3.3.29 \(\int \frac {x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\) [229]

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

[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

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Rubi [A]
time = 0.05, 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 = {6131, 6055, 2449, 2352} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*ArcTanh[a*x])/(1 - a^2*x^2),x]

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

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 6055

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 6131

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 {\text {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*}

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Mathematica [A]
time = 0.04, size = 44, normalized size = 0.81 \begin {gather*} -\frac {-\tanh ^{-1}(a x) \left (\tanh ^{-1}(a x)+2 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [A]
time = 0.27, size = 99, normalized size = 1.83

method	result	size
risch	\(\frac {\ln \left (-a x +1\right )^{2}}{8 a^{2}}+\frac {\ln \left (\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (-a x +1\right )}{4 a^{2}}-\frac {\dilog \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4 a^{2}}-\frac {\ln \left (a x +1\right )^{2}}{8 a^{2}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{4 a^{2}}+\frac {\dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{4 a^{2}}\)	\(90\)
derivativedivides	\(\frac {-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2}+\frac {\dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x -1\right )^{2}}{8}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (a x +1\right )^{2}}{8}}{a^{2}}\)	\(99\)
default	\(\frac {-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2}+\frac {\dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x -1\right )^{2}}{8}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (a x +1\right )^{2}}{8}}{a^{2}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (47) = 94\).
time = 0.25, size = 125, normalized size = 2.31 \begin {gather*} -\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}} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

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)

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