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

Optimal. Leaf size=55 \[ -\frac{x}{4 a \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{4 a^2} \]

[Out]

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

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Rubi [A]  time = 0.0376937, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5994, 199, 206} \[ -\frac{x}{4 a \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5994

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0542891, size = 66, normalized size = 1.2 \[ \frac{-a^2 x^2 \log (a x+1)+\left (a^2 x^2-1\right ) \log (1-a x)+2 a x+\log (a x+1)-4 \tanh ^{-1}(a x)}{8 a^2 \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*x - 4*ArcTanh[a*x] + (-1 + a^2*x^2)*Log[1 - a*x] + Log[1 + a*x] - a^2*x^2*Log[1 + a*x])/(8*a^2*(-1 + a^2*
x^2))

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Maple [A]  time = 0.034, size = 68, normalized size = 1.2 \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) }{2\,{a}^{2} \left ({a}^{2}{x}^{2}-1 \right ) }}+{\frac{1}{8\,{a}^{2} \left ( ax-1 \right ) }}+{\frac{\ln \left ( ax-1 \right ) }{8\,{a}^{2}}}+{\frac{1}{8\,{a}^{2} \left ( ax+1 \right ) }}-{\frac{\ln \left ( ax+1 \right ) }{8\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.947538, size = 84, normalized size = 1.53 \begin{align*} \frac{\frac{2 \, x}{a^{2} x^{2} - 1} - \frac{\log \left (a x + 1\right )}{a} + \frac{\log \left (a x - 1\right )}{a}}{8 \, a} - \frac{\operatorname{artanh}\left (a x\right )}{2 \,{\left (a^{2} x^{2} - 1\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.9274, size = 96, normalized size = 1.75 \begin{align*} \frac{2 \, a x -{\left (a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{8 \,{\left (a^{4} x^{2} - a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.91601, size = 61, normalized size = 1.11 \begin{align*} \begin{cases} - \frac{a^{2} x^{2} \operatorname{atanh}{\left (a x \right )}}{4 a^{4} x^{2} - 4 a^{2}} + \frac{a x}{4 a^{4} x^{2} - 4 a^{2}} - \frac{\operatorname{atanh}{\left (a x \right )}}{4 a^{4} x^{2} - 4 a^{2}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x**2*atanh(a*x)/(4*a**4*x**2 - 4*a**2) + a*x/(4*a**4*x**2 - 4*a**2) - atanh(a*x)/(4*a**4*x**2
- 4*a**2), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.11388, size = 99, normalized size = 1.8 \begin{align*} \frac{x}{4 \,{\left (a^{2} x^{2} - 1\right )} a} - \frac{\log \left ({\left | a x + 1 \right |}\right )}{8 \, a^{2}} + \frac{\log \left ({\left | a x - 1 \right |}\right )}{8 \, a^{2}} - \frac{\log \left (-\frac{a x + 1}{a x - 1}\right )}{4 \,{\left (a^{2} x^{2} - 1\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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