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

Optimal. Leaf size=42 \[ \frac {\tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \tanh ^{-1}(a x)}{a^2}-\frac {\log \left (1-a^2 x^2\right )}{2 a^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5980, 5910, 260, 5948} \[ -\frac {\log \left (1-a^2 x^2\right )}{2 a^3}+\frac {\tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \tanh ^{-1}(a x)}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 42, normalized size = 1.00 \[ \frac {\tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \tanh ^{-1}(a x)}{a^2}-\frac {\log \left (1-a^2 x^2\right )}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 56, normalized size = 1.33 \[ -\frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, \log \left (a^{2} x^{2} - 1\right )}{8 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.05, size = 145, normalized size = 3.45 \[ -\frac {x \arctanh \left (a x \right )}{a^{2}}-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2 a^{3}}+\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2 a^{3}}-\frac {\ln \left (a x -1\right )^{2}}{8 a^{3}}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{3}}-\frac {\ln \left (a x -1\right )}{2 a^{3}}-\frac {\ln \left (a x +1\right )}{2 a^{3}}-\frac {\ln \left (a x +1\right )^{2}}{8 a^{3}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{4 a^{3}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.32, size = 85, normalized size = 2.02 \[ -\frac {1}{2} \, {\left (\frac {2 \, x}{a^{2}} - \frac {\log \left (a x + 1\right )}{a^{3}} + \frac {\log \left (a x - 1\right )}{a^{3}}\right )} \operatorname {artanh}\left (a x\right ) + \frac {2 \, {\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} - \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right )}{8 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

mupad [B]  time = 0.96, size = 82, normalized size = 1.95 \[ \frac {{\ln \left (a\,x+1\right )}^2}{8\,a^3}-\ln \left (1-a\,x\right )\,\left (\frac {\ln \left (a\,x+1\right )}{4\,a^3}-\frac {x}{2\,a^2}\right )+\frac {{\ln \left (1-a\,x\right )}^2}{8\,a^3}-\frac {\ln \left (a^2\,x^2-1\right )}{2\,a^3}-\frac {x\,\ln \left (a\,x+1\right )}{2\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.99, size = 41, normalized size = 0.98 \[ \begin {cases} - \frac {x \operatorname {atanh}{\left (a x \right )}}{a^{2}} - \frac {\log {\left (x - \frac {1}{a} \right )}}{a^{3}} + \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{2 a^{3}} - \frac {\operatorname {atanh}{\left (a x \right )}}{a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x*atanh(a*x)/a**2 - log(x - 1/a)/a**3 + atanh(a*x)**2/(2*a**3) - atanh(a*x)/a**3, Ne(a, 0)), (0, T
rue))

________________________________________________________________________________________