3.34 \(\int \frac{\tan ^{-1}(x) \log (x)}{x} \, dx\)

Optimal. Leaf size=57 \[ -\frac{1}{2} i \text{PolyLog}(3,-i x)+\frac{1}{2} i \text{PolyLog}(3,i x)+\frac{1}{2} i \log (x) \text{PolyLog}(2,-i x)-\frac{1}{2} i \log (x) \text{PolyLog}(2,i x) \]

[Out]

(I/2)*Log[x]*PolyLog[2, (-I)*x] - (I/2)*Log[x]*PolyLog[2, I*x] - (I/2)*PolyLog[3, (-I)*x] + (I/2)*PolyLog[3, I
*x]

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Rubi [A]  time = 0.075564, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4848, 2391, 5005, 2374, 6589} \[ -\frac{1}{2} i \text{PolyLog}(3,-i x)+\frac{1}{2} i \text{PolyLog}(3,i x)+\frac{1}{2} i \log (x) \text{PolyLog}(2,-i x)-\frac{1}{2} i \log (x) \text{PolyLog}(2,i x) \]

Antiderivative was successfully verified.

[In]

Int[(ArcTan[x]*Log[x])/x,x]

[Out]

(I/2)*Log[x]*PolyLog[2, (-I)*x] - (I/2)*Log[x]*PolyLog[2, I*x] - (I/2)*PolyLog[3, (-I)*x] + (I/2)*PolyLog[3, I
*x]

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]

Rule 5005

Int[(ArcTan[(c_.)*(x_)^(n_.)]*Log[(d_.)*(x_)^(m_.)])/(x_), x_Symbol] :> Dist[I/2, Int[(Log[d*x^m]*Log[1 - I*c*
x^n])/x, x], x] - Dist[I/2, Int[(Log[d*x^m]*Log[1 + I*c*x^n])/x, x], x] /; FreeQ[{c, d, m, n}, x]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(x) \log (x)}{x} \, dx &=\frac{1}{2} i \int \frac{\log (1-i x) \log (x)}{x} \, dx-\frac{1}{2} i \int \frac{\log (1+i x) \log (x)}{x} \, dx\\ &=\frac{1}{2} i \log (x) \text{Li}_2(-i x)-\frac{1}{2} i \log (x) \text{Li}_2(i x)-\frac{1}{2} i \int \frac{\text{Li}_2(-i x)}{x} \, dx+\frac{1}{2} i \int \frac{\text{Li}_2(i x)}{x} \, dx\\ &=\frac{1}{2} i \log (x) \text{Li}_2(-i x)-\frac{1}{2} i \log (x) \text{Li}_2(i x)-\frac{1}{2} i \text{Li}_3(-i x)+\frac{1}{2} i \text{Li}_3(i x)\\ \end{align*}

Mathematica [A]  time = 0.0536139, size = 44, normalized size = 0.77 \[ \frac{1}{2} i (-\text{PolyLog}(3,-i x)+\text{PolyLog}(3,i x)+\log (x) \text{PolyLog}(2,-i x)-\log (x) \text{PolyLog}(2,i x)) \]

Antiderivative was successfully verified.

[In]

Integrate[(ArcTan[x]*Log[x])/x,x]

[Out]

(I/2)*(Log[x]*PolyLog[2, (-I)*x] - Log[x]*PolyLog[2, I*x] - PolyLog[3, (-I)*x] + PolyLog[3, I*x])

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Maple [F]  time = 0.35, size = 0, normalized size = 0. \begin{align*} \int{\frac{\arctan \left ( x \right ) \ln \left ( x \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x)*ln(x)/x,x)

[Out]

int(arctan(x)*ln(x)/x,x)

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Maxima [A]  time = 1.62968, size = 42, normalized size = 0.74 \begin{align*} -\frac{1}{2} i \,{\rm Li}_2\left (i \, x\right ) \log \left (x\right ) + \frac{1}{2} i \,{\rm Li}_2\left (-i \, x\right ) \log \left (x\right ) + \frac{1}{2} i \,{\rm Li}_{3}(i \, x) - \frac{1}{2} i \,{\rm Li}_{3}(-i \, x) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*dilog(I*x)*log(x) + 1/2*I*dilog(-I*x)*log(x) + 1/2*I*polylog(3, I*x) - 1/2*I*polylog(3, -I*x)

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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\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(x)*log(x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(x)*ln(x)/x,x)

[Out]

Integral(log(x)*atan(x)/x, 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\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctan(x)*log(x)/x, x)