3.304 \(\int \frac{x \tanh ^{-1}(a x)}{(1-a^2 x^2)^3} \, dx\)

Optimal. Leaf size=75 \[ -\frac{3 x}{32 a \left (1-a^2 x^2\right )}-\frac{x}{16 a \left (1-a^2 x^2\right )^2}+\frac{\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)}{32 a^2} \]

[Out]

-x/(16*a*(1 - a^2*x^2)^2) - (3*x)/(32*a*(1 - a^2*x^2)) - (3*ArcTanh[a*x])/(32*a^2) + ArcTanh[a*x]/(4*a^2*(1 -
a^2*x^2)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0469405, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5994, 199, 206} \[ -\frac{3 x}{32 a \left (1-a^2 x^2\right )}-\frac{x}{16 a \left (1-a^2 x^2\right )^2}+\frac{\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)}{32 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTanh[a*x])/(1 - a^2*x^2)^3,x]

[Out]

-x/(16*a*(1 - a^2*x^2)^2) - (3*x)/(32*a*(1 - a^2*x^2)) - (3*ArcTanh[a*x])/(32*a^2) + ArcTanh[a*x]/(4*a^2*(1 -
a^2*x^2)^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 )^3} \, dx &=\frac{\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{\int \frac{1}{\left (1-a^2 x^2\right )^3} \, dx}{4 a}\\ &=-\frac{x}{16 a \left (1-a^2 x^2\right )^2}+\frac{\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{3 \int \frac{1}{\left (1-a^2 x^2\right )^2} \, dx}{16 a}\\ &=-\frac{x}{16 a \left (1-a^2 x^2\right )^2}-\frac{3 x}{32 a \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{3 \int \frac{1}{1-a^2 x^2} \, dx}{32 a}\\ &=-\frac{x}{16 a \left (1-a^2 x^2\right )^2}-\frac{3 x}{32 a \left (1-a^2 x^2\right )}-\frac{3 \tanh ^{-1}(a x)}{32 a^2}+\frac{\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.0566228, size = 88, normalized size = 1.17 \[ \frac{3 x}{32 a \left (a^2 x^2-1\right )}-\frac{x}{16 a \left (a^2 x^2-1\right )^2}+\frac{\tanh ^{-1}(a x)}{4 a^2 \left (a^2 x^2-1\right )^2}+\frac{3 \log (1-a x)}{64 a^2}-\frac{3 \log (a x+1)}{64 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcTanh[a*x])/(1 - a^2*x^2)^3,x]

[Out]

-x/(16*a*(-1 + a^2*x^2)^2) + (3*x)/(32*a*(-1 + a^2*x^2)) + ArcTanh[a*x]/(4*a^2*(-1 + a^2*x^2)^2) + (3*Log[1 -
a*x])/(64*a^2) - (3*Log[1 + a*x])/(64*a^2)

________________________________________________________________________________________

Maple [A]  time = 0.035, size = 92, normalized size = 1.2 \begin{align*}{\frac{{\it Artanh} \left ( ax \right ) }{4\,{a}^{2} \left ({a}^{2}{x}^{2}-1 \right ) ^{2}}}-{\frac{1}{64\,{a}^{2} \left ( ax-1 \right ) ^{2}}}+{\frac{3}{64\,{a}^{2} \left ( ax-1 \right ) }}+{\frac{3\,\ln \left ( ax-1 \right ) }{64\,{a}^{2}}}+{\frac{1}{64\,{a}^{2} \left ( ax+1 \right ) ^{2}}}+{\frac{3}{64\,{a}^{2} \left ( ax+1 \right ) }}-{\frac{3\,\ln \left ( ax+1 \right ) }{64\,{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)^3,x)

[Out]

1/4/a^2/(a^2*x^2-1)^2*arctanh(a*x)-1/64/a^2/(a*x-1)^2+3/64/a^2/(a*x-1)+3/64/a^2*ln(a*x-1)+1/64/a^2/(a*x+1)^2+3
/64/a^2/(a*x+1)-3/64/a^2*ln(a*x+1)

________________________________________________________________________________________

Maxima [A]  time = 0.968843, size = 111, normalized size = 1.48 \begin{align*} \frac{\frac{2 \,{\left (3 \, a^{2} x^{3} - 5 \, x\right )}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - \frac{3 \, \log \left (a x + 1\right )}{a} + \frac{3 \, \log \left (a x - 1\right )}{a}}{64 \, a} + \frac{\operatorname{artanh}\left (a x\right )}{4 \,{\left (a^{2} x^{2} - 1\right )}^{2} 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)^3,x, algorithm="maxima")

[Out]

1/64*(2*(3*a^2*x^3 - 5*x)/(a^4*x^4 - 2*a^2*x^2 + 1) - 3*log(a*x + 1)/a + 3*log(a*x - 1)/a)/a + 1/4*arctanh(a*x
)/((a^2*x^2 - 1)^2*a^2)

________________________________________________________________________________________

Fricas [A]  time = 1.99264, size = 150, normalized size = 2. \begin{align*} \frac{6 \, a^{3} x^{3} - 10 \, a x -{\left (3 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 5\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{64 \,{\left (a^{6} x^{4} - 2 \, 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)^3,x, algorithm="fricas")

[Out]

1/64*(6*a^3*x^3 - 10*a*x - (3*a^4*x^4 - 6*a^2*x^2 - 5)*log(-(a*x + 1)/(a*x - 1)))/(a^6*x^4 - 2*a^4*x^2 + a^2)

________________________________________________________________________________________

Sympy [A]  time = 4.02812, size = 158, normalized size = 2.11 \begin{align*} \begin{cases} - \frac{3 a^{4} x^{4} \operatorname{atanh}{\left (a x \right )}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} + \frac{3 a^{3} x^{3}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} + \frac{6 a^{2} x^{2} \operatorname{atanh}{\left (a x \right )}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} - \frac{5 a x}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} + \frac{5 \operatorname{atanh}{\left (a x \right )}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 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)**3,x)

[Out]

Piecewise((-3*a**4*x**4*atanh(a*x)/(32*a**6*x**4 - 64*a**4*x**2 + 32*a**2) + 3*a**3*x**3/(32*a**6*x**4 - 64*a*
*4*x**2 + 32*a**2) + 6*a**2*x**2*atanh(a*x)/(32*a**6*x**4 - 64*a**4*x**2 + 32*a**2) - 5*a*x/(32*a**6*x**4 - 64
*a**4*x**2 + 32*a**2) + 5*atanh(a*x)/(32*a**6*x**4 - 64*a**4*x**2 + 32*a**2), Ne(a, 0)), (0, True))

________________________________________________________________________________________

Giac [A]  time = 1.18839, size = 113, normalized size = 1.51 \begin{align*} -\frac{3 \, \log \left ({\left | a x + 1 \right |}\right )}{64 \, a^{2}} + \frac{3 \, \log \left ({\left | a x - 1 \right |}\right )}{64 \, a^{2}} + \frac{3 \, a^{2} x^{3} - 5 \, x}{32 \,{\left (a^{2} x^{2} - 1\right )}^{2} a} + \frac{\log \left (-\frac{a x + 1}{a x - 1}\right )}{8 \,{\left (a^{2} x^{2} - 1\right )}^{2} 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)^3,x, algorithm="giac")

[Out]

-3/64*log(abs(a*x + 1))/a^2 + 3/64*log(abs(a*x - 1))/a^2 + 1/32*(3*a^2*x^3 - 5*x)/((a^2*x^2 - 1)^2*a) + 1/8*lo
g(-(a*x + 1)/(a*x - 1))/((a^2*x^2 - 1)^2*a^2)