3.171 \(\int \frac{(1-a^2 x^2) \tanh ^{-1}(a x)}{x^6} \, dx\)

Optimal. Leaf size=71 \[ \frac{a^3}{15 x^2}+\frac{1}{15} a^5 \log \left (1-a^2 x^2\right )+\frac{a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac{2}{15} a^5 \log (x)-\frac{a}{20 x^4}-\frac{\tanh ^{-1}(a x)}{5 x^5} \]

[Out]

-a/(20*x^4) + a^3/(15*x^2) - ArcTanh[a*x]/(5*x^5) + (a^2*ArcTanh[a*x])/(3*x^3) - (2*a^5*Log[x])/15 + (a^5*Log[
1 - a^2*x^2])/15

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Rubi [A]  time = 0.0876415, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {6014, 5916, 266, 44} \[ \frac{a^3}{15 x^2}+\frac{1}{15} a^5 \log \left (1-a^2 x^2\right )+\frac{a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac{2}{15} a^5 \log (x)-\frac{a}{20 x^4}-\frac{\tanh ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)*ArcTanh[a*x])/x^6,x]

[Out]

-a/(20*x^4) + a^3/(15*x^2) - ArcTanh[a*x]/(5*x^5) + (a^2*ArcTanh[a*x])/(3*x^3) - (2*a^5*Log[x])/15 + (a^5*Log[
1 - a^2*x^2])/15

Rule 6014

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

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{x^6} \, dx &=-\left (a^2 \int \frac{\tanh ^{-1}(a x)}{x^4} \, dx\right )+\int \frac{\tanh ^{-1}(a x)}{x^6} \, dx\\ &=-\frac{\tanh ^{-1}(a x)}{5 x^5}+\frac{a^2 \tanh ^{-1}(a x)}{3 x^3}+\frac{1}{5} a \int \frac{1}{x^5 \left (1-a^2 x^2\right )} \, dx-\frac{1}{3} a^3 \int \frac{1}{x^3 \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{\tanh ^{-1}(a x)}{5 x^5}+\frac{a^2 \tanh ^{-1}(a x)}{3 x^3}+\frac{1}{10} a \operatorname{Subst}\left (\int \frac{1}{x^3 \left (1-a^2 x\right )} \, dx,x,x^2\right )-\frac{1}{6} a^3 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{\tanh ^{-1}(a x)}{5 x^5}+\frac{a^2 \tanh ^{-1}(a x)}{3 x^3}+\frac{1}{10} a \operatorname{Subst}\left (\int \left (\frac{1}{x^3}+\frac{a^2}{x^2}+\frac{a^4}{x}-\frac{a^6}{-1+a^2 x}\right ) \, dx,x,x^2\right )-\frac{1}{6} a^3 \operatorname{Subst}\left (\int \left (\frac{1}{x^2}+\frac{a^2}{x}-\frac{a^4}{-1+a^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac{a}{20 x^4}+\frac{a^3}{15 x^2}-\frac{\tanh ^{-1}(a x)}{5 x^5}+\frac{a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac{2}{15} a^5 \log (x)+\frac{1}{15} a^5 \log \left (1-a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0160814, size = 71, normalized size = 1. \[ \frac{a^3}{15 x^2}+\frac{1}{15} a^5 \log \left (1-a^2 x^2\right )+\frac{a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac{2}{15} a^5 \log (x)-\frac{a}{20 x^4}-\frac{\tanh ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)*ArcTanh[a*x])/x^6,x]

[Out]

-a/(20*x^4) + a^3/(15*x^2) - ArcTanh[a*x]/(5*x^5) + (a^2*ArcTanh[a*x])/(3*x^3) - (2*a^5*Log[x])/15 + (a^5*Log[
1 - a^2*x^2])/15

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Maple [A]  time = 0.039, size = 68, normalized size = 1. \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) }{5\,{x}^{5}}}+{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) }{3\,{x}^{3}}}+{\frac{{a}^{5}\ln \left ( ax-1 \right ) }{15}}-{\frac{a}{20\,{x}^{4}}}+{\frac{{a}^{3}}{15\,{x}^{2}}}-{\frac{2\,{a}^{5}\ln \left ( ax \right ) }{15}}+{\frac{{a}^{5}\ln \left ( ax+1 \right ) }{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)*arctanh(a*x)/x^6,x)

[Out]

-1/5*arctanh(a*x)/x^5+1/3*a^2*arctanh(a*x)/x^3+1/15*a^5*ln(a*x-1)-1/20*a/x^4+1/15*a^3/x^2-2/15*a^5*ln(a*x)+1/1
5*a^5*ln(a*x+1)

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Maxima [A]  time = 0.956459, size = 84, normalized size = 1.18 \begin{align*} \frac{1}{60} \,{\left (4 \, a^{4} \log \left (a^{2} x^{2} - 1\right ) - 4 \, a^{4} \log \left (x^{2}\right ) + \frac{4 \, a^{2} x^{2} - 3}{x^{4}}\right )} a + \frac{{\left (5 \, a^{2} x^{2} - 3\right )} \operatorname{artanh}\left (a x\right )}{15 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^6,x, algorithm="maxima")

[Out]

1/60*(4*a^4*log(a^2*x^2 - 1) - 4*a^4*log(x^2) + (4*a^2*x^2 - 3)/x^4)*a + 1/15*(5*a^2*x^2 - 3)*arctanh(a*x)/x^5

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Fricas [A]  time = 2.24535, size = 167, normalized size = 2.35 \begin{align*} \frac{4 \, a^{5} x^{5} \log \left (a^{2} x^{2} - 1\right ) - 8 \, a^{5} x^{5} \log \left (x\right ) + 4 \, a^{3} x^{3} - 3 \, a x + 2 \,{\left (5 \, a^{2} x^{2} - 3\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^6,x, algorithm="fricas")

[Out]

1/60*(4*a^5*x^5*log(a^2*x^2 - 1) - 8*a^5*x^5*log(x) + 4*a^3*x^3 - 3*a*x + 2*(5*a^2*x^2 - 3)*log(-(a*x + 1)/(a*
x - 1)))/x^5

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Sympy [A]  time = 2.90935, size = 75, normalized size = 1.06 \begin{align*} \begin{cases} - \frac{2 a^{5} \log{\left (x \right )}}{15} + \frac{2 a^{5} \log{\left (x - \frac{1}{a} \right )}}{15} + \frac{2 a^{5} \operatorname{atanh}{\left (a x \right )}}{15} + \frac{a^{3}}{15 x^{2}} + \frac{a^{2} \operatorname{atanh}{\left (a x \right )}}{3 x^{3}} - \frac{a}{20 x^{4}} - \frac{\operatorname{atanh}{\left (a x \right )}}{5 x^{5}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)*atanh(a*x)/x**6,x)

[Out]

Piecewise((-2*a**5*log(x)/15 + 2*a**5*log(x - 1/a)/15 + 2*a**5*atanh(a*x)/15 + a**3/(15*x**2) + a**2*atanh(a*x
)/(3*x**3) - a/(20*x**4) - atanh(a*x)/(5*x**5), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.18245, size = 109, normalized size = 1.54 \begin{align*} -\frac{1}{15} \, a^{5} \log \left (x^{2}\right ) + \frac{1}{15} \, a^{5} \log \left ({\left | a^{2} x^{2} - 1 \right |}\right ) + \frac{6 \, a^{5} x^{4} + 4 \, a^{3} x^{2} - 3 \, a}{60 \, x^{4}} + \frac{{\left (5 \, a^{2} x^{2} - 3\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{30 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^6,x, algorithm="giac")

[Out]

-1/15*a^5*log(x^2) + 1/15*a^5*log(abs(a^2*x^2 - 1)) + 1/60*(6*a^5*x^4 + 4*a^3*x^2 - 3*a)/x^4 + 1/30*(5*a^2*x^2
- 3)*log(-(a*x + 1)/(a*x - 1))/x^5