3.229 \(\int \frac {x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\)

Optimal. Leaf size=54 \[ \frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^2}-\frac {\tanh ^{-1}(a x)^2}{2 a^2}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^2} \]

[Out]

-1/2*arctanh(a*x)^2/a^2+arctanh(a*x)*ln(2/(-a*x+1))/a^2+1/2*polylog(2,1-2/(-a*x+1))/a^2

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5984, 5918, 2402, 2315} \[ \frac {\text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{2 a^2}-\frac {\tanh ^{-1}(a x)^2}{2 a^2}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTanh[a*x])/(1 - a^2*x^2),x]

[Out]

-ArcTanh[a*x]^2/(2*a^2) + (ArcTanh[a*x]*Log[2/(1 - a*x)])/a^2 + PolyLog[2, 1 - 2/(1 - a*x)]/(2*a^2)

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 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 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[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 5984

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 44, normalized size = 0.81 \[ -\frac {\text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x) \left (\tanh ^{-1}(a x)+2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )\right )}{2 a^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*ArcTanh[a*x])/(1 - a^2*x^2),x]

[Out]

-1/2*(-(ArcTanh[a*x]*(ArcTanh[a*x] + 2*Log[1 + E^(-2*ArcTanh[a*x])])) + PolyLog[2, -E^(-2*ArcTanh[a*x])])/a^2

________________________________________________________________________________________

fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x \operatorname {artanh}\left (a x\right )}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(a*x)/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-x*arctanh(a*x)/(a^2*x^2 - 1), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x \operatorname {artanh}\left (a x\right )}{a^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(a*x)/(-a^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-x*arctanh(a*x)/(a^2*x^2 - 1), x)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 125, normalized size = 2.31 \[ -\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2 a^{2}}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2 a^{2}}-\frac {\ln \left (a x -1\right )^{2}}{8 a^{2}}+\frac {\dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{2 a^{2}}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{2}}+\frac {\ln \left (a x +1\right )^{2}}{8 a^{2}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{4 a^{2}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(a*x)/(-a^2*x^2+1),x)

[Out]

-1/2/a^2*arctanh(a*x)*ln(a*x-1)-1/2/a^2*arctanh(a*x)*ln(a*x+1)-1/8/a^2*ln(a*x-1)^2+1/2/a^2*dilog(1/2+1/2*a*x)+
1/4/a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+1/8/a^2*ln(a*x+1)^2-1/4/a^2*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/4/a^2*ln(-1/2*a*x+1
/2)*ln(1/2+1/2*a*x)

________________________________________________________________________________________

maxima [B]  time = 0.32, size = 125, normalized size = 2.31 \[ -\frac {1}{8} \, a {\left (\frac {\log \left (a x + 1\right )^{2} + 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) - \log \left (a x - 1\right )^{2}}{a^{3}} - \frac {4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{3}}\right )} + \frac {{\left (\frac {\log \left (a x + 1\right )}{a} - \frac {\log \left (a x - 1\right )}{a}\right )} \log \left (a^{2} x^{2} - 1\right )}{4 \, a} - \frac {\operatorname {artanh}\left (a x\right ) \log \left (a^{2} x^{2} - 1\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(a*x)/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-1/8*a*((log(a*x + 1)^2 + 2*log(a*x + 1)*log(a*x - 1) - log(a*x - 1)^2)/a^3 - 4*(log(a*x - 1)*log(1/2*a*x + 1/
2) + dilog(-1/2*a*x + 1/2))/a^3) + 1/4*(log(a*x + 1)/a - log(a*x - 1)/a)*log(a^2*x^2 - 1)/a - 1/2*arctanh(a*x)
*log(a^2*x^2 - 1)/a^2

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int \frac {x\,\mathrm {atanh}\left (a\,x\right )}{a^2\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*atanh(a*x))/(a^2*x^2 - 1),x)

[Out]

-int((x*atanh(a*x))/(a^2*x^2 - 1), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x \operatorname {atanh}{\left (a x \right )}}{a^{2} x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(a*x)/(-a**2*x**2+1),x)

[Out]

-Integral(x*atanh(a*x)/(a**2*x**2 - 1), x)

________________________________________________________________________________________