3.674 \(\int \frac{x^3 \tan ^{-1}(x)}{(1+x^2)^2} \, dx\)

Optimal. Leaf size=79 \[ -\frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2}{1+i x}\right )-\frac{x}{4 \left (x^2+1\right )}+\frac{\tan ^{-1}(x)}{2 \left (x^2+1\right )}-\frac{1}{2} i \tan ^{-1}(x)^2-\frac{1}{4} \tan ^{-1}(x)-\log \left (\frac{2}{1+i x}\right ) \tan ^{-1}(x) \]

[Out]

-x/(4*(1 + x^2)) - ArcTan[x]/4 + ArcTan[x]/(2*(1 + x^2)) - (I/2)*ArcTan[x]^2 - ArcTan[x]*Log[2/(1 + I*x)] - (I
/2)*PolyLog[2, 1 - 2/(1 + I*x)]

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Rubi [A]  time = 0.11268, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {4964, 4920, 4854, 2402, 2315, 4930, 199, 203} \[ -\frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2}{1+i x}\right )-\frac{x}{4 \left (x^2+1\right )}+\frac{\tan ^{-1}(x)}{2 \left (x^2+1\right )}-\frac{1}{2} i \tan ^{-1}(x)^2-\frac{1}{4} \tan ^{-1}(x)-\log \left (\frac{2}{1+i x}\right ) \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[x])/(1 + x^2)^2,x]

[Out]

-x/(4*(1 + x^2)) - ArcTan[x]/4 + ArcTan[x]/(2*(1 + x^2)) - (I/2)*ArcTan[x]^2 - ArcTan[x]*Log[2/(1 + I*x)] - (I
/2)*PolyLog[2, 1 - 2/(1 + I*x)]

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int[
x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*Arc
Tan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && IGtQ[m
, 1] && NeQ[p, -1]

Rule 4920

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

Rule 4854

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

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 4930

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]

Rule 199

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

Rule 203

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

Rubi steps

\begin{align*} \int \frac{x^3 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx &=-\int \frac{x \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx+\int \frac{x \tan ^{-1}(x)}{1+x^2} \, dx\\ &=\frac{\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac{1}{2} i \tan ^{-1}(x)^2-\frac{1}{2} \int \frac{1}{\left (1+x^2\right )^2} \, dx-\int \frac{\tan ^{-1}(x)}{i-x} \, dx\\ &=-\frac{x}{4 \left (1+x^2\right )}+\frac{\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac{1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac{2}{1+i x}\right )-\frac{1}{4} \int \frac{1}{1+x^2} \, dx+\int \frac{\log \left (\frac{2}{1+i x}\right )}{1+x^2} \, dx\\ &=-\frac{x}{4 \left (1+x^2\right )}-\frac{1}{4} \tan ^{-1}(x)+\frac{\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac{1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac{2}{1+i x}\right )-i \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i x}\right )\\ &=-\frac{x}{4 \left (1+x^2\right )}-\frac{1}{4} \tan ^{-1}(x)+\frac{\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac{1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac{2}{1+i x}\right )-\frac{1}{2} i \text{Li}_2\left (1-\frac{2}{1+i x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0673323, size = 64, normalized size = 0.81 \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(x)}\right )+\frac{1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (1+e^{2 i \tan ^{-1}(x)}\right )-\frac{1}{8} \sin \left (2 \tan ^{-1}(x)\right )+\frac{1}{4} \tan ^{-1}(x) \cos \left (2 \tan ^{-1}(x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I/2)*ArcTan[x]^2 + (ArcTan[x]*Cos[2*ArcTan[x]])/4 - ArcTan[x]*Log[1 + E^((2*I)*ArcTan[x])] + (I/2)*PolyLog[2,
-E^((2*I)*ArcTan[x])] - Sin[2*ArcTan[x]]/8

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Maple [B]  time = 0.013, size = 139, normalized size = 1.8 \begin{align*}{\frac{\arctan \left ( x \right ) \ln \left ({x}^{2}+1 \right ) }{2}}+{\frac{\arctan \left ( x \right ) }{2\,{x}^{2}+2}}-{\frac{x}{4\,{x}^{2}+4}}-{\frac{\arctan \left ( x \right ) }{4}}-{\frac{i}{4}}\ln \left ( -{\frac{i}{2}} \left ( x+i \right ) \right ) \ln \left ( x-i \right ) -{\frac{i}{8}} \left ( \ln \left ( x-i \right ) \right ) ^{2}+{\frac{i}{4}}\ln \left ( x-i \right ) \ln \left ({x}^{2}+1 \right ) -{\frac{i}{4}}{\it dilog} \left ( -{\frac{i}{2}} \left ( x+i \right ) \right ) +{\frac{i}{4}}\ln \left ({\frac{i}{2}} \left ( x-i \right ) \right ) \ln \left ( x+i \right ) +{\frac{i}{8}} \left ( \ln \left ( x+i \right ) \right ) ^{2}-{\frac{i}{4}}\ln \left ( x+i \right ) \ln \left ({x}^{2}+1 \right ) +{\frac{i}{4}}{\it dilog} \left ({\frac{i}{2}} \left ( x-i \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(x)*ln(x^2+1)+1/2*arctan(x)/(x^2+1)-1/4*x/(x^2+1)-1/4*arctan(x)-1/4*I*ln(-1/2*I*(x+I))*ln(x-I)-1/8*I
*ln(x-I)^2+1/4*I*ln(x-I)*ln(x^2+1)-1/4*I*dilog(-1/2*I*(x+I))+1/4*I*ln(1/2*I*(x-I))*ln(x+I)+1/8*I*ln(x+I)^2-1/4
*I*ln(x+I)*ln(x^2+1)+1/4*I*dilog(1/2*I*(x-I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \arctan \left (x\right )}{{\left (x^{2} + 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arctan(x)/(x^2 + 1)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{3} \arctan \left (x\right )}{x^{4} + 2 \, x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*arctan(x)/(x^4 + 2*x^2 + 1), x)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RecursionError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RecursionError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \arctan \left (x\right )}{{\left (x^{2} + 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arctan(x)/(x^2 + 1)^2, x)