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

Optimal. Leaf size=78 \[ -\frac {\text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^2}+\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^2}-\frac {\tanh ^{-1}(a x)^3}{3 a^2}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^2} \]

[Out]

-1/3*arctanh(a*x)^3/a^2+arctanh(a*x)^2*ln(2/(-a*x+1))/a^2+arctanh(a*x)*polylog(2,1-2/(-a*x+1))/a^2-1/2*polylog
(3,1-2/(-a*x+1))/a^2

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Rubi [A]  time = 0.16, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5984, 5918, 5948, 6058, 6610} \[ -\frac {\text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{2 a^2}+\frac {\tanh ^{-1}(a x) \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a^2}-\frac {\tanh ^{-1}(a x)^3}{3 a^2}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[a*x]^3/(3*a^2) + (ArcTanh[a*x]^2*Log[2/(1 - a*x)])/a^2 + (ArcTanh[a*x]*PolyLog[2, 1 - 2/(1 - a*x)])/a
^2 - PolyLog[3, 1 - 2/(1 - a*x)]/(2*a^2)

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 5948

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

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]

Rule 6058

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {x \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx &=-\frac {\tanh ^{-1}(a x)^3}{3 a^2}+\frac {\int \frac {\tanh ^{-1}(a x)^2}{1-a x} \, dx}{a}\\ &=-\frac {\tanh ^{-1}(a x)^3}{3 a^2}+\frac {\tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{a^2}-\frac {2 \int \frac {\tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a}\\ &=-\frac {\tanh ^{-1}(a x)^3}{3 a^2}+\frac {\tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{a^2}+\frac {\tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2}-\frac {\int \frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a}\\ &=-\frac {\tanh ^{-1}(a x)^3}{3 a^2}+\frac {\tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{a^2}+\frac {\tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2}-\frac {\text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 68, normalized size = 0.87 \[ -\frac {\tanh ^{-1}(a x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{-2 \tanh ^{-1}(a x)}\right )-\frac {1}{3} \tanh ^{-1}(a x)^3-\tanh ^{-1}(a x)^2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((-1/3*ArcTanh[a*x]^3 - ArcTanh[a*x]^2*Log[1 + E^(-2*ArcTanh[a*x])] + ArcTanh[a*x]*PolyLog[2, -E^(-2*ArcTanh[
a*x])] + PolyLog[3, -E^(-2*ArcTanh[a*x])]/2)/a^2)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x \operatorname {artanh}\left (a x\right )^{2}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.52, size = 741, normalized size = 9.50 \[ -\frac {\arctanh \left (a x \right )^{2} \ln \left (a x -1\right )}{2 a^{2}}-\frac {\arctanh \left (a x \right )^{2} \ln \left (a x +1\right )}{2 a^{2}}+\frac {\arctanh \left (a x \right )^{2} \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2}}-\frac {\arctanh \left (a x \right )^{3}}{3 a^{2}}+\frac {\arctanh \left (a x \right ) \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{2}}-\frac {\polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 a^{2}}+\frac {i \arctanh \left (a x \right )^{2} \pi }{2 a^{2}}-\frac {i \arctanh \left (a x \right )^{2} \pi \,\mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{\left (a^{2} x^{2}-1\right ) \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}\right )}{4 a^{2}}+\frac {i \arctanh \left (a x \right )^{2} \pi \,\mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{\left (a^{2} x^{2}-1\right ) \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}\right )^{2}}{4 a^{2}}+\frac {i \arctanh \left (a x \right )^{2} \pi \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right )^{2} \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2}}+\frac {i \arctanh \left (a x \right )^{2} \pi \,\mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )^{2}}{4 a^{2}}-\frac {i \arctanh \left (a x \right )^{2} \pi \,\mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{\left (a^{2} x^{2}-1\right ) \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}\right )^{2}}{4 a^{2}}+\frac {i \arctanh \left (a x \right )^{2} \pi \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right )^{3}}{4 a^{2}}-\frac {i \arctanh \left (a x \right )^{2} \pi \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{2}}{2 a^{2}}+\frac {i \arctanh \left (a x \right )^{2} \pi \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{3}}{2 a^{2}}+\frac {i \arctanh \left (a x \right )^{2} \pi \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{\left (a^{2} x^{2}-1\right ) \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}\right )^{3}}{4 a^{2}}+\frac {\arctanh \left (a x \right )^{2} \ln \relax (2)}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {3 \, \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{2} + \log \left (-a x + 1\right )^{3}}{24 \, a^{2}} + \frac {1}{4} \, \int -\frac {a x \log \left (a x + 1\right )^{2} - {\left (3 \, a x + 1\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{3} x^{2} - a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*log(a*x + 1)*log(-a*x + 1)^2 + log(-a*x + 1)^3)/a^2 + 1/4*integrate(-(a*x*log(a*x + 1)^2 - (3*a*x + 1
)*log(a*x + 1)*log(-a*x + 1))/(a^3*x^2 - a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {x\,{\mathrm {atanh}\left (a\,x\right )}^2}{a^2\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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