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

Optimal. Leaf size=57 \[ -\frac {\tanh ^{-1}(a x)^2}{4 a^3}+\frac {x \tanh ^{-1}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {1}{4 a^3 \left (1-a^2 x^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5998, 5948} \[ -\frac {1}{4 a^3 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^2}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(4*a^3*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(2*a^2*(1 - a^2*x^2)) - ArcTanh[a*x]^2/(4*a^3)

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 5998

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(
q + 1))/(4*c^3*d*(q + 1)^2), x] + (Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x],
x] - Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]))/(2*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q, -5/2]

Rubi steps

\begin {align*} \int \frac {x^2 \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx &=-\frac {1}{4 a^3 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\int \frac {\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{2 a^2}\\ &=-\frac {1}{4 a^3 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^2}{4 a^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 45, normalized size = 0.79 \[ \frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-2 a x \tanh ^{-1}(a x)+1}{4 a^3 \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - 2*a*x*ArcTanh[a*x] + (1 - a^2*x^2)*ArcTanh[a*x]^2)/(4*a^3*(-1 + a^2*x^2))

________________________________________________________________________________________

fricas [A]  time = 0.72, size = 65, normalized size = 1.14 \[ -\frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4}{16 \, {\left (a^{5} x^{2} - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*a*x*log(-(a*x + 1)/(a*x - 1)) + (a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^2 - 4)/(a^5*x^2 - a^3)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.06, size = 169, normalized size = 2.96 \[ -\frac {\arctanh \left (a x \right )}{4 a^{3} \left (a x -1\right )}+\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{4 a^{3}}-\frac {\arctanh \left (a x \right )}{4 a^{3} \left (a x +1\right )}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{4 a^{3}}+\frac {\ln \left (a x +1\right )^{2}}{16 a^{3}}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a^{3}}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{8 a^{3}}+\frac {\ln \left (a x -1\right )^{2}}{16 a^{3}}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8 a^{3}}+\frac {1}{8 a^{3} \left (a x -1\right )}-\frac {1}{8 a^{3} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^3*arctanh(a*x)/(a*x-1)+1/4/a^3*arctanh(a*x)*ln(a*x-1)-1/4/a^3*arctanh(a*x)/(a*x+1)-1/4/a^3*arctanh(a*x)
*ln(a*x+1)+1/16/a^3*ln(a*x+1)^2+1/8/a^3*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/8/a^3*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/
16/a^3*ln(a*x-1)^2-1/8/a^3*ln(a*x-1)*ln(1/2+1/2*a*x)+1/8/a^3/(a*x-1)-1/8/a^3/(a*x+1)

________________________________________________________________________________________

maxima [B]  time = 0.32, size = 126, normalized size = 2.21 \[ -\frac {1}{4} \, {\left (\frac {2 \, x}{a^{4} x^{2} - a^{2}} + \frac {\log \left (a x + 1\right )}{a^{3}} - \frac {\log \left (a x - 1\right )}{a^{3}}\right )} \operatorname {artanh}\left (a x\right ) + \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} + 4\right )} a}{16 \, {\left (a^{6} x^{2} - a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x/(a^4*x^2 - a^2) + log(a*x + 1)/a^3 - log(a*x - 1)/a^3)*arctanh(a*x) + 1/16*((a^2*x^2 - 1)*log(a*x +
1)^2 - 2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) + (a^2*x^2 - 1)*log(a*x - 1)^2 + 4)*a/(a^6*x^2 - a^4)

________________________________________________________________________________________

mupad [B]  time = 0.96, size = 110, normalized size = 1.93 \[ \ln \left (1-a\,x\right )\,\left (\frac {\ln \left (a\,x+1\right )}{8\,a^3}+\frac {x}{2\,a^2\,\left (2\,a^2\,x^2-2\right )}\right )-\frac {{\ln \left (a\,x+1\right )}^2}{16\,a^3}-\frac {{\ln \left (1-a\,x\right )}^2}{16\,a^3}-\frac {1}{2\,a^2\,\left (2\,a-2\,a^3\,x^2\right )}-\frac {x\,\ln \left (a\,x+1\right )}{4\,a^3\,\left (a\,x^2-\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1 - a*x)*(log(a*x + 1)/(8*a^3) + x/(2*a^2*(2*a^2*x^2 - 2))) - log(a*x + 1)^2/(16*a^3) - log(1 - a*x)^2/(16
*a^3) - 1/(2*a^2*(2*a - 2*a^3*x^2)) - (x*log(a*x + 1))/(4*a^3*(a*x^2 - 1/a))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________