}{\left (1+x^2\right )^2} \, dx \\ & = \frac {x}{16 \left (1+x^2\right )^2}+\frac {3 x}{32 \left (1+x^2\right )}-\frac {\arctan (x)}{4 \left (1+x^2\right )^2}+\frac {3}{32} \int \frac {1}{1+x^2} \, dx \\ & = \frac {x}{16 \left (1+x^2\right )^2}+\frac {3 x}{32 \left (1+x^2\right )}+\frac {3 \arctan (x)}{32}-\frac {\arctan (x)}{4 \left (1+x^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^3} \, dx=\frac {x \left (5+3 x^2\right )+\left (-5+6 x^2+3 x^4\right ) \arctan (x)}{32 \left (1+x^2\right )^2} \]

[In]

Integrate[(x*ArcTan[x])/(1 + x^2)^3,x]

[Out]

(x*(5 + 3*x^2) + (-5 + 6*x^2 + 3*x^4)*ArcTan[x])/(32*(1 + x^2)^2)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {x}{16 \left (x^{2}+1\right )^{2}}+\frac {3 x}{32 \left (x^{2}+1\right )}+\frac {3 \arctan \left (x \right )}{32}-\frac {\arctan \left (x \right )}{4 \left (x^{2}+1\right )^{2}}\)	\(37\)
parallelrisch	\(\frac {3 \arctan \left (x \right ) x^{4}+3 x^{3}+6 x^{2} \arctan \left (x \right )+5 x -5 \arctan \left (x \right )}{32 \left (x^{2}+1\right )^{2}}\)	\(37\)
parts	\(\frac {x}{16 \left (x^{2}+1\right )^{2}}+\frac {3 x}{32 \left (x^{2}+1\right )}+\frac {3 \arctan \left (x \right )}{32}-\frac {\arctan \left (x \right )}{4 \left (x^{2}+1\right )^{2}}\)	\(37\)
risch	\(\frac {i \ln \left (i x +1\right )}{8 \left (x^{2}+1\right )^{2}}-\frac {i \left (8 \ln \left (-i x +1\right )+3 \ln \left (x -i\right ) x^{4}+6 x^{2} \ln \left (x -i\right )+3 \ln \left (x -i\right )-3 \ln \left (x +i\right ) x^{4}-6 \ln \left (x +i\right ) x^{2}-3 \ln \left (x +i\right )+6 i x^{3}+10 i x \right )}{64 \left (x +i\right )^{2} \left (x -i\right )^{2}}\)	\(108\)

[In]

int(x*arctan(x)/(x^2+1)^3,x,method=_RETURNVERBOSE)

[Out]

1/16*x/(x^2+1)^2+3/32*x/(x^2+1)+3/32*arctan(x)-1/4*arctan(x)/(x^2+1)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^3} \, dx=\frac {3 \, x^{3} + {\left (3 \, x^{4} + 6 \, x^{2} - 5\right )} \arctan \left (x\right ) + 5 \, x}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]

[In]

integrate(x*arctan(x)/(x^2+1)^3,x, algorithm="fricas")

[Out]

1/32*(3*x^3 + (3*x^4 + 6*x^2 - 5)*arctan(x) + 5*x)/(x^4 + 2*x^2 + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (37) = 74\).

Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.00 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^3} \, dx=\frac {3 x^{4} \operatorname {atan}{\left (x \right )}}{32 x^{4} + 64 x^{2} + 32} + \frac {3 x^{3}}{32 x^{4} + 64 x^{2} + 32} + \frac {6 x^{2} \operatorname {atan}{\left (x \right )}}{32 x^{4} + 64 x^{2} + 32} + \frac {5 x}{32 x^{4} + 64 x^{2} + 32} - \frac {5 \operatorname {atan}{\left (x \right )}}{32 x^{4} + 64 x^{2} + 32} \]

[In]

integrate(x*atan(x)/(x**2+1)**3,x)

[Out]

3*x**4*atan(x)/(32*x**4 + 64*x**2 + 32) + 3*x**3/(32*x**4 + 64*x**2 + 32) + 6*x**2*atan(x)/(32*x**4 + 64*x**2
+ 32) + 5*x/(32*x**4 + 64*x**2 + 32) - 5*atan(x)/(32*x**4 + 64*x**2 + 32)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^3} \, dx=\frac {3 \, x^{3} + 5 \, x}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} - \frac {\arctan \left (x\right )}{4 \, {\left (x^{2} + 1\right )}^{2}} + \frac {3}{32} \, \arctan \left (x\right ) \]

[In]

integrate(x*arctan(x)/(x^2+1)^3,x, algorithm="maxima")

[Out]

1/32*(3*x^3 + 5*x)/(x^4 + 2*x^2 + 1) - 1/4*arctan(x)/(x^2 + 1)^2 + 3/32*arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^3} \, dx=\frac {3 \, x^{3} + 5 \, x}{32 \, {\left (x^{2} + 1\right )}^{2}} - \frac {\arctan \left (x\right )}{4 \, {\left (x^{2} + 1\right )}^{2}} + \frac {3}{32} \, \arctan \left (x\right ) \]

[In]

integrate(x*arctan(x)/(x^2+1)^3,x, algorithm="giac")

[Out]

1/32*(3*x^3 + 5*x)/(x^2 + 1)^2 - 1/4*arctan(x)/(x^2 + 1)^2 + 3/32*arctan(x)

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.59 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^3} \, dx=\frac {3\,\mathrm {atan}\left (x\right )}{32}+\frac {\frac {5\,x}{32}-\frac {\mathrm {atan}\left (x\right )}{4}+\frac {3\,x^3}{32}}{{\left (x^2+1\right )}^2} \]

[In]

int((x*atan(x))/(x^2 + 1)^3,x)

[Out]

(3*atan(x))/32 + ((5*x)/32 - atan(x)/4 + (3*x^3)/32)/(x^2 + 1)^2