3.170 \(\int \frac{(1-a^2 x^2) \tanh ^{-1}(a x)}{x^5} \, dx\)

Optimal. Leaf size=42 \[ -\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}+\frac{a^3}{4 x}-\frac{a}{12 x^3} \]

[Out]

-a/(12*x^3) + a^3/(4*x) - ((1 - a^2*x^2)^2*ArcTanh[a*x])/(4*x^4)

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Rubi [A]  time = 0.0302711, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6008, 14} \[ -\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}+\frac{a^3}{4 x}-\frac{a}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)*ArcTanh[a*x])/x^5,x]

[Out]

-a/(12*x^3) + a^3/(4*x) - ((1 - a^2*x^2)^2*ArcTanh[a*x])/(4*x^4)

Rule 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{x^5} \, dx &=-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}+\frac{1}{4} a \int \frac{1-a^2 x^2}{x^4} \, dx\\ &=-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}+\frac{1}{4} a \int \left (\frac{1}{x^4}-\frac{a^2}{x^2}\right ) \, dx\\ &=-\frac{a}{12 x^3}+\frac{a^3}{4 x}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.0158646, size = 71, normalized size = 1.69 \[ \frac{a^2 \tanh ^{-1}(a x)}{2 x^2}+\frac{a^3}{4 x}+\frac{1}{8} a^4 \log (1-a x)-\frac{1}{8} a^4 \log (a x+1)-\frac{a}{12 x^3}-\frac{\tanh ^{-1}(a x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)*ArcTanh[a*x])/x^5,x]

[Out]

-a/(12*x^3) + a^3/(4*x) - ArcTanh[a*x]/(4*x^4) + (a^2*ArcTanh[a*x])/(2*x^2) + (a^4*Log[1 - a*x])/8 - (a^4*Log[
1 + a*x])/8

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Maple [A]  time = 0.036, size = 59, normalized size = 1.4 \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) }{4\,{x}^{4}}}+{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) }{2\,{x}^{2}}}+{\frac{{a}^{4}\ln \left ( ax-1 \right ) }{8}}+{\frac{{a}^{3}}{4\,x}}-{\frac{a}{12\,{x}^{3}}}-{\frac{{a}^{4}\ln \left ( ax+1 \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)*arctanh(a*x)/x^5,x)

[Out]

-1/4*arctanh(a*x)/x^4+1/2*a^2*arctanh(a*x)/x^2+1/8*a^4*ln(a*x-1)+1/4*a^3/x-1/12*a/x^3-1/8*a^4*ln(a*x+1)

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Maxima [A]  time = 0.956346, size = 82, normalized size = 1.95 \begin{align*} -\frac{1}{24} \,{\left (3 \, a^{3} \log \left (a x + 1\right ) - 3 \, a^{3} \log \left (a x - 1\right ) - \frac{2 \,{\left (3 \, a^{2} x^{2} - 1\right )}}{x^{3}}\right )} a + \frac{{\left (2 \, a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.24379, size = 116, normalized size = 2.76 \begin{align*} \frac{6 \, a^{3} x^{3} - 2 \, a x - 3 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{24 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.23657, size = 46, normalized size = 1.1 \begin{align*} - \frac{a^{4} \operatorname{atanh}{\left (a x \right )}}{4} + \frac{a^{3}}{4 x} + \frac{a^{2} \operatorname{atanh}{\left (a x \right )}}{2 x^{2}} - \frac{a}{12 x^{3}} - \frac{\operatorname{atanh}{\left (a x \right )}}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)*atanh(a*x)/x**5,x)

[Out]

-a**4*atanh(a*x)/4 + a**3/(4*x) + a**2*atanh(a*x)/(2*x**2) - a/(12*x**3) - atanh(a*x)/(4*x**4)

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Giac [B]  time = 1.23577, size = 97, normalized size = 2.31 \begin{align*} -\frac{1}{8} \, a^{4} \log \left ({\left | a x + 1 \right |}\right ) + \frac{1}{8} \, a^{4} \log \left ({\left | a x - 1 \right |}\right ) + \frac{3 \, a^{3} x^{2} - a}{12 \, x^{3}} + \frac{{\left (2 \, a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{8 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^5,x, algorithm="giac")

[Out]

-1/8*a^4*log(abs(a*x + 1)) + 1/8*a^4*log(abs(a*x - 1)) + 1/12*(3*a^3*x^2 - a)/x^3 + 1/8*(2*a^2*x^2 - 1)*log(-(
a*x + 1)/(a*x - 1))/x^4