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

Optimal. Leaf size=83 \[ \frac{7 a^3}{30 x^2}-\frac{4}{15} a^5 \log \left (1-a^2 x^2\right )+\frac{2 a^2 \tanh ^{-1}(a x)}{3 x^3}+\frac{8}{15} a^5 \log (x)-\frac{a^4 \tanh ^{-1}(a x)}{x}-\frac{a}{20 x^4}-\frac{\tanh ^{-1}(a x)}{5 x^5} \]

[Out]

-a/(20*x^4) + (7*a^3)/(30*x^2) - ArcTanh[a*x]/(5*x^5) + (2*a^2*ArcTanh[a*x])/(3*x^3) - (a^4*ArcTanh[a*x])/x +
(8*a^5*Log[x])/15 - (4*a^5*Log[1 - a^2*x^2])/15

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Rubi [A]  time = 0.137952, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {6012, 5916, 266, 44, 36, 29, 31} \[ \frac{7 a^3}{30 x^2}-\frac{4}{15} a^5 \log \left (1-a^2 x^2\right )+\frac{2 a^2 \tanh ^{-1}(a x)}{3 x^3}+\frac{8}{15} a^5 \log (x)-\frac{a^4 \tanh ^{-1}(a x)}{x}-\frac{a}{20 x^4}-\frac{\tanh ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(20*x^4) + (7*a^3)/(30*x^2) - ArcTanh[a*x]/(5*x^5) + (2*a^2*ArcTanh[a*x])/(3*x^3) - (a^4*ArcTanh[a*x])/x +
(8*a^5*Log[x])/15 - (4*a^5*Log[1 - a^2*x^2])/15

Rule 6012

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

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])

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{x^6} \, dx &=\int \left (\frac{\tanh ^{-1}(a x)}{x^6}-\frac{2 a^2 \tanh ^{-1}(a x)}{x^4}+\frac{a^4 \tanh ^{-1}(a x)}{x^2}\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int \frac{\tanh ^{-1}(a x)}{x^4} \, dx\right )+a^4 \int \frac{\tanh ^{-1}(a x)}{x^2} \, dx+\int \frac{\tanh ^{-1}(a x)}{x^6} \, dx\\ &=-\frac{\tanh ^{-1}(a x)}{5 x^5}+\frac{2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac{a^4 \tanh ^{-1}(a x)}{x}+\frac{1}{5} a \int \frac{1}{x^5 \left (1-a^2 x^2\right )} \, dx-\frac{1}{3} \left (2 a^3\right ) \int \frac{1}{x^3 \left (1-a^2 x^2\right )} \, dx+a^5 \int \frac{1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{\tanh ^{-1}(a x)}{5 x^5}+\frac{2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac{a^4 \tanh ^{-1}(a x)}{x}+\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}{3} a^3 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-a^2 x\right )} \, dx,x,x^2\right )+\frac{1}{2} a^5 \operatorname{Subst}\left (\int \frac{1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{\tanh ^{-1}(a x)}{5 x^5}+\frac{2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac{a^4 \tanh ^{-1}(a x)}{x}+\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}{3} 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{1}{2} a^5 \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{2} a^7 \operatorname{Subst}\left (\int \frac{1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac{a}{20 x^4}+\frac{7 a^3}{30 x^2}-\frac{\tanh ^{-1}(a x)}{5 x^5}+\frac{2 a^2 \tanh ^{-1}(a x)}{3 x^3}-\frac{a^4 \tanh ^{-1}(a x)}{x}+\frac{8}{15} a^5 \log (x)-\frac{4}{15} a^5 \log \left (1-a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0228139, size = 83, normalized size = 1. \[ \frac{7 a^3}{30 x^2}-\frac{4}{15} a^5 \log \left (1-a^2 x^2\right )+\frac{2 a^2 \tanh ^{-1}(a x)}{3 x^3}+\frac{8}{15} a^5 \log (x)-\frac{a^4 \tanh ^{-1}(a x)}{x}-\frac{a}{20 x^4}-\frac{\tanh ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(20*x^4) + (7*a^3)/(30*x^2) - ArcTanh[a*x]/(5*x^5) + (2*a^2*ArcTanh[a*x])/(3*x^3) - (a^4*ArcTanh[a*x])/x +
(8*a^5*Log[x])/15 - (4*a^5*Log[1 - a^2*x^2])/15

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Maple [A]  time = 0.038, size = 80, normalized size = 1. \begin{align*} -{\frac{{a}^{4}{\it Artanh} \left ( ax \right ) }{x}}-{\frac{{\it Artanh} \left ( ax \right ) }{5\,{x}^{5}}}+{\frac{2\,{a}^{2}{\it Artanh} \left ( ax \right ) }{3\,{x}^{3}}}-{\frac{4\,{a}^{5}\ln \left ( ax-1 \right ) }{15}}-{\frac{a}{20\,{x}^{4}}}+{\frac{7\,{a}^{3}}{30\,{x}^{2}}}+{\frac{8\,{a}^{5}\ln \left ( ax \right ) }{15}}-{\frac{4\,{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)^2*arctanh(a*x)/x^6,x)

[Out]

-a^4*arctanh(a*x)/x-1/5*arctanh(a*x)/x^5+2/3*a^2*arctanh(a*x)/x^3-4/15*a^5*ln(a*x-1)-1/20*a/x^4+7/30*a^3/x^2+8
/15*a^5*ln(a*x)-4/15*a^5*ln(a*x+1)

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Maxima [A]  time = 0.949418, size = 96, normalized size = 1.16 \begin{align*} -\frac{1}{60} \,{\left (16 \, a^{4} \log \left (a^{2} x^{2} - 1\right ) - 16 \, a^{4} \log \left (x^{2}\right ) - \frac{14 \, a^{2} x^{2} - 3}{x^{4}}\right )} a - \frac{{\left (15 \, a^{4} x^{4} - 10 \, 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)^2*arctanh(a*x)/x^6,x, algorithm="maxima")

[Out]

-1/60*(16*a^4*log(a^2*x^2 - 1) - 16*a^4*log(x^2) - (14*a^2*x^2 - 3)/x^4)*a - 1/15*(15*a^4*x^4 - 10*a^2*x^2 + 3
)*arctanh(a*x)/x^5

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Fricas [A]  time = 1.96243, size = 192, normalized size = 2.31 \begin{align*} -\frac{16 \, a^{5} x^{5} \log \left (a^{2} x^{2} - 1\right ) - 32 \, a^{5} x^{5} \log \left (x\right ) - 14 \, a^{3} x^{3} + 3 \, a x + 2 \,{\left (15 \, a^{4} x^{4} - 10 \, 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)^2*arctanh(a*x)/x^6,x, algorithm="fricas")

[Out]

-1/60*(16*a^5*x^5*log(a^2*x^2 - 1) - 32*a^5*x^5*log(x) - 14*a^3*x^3 + 3*a*x + 2*(15*a^4*x^4 - 10*a^2*x^2 + 3)*
log(-(a*x + 1)/(a*x - 1)))/x^5

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Sympy [A]  time = 3.41173, size = 88, normalized size = 1.06 \begin{align*} \begin{cases} \frac{8 a^{5} \log{\left (x \right )}}{15} - \frac{8 a^{5} \log{\left (x - \frac{1}{a} \right )}}{15} - \frac{8 a^{5} \operatorname{atanh}{\left (a x \right )}}{15} - \frac{a^{4} \operatorname{atanh}{\left (a x \right )}}{x} + \frac{7 a^{3}}{30 x^{2}} + \frac{2 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)**2*atanh(a*x)/x**6,x)

[Out]

Piecewise((8*a**5*log(x)/15 - 8*a**5*log(x - 1/a)/15 - 8*a**5*atanh(a*x)/15 - a**4*atanh(a*x)/x + 7*a**3/(30*x
**2) + 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 = 1.17547, size = 120, normalized size = 1.45 \begin{align*} \frac{4}{15} \, a^{5} \log \left (x^{2}\right ) - \frac{4}{15} \, a^{5} \log \left ({\left | a^{2} x^{2} - 1 \right |}\right ) - \frac{24 \, a^{5} x^{4} - 14 \, a^{3} x^{2} + 3 \, a}{60 \, x^{4}} - \frac{{\left (15 \, a^{4} x^{4} - 10 \, 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)^2*arctanh(a*x)/x^6,x, algorithm="giac")

[Out]

4/15*a^5*log(x^2) - 4/15*a^5*log(abs(a^2*x^2 - 1)) - 1/60*(24*a^5*x^4 - 14*a^3*x^2 + 3*a)/x^4 - 1/30*(15*a^4*x
^4 - 10*a^2*x^2 + 3)*log(-(a*x + 1)/(a*x - 1))/x^5