3.40 \(\int \frac{x \cot ^{-1}(x)}{1+x^2} \, dx\)

Optimal. Leaf size=48 \[ \frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2}{1+i x}\right )+\frac{1}{2} i \cot ^{-1}(x)^2-\log \left (\frac{2}{1+i x}\right ) \cot ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0540192, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {4921, 4855, 2402, 2315} \[ \frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2}{1+i x}\right )+\frac{1}{2} i \cot ^{-1}(x)^2-\log \left (\frac{2}{1+i x}\right ) \cot ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcCot[x])/(1 + x^2),x]

[Out]

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

Rule 4921

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

Rule 4855

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCot[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] - Dist[(b*c*p)/e, Int[((a + b*ArcCot[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]

Rubi steps

\begin{align*} \int \frac{x \cot ^{-1}(x)}{1+x^2} \, dx &=\frac{1}{2} i \cot ^{-1}(x)^2-\int \frac{\cot ^{-1}(x)}{i-x} \, dx\\ &=\frac{1}{2} i \cot ^{-1}(x)^2-\cot ^{-1}(x) \log \left (\frac{2}{1+i x}\right )-\int \frac{\log \left (\frac{2}{1+i x}\right )}{1+x^2} \, dx\\ &=\frac{1}{2} i \cot ^{-1}(x)^2-\cot ^{-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{1}{2} i \cot ^{-1}(x)^2-\cot ^{-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 [B]  time = 0.0472643, size = 221, normalized size = 4.6 \[ \frac{1}{4} i \text{PolyLog}\left (2,-\frac{1}{2} i (-x+i)\right )-\frac{1}{4} i \text{PolyLog}\left (2,-\frac{1}{2} i (x+i)\right )-\frac{1}{8} i \log ^2(-x+i)+\frac{1}{8} i \log ^2(x+i)+\frac{1}{4} i \log \left (-\frac{-x+i}{x}\right ) \log (-x+i)+\frac{1}{4} i \log \left (-\frac{1}{2} i (x+i)\right ) \log (-x+i)-\frac{1}{4} i \log \left (\frac{x+i}{x}\right ) \log (-x+i)-\frac{1}{4} i \log \left (-\frac{1}{2} i (-x+i)\right ) \log (x+i)+\frac{1}{4} i \log \left (-\frac{-x+i}{x}\right ) \log (x+i)-\frac{1}{4} i \log (x+i) \log \left (\frac{x+i}{x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*ArcCot[x])/(1 + x^2),x]

[Out]

(-I/8)*Log[I - x]^2 + (I/4)*Log[I - x]*Log[-((I - x)/x)] + (I/4)*Log[I - x]*Log[(-I/2)*(I + x)] - (I/4)*Log[(-
I/2)*(I - x)]*Log[I + x] + (I/4)*Log[-((I - x)/x)]*Log[I + x] + (I/8)*Log[I + x]^2 - (I/4)*Log[I - x]*Log[(I +
x)/x] - (I/4)*Log[I + x]*Log[(I + x)/x] + (I/4)*PolyLog[2, (-I/2)*(I - x)] - (I/4)*PolyLog[2, (-I/2)*(I + x)]

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Maple [B]  time = 0.104, size = 114, normalized size = 2.4 \begin{align*}{\frac{{\rm arccot} \left (x\right )\ln \left ({x}^{2}+1 \right ) }{2}}-{\frac{i}{4}}\ln \left ( x-i \right ) \ln \left ({x}^{2}+1 \right ) +{\frac{i}{8}} \left ( \ln \left ( x-i \right ) \right ) ^{2}+{\frac{i}{4}}\ln \left ( x-i \right ) \ln \left ( -{\frac{i}{2}} \left ( x+i \right ) \right ) +{\frac{i}{4}}{\it dilog} \left ( -{\frac{i}{2}} \left ( x+i \right ) \right ) +{\frac{i}{4}}\ln \left ( x+i \right ) \ln \left ({x}^{2}+1 \right ) -{\frac{i}{8}} \left ( \ln \left ( x+i \right ) \right ) ^{2}-{\frac{i}{4}}\ln \left ( x+i \right ) \ln \left ({\frac{i}{2}} \left ( x-i \right ) \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*arccot(x)/(x^2+1),x)

[Out]

1/2*arccot(x)*ln(x^2+1)-1/4*I*ln(x-I)*ln(x^2+1)+1/8*I*ln(x-I)^2+1/4*I*ln(x-I)*ln(-1/2*I*(x+I))+1/4*I*dilog(-1/
2*I*(x+I))+1/4*I*ln(x+I)*ln(x^2+1)-1/8*I*ln(x+I)^2-1/4*I*ln(x+I)*ln(1/2*I*(x-I))-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 \operatorname{arccot}\left (x\right )}{x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccot(x)/(x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*arccot(x)/(x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*acot(x)/(x**2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccot(x)/(x^2 + 1), x)