3.1282 \(\int \frac{\tan ^{-1}(x) \log (1+x^2)}{x^3} \, dx\)

Optimal. Leaf size=69 \[ \frac{1}{2} i \text{PolyLog}(2,-i x)-\frac{1}{2} i \text{PolyLog}(2,i x)-\frac{\log \left (x^2+1\right )}{2 x}-\frac{\log \left (x^2+1\right ) \tan ^{-1}(x)}{2 x^2}-\frac{1}{2} \log \left (x^2+1\right ) \tan ^{-1}(x)+\tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0752657, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4852, 325, 203, 5021, 4848, 2391} \[ \frac{1}{2} i \text{PolyLog}(2,-i x)-\frac{1}{2} i \text{PolyLog}(2,i x)-\frac{\log \left (x^2+1\right )}{2 x}-\frac{\log \left (x^2+1\right ) \tan ^{-1}(x)}{2 x^2}-\frac{1}{2} \log \left (x^2+1\right ) \tan ^{-1}(x)+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(ArcTan[x]*Log[1 + x^2])/x^3,x]

[Out]

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

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 325

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

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])

Rule 5021

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> With
[{u = IntHide[x^m*(a + b*ArcTan[c*x]), x]}, Dist[d + e*Log[f + g*x^2], u, x] - Dist[2*e*g, Int[ExpandIntegrand
[(x*u)/(f + g*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[m] && NeQ[m, -1]

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

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

Rubi steps

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

Mathematica [A]  time = 0.0286804, size = 49, normalized size = 0.71 \[ \frac{1}{2} i (\text{PolyLog}(2,-i x)-\text{PolyLog}(2,i x))-\frac{\log \left (x^2+1\right ) \left (x^2 \tan ^{-1}(x)+x+\tan ^{-1}(x)\right )}{2 x^2}+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 2.244, size = 0, normalized size = 0. \begin{align*} \int{\frac{\arctan \left ( x \right ) \ln \left ({x}^{2}+1 \right ) }{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.63935, size = 95, normalized size = 1.38 \begin{align*} \frac{4 \, x^{2} \arctan \left (x\right ) \log \left (x\right ) + 4 \, x^{2} \arctan \left (x\right ) - 2 i \, x^{2}{\rm Li}_2\left (i \, x + 1\right ) + 2 i \, x^{2}{\rm Li}_2\left (-i \, x + 1\right ) -{\left (\pi x^{2} + 2 \,{\left (x^{2} + 1\right )} \arctan \left (x\right ) + 2 \, x\right )} \log \left (x^{2} + 1\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*x^2*arctan(x)*log(x) + 4*x^2*arctan(x) - 2*I*x^2*dilog(I*x + 1) + 2*I*x^2*dilog(-I*x + 1) - (pi*x^2 + 2
*(x^2 + 1)*arctan(x) + 2*x)*log(x^2 + 1))/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log{\left (x^{2} + 1 \right )} \operatorname{atan}{\left (x \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(x**2 + 1)*atan(x)/x**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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