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

Optimal. Leaf size=75 \[ \frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^3}+\frac {\tanh ^{-1}(a x)^3}{3 a^3}-\frac {\tanh ^{-1}(a x)^2}{a^3}+\frac {2 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^3}-\frac {x \tanh ^{-1}(a x)^2}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {5980, 5910, 5984, 5918, 2402, 2315, 5948} \[ \frac {\text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a^3}+\frac {\tanh ^{-1}(a x)^3}{3 a^3}-\frac {x \tanh ^{-1}(a x)^2}{a^2}-\frac {\tanh ^{-1}(a x)^2}{a^3}+\frac {2 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 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 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 5980

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

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 59, normalized size = 0.79 \[ -\frac {\text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )-\frac {1}{3} \tanh ^{-1}(a x) \left (-3 a x \tanh ^{-1}(a x)+\left (\tanh ^{-1}(a x)+3\right ) \tanh ^{-1}(a x)+6 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )\right )}{a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((-1/3*(ArcTanh[a*x]*(-3*a*x*ArcTanh[a*x] + ArcTanh[a*x]*(3 + ArcTanh[a*x]) + 6*Log[1 + E^(-2*ArcTanh[a*x])])
) + PolyLog[2, -E^(-2*ArcTanh[a*x])])/a^3)

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x^{2} \operatorname {artanh}\left (a x\right )^{2}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-x^2*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 \[ \int -\frac {x^{2} \operatorname {artanh}\left (a x\right )^{2}}{a^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.55, size = 5573, normalized size = 74.31 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [B]  time = 0.33, size = 200, normalized size = 2.67 \[ -\frac {1}{2} \, {\left (\frac {2 \, x}{a^{2}} - \frac {\log \left (a x + 1\right )}{a^{3}} + \frac {\log \left (a x - 1\right )}{a^{3}}\right )} \operatorname {artanh}\left (a x\right )^{2} - \frac {\frac {3 \, {\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right )^{2} - \log \left (a x + 1\right )^{3} + \log \left (a x - 1\right )^{3} - 3 \, {\left (\log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right )\right )} \log \left (a x + 1\right ) + 6 \, \log \left (a x - 1\right )^{2}}{a} - \frac {24 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a}}{24 \, a^{2}} + \frac {{\left (2 \, {\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} - \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right )\right )} \operatorname {artanh}\left (a x\right )}{4 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*x/a^2 - log(a*x + 1)/a^3 + log(a*x - 1)/a^3)*arctanh(a*x)^2 - 1/24*((3*(log(a*x - 1) - 2)*log(a*x + 1)
^2 - log(a*x + 1)^3 + log(a*x - 1)^3 - 3*(log(a*x - 1)^2 - 4*log(a*x - 1))*log(a*x + 1) + 6*log(a*x - 1)^2)/a
- 24*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))/a)/a^2 + 1/4*(2*(log(a*x - 1) - 2)*log(a*x + 1)
- log(a*x + 1)^2 - log(a*x - 1)^2 - 4*log(a*x - 1))*arctanh(a*x)/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-int((x^2*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 \[ - \int \frac {x^{2} \operatorname {atanh}^{2}{\left (a x \right )}}{a^{2} x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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