\(\int \frac {\arctan (x) \log (x)}{x} \, dx\) [34]

Optimal result

Integrand size = 8, antiderivative size = 57 \[ \int \frac {\arctan (x) \log (x)}{x} \, dx=\frac {1}{2} i \log (x) \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \log (x) \operatorname {PolyLog}(2,i x)-\frac {1}{2} i \operatorname {PolyLog}(3,-i x)+\frac {1}{2} i \operatorname {PolyLog}(3,i x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4940, 2438, 5125, 2421, 6724} \[ \int \frac {\arctan (x) \log (x)}{x} \, dx=-\frac {1}{2} i \operatorname {PolyLog}(3,-i x)+\frac {1}{2} i \operatorname {PolyLog}(3,i x)+\frac {1}{2} i \operatorname {PolyLog}(2,-i x) \log (x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x) \log (x) \]

[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 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-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*L
og[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 2438

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 4940

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 5125

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 6724

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*} \text {integral}& = \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) \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \log (x) \operatorname {PolyLog}(2,i x)-\frac {1}{2} i \int \frac {\operatorname {PolyLog}(2,-i x)}{x} \, dx+\frac {1}{2} i \int \frac {\operatorname {PolyLog}(2,i x)}{x} \, dx \\ & = \frac {1}{2} i \log (x) \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \log (x) \operatorname {PolyLog}(2,i x)-\frac {1}{2} i \operatorname {PolyLog}(3,-i x)+\frac {1}{2} i \operatorname {PolyLog}(3,i x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int \frac {\arctan (x) \log (x)}{x} \, dx=\frac {1}{2} i (\log (x) \operatorname {PolyLog}(2,-i x)-\log (x) \operatorname {PolyLog}(2,i x)-\operatorname {PolyLog}(3,-i x)+\operatorname {PolyLog}(3,i x)) \]

[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])

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.25

method	result	size
risch	\(\frac {i \ln \left (x \right )^{2} \ln \left (-i \left (x +i\right )\right )}{4}-\frac {i \ln \left (x \right )^{2} \ln \left (-i x +1\right )}{4}-\frac {i \ln \left (x \right ) \operatorname {Li}_{2}\left (i x \right )}{2}+\frac {i \operatorname {Li}_{3}\left (i x \right )}{2}+\frac {i \ln \left (x \right ) \operatorname {Li}_{2}\left (-i x \right )}{2}-\frac {i \operatorname {Li}_{3}\left (-i x \right )}{2}\)	\(71\)

[In]

int(arctan(x)*ln(x)/x,x,method=_RETURNVERBOSE)

[Out]

1/4*I*ln(x)^2*ln(-I*(x+I))-1/4*I*ln(x)^2*ln(1-I*x)-1/2*I*ln(x)*polylog(2,I*x)+1/2*I*polylog(3,I*x)+1/2*I*ln(x)
*polylog(2,-I*x)-1/2*I*polylog(3,-I*x)

Fricas [F]

\[ \int \frac {\arctan (x) \log (x)}{x} \, dx=\int { \frac {\arctan \left (x\right ) \log \left (x\right )}{x} \,d x } \]

[In]

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

[Out]

integral(arctan(x)*log(x)/x, x)

Sympy [F]

\[ \int \frac {\arctan (x) \log (x)}{x} \, dx=\int \frac {\log {\left (x \right )} \operatorname {atan}{\left (x \right )}}{x}\, dx \]

[In]

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

[Out]

Integral(log(x)*atan(x)/x, x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.54 \[ \int \frac {\arctan (x) \log (x)}{x} \, dx=-\frac {1}{2} i \, {\rm Li}_2\left (i \, x\right ) \log \left (x\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-i \, x\right ) \log \left (x\right ) + \frac {1}{2} i \, {\rm Li}_{3}(i \, x) - \frac {1}{2} i \, {\rm Li}_{3}(-i \, x) \]

[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)

Giac [F]

\[ \int \frac {\arctan (x) \log (x)}{x} \, dx=\int { \frac {\arctan \left (x\right ) \log \left (x\right )}{x} \,d x } \]

[In]

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

[Out]

integrate(arctan(x)*log(x)/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (x) \log (x)}{x} \, dx=\int \frac {\mathrm {atan}\left (x\right )\,\ln \left (x\right )}{x} \,d x \]

[In]

int((atan(x)*log(x))/x,x)

[Out]

int((atan(x)*log(x))/x, x)