3.169 \(\int \frac{(1-a^2 x^2) \tanh ^{-1}(a x)}{x^4} \, dx\)

Optimal. Leaf size=58 \[ \frac{1}{3} a^3 \log \left (1-a^2 x^2\right )-\frac{2}{3} a^3 \log (x)+\frac{a^2 \tanh ^{-1}(a x)}{x}-\frac{a}{6 x^2}-\frac{\tanh ^{-1}(a x)}{3 x^3} \]

[Out]

-a/(6*x^2) - ArcTanh[a*x]/(3*x^3) + (a^2*ArcTanh[a*x])/x - (2*a^3*Log[x])/3 + (a^3*Log[1 - a^2*x^2])/3

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Rubi [A]  time = 0.0763112, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {6014, 5916, 266, 44, 36, 29, 31} \[ \frac{1}{3} a^3 \log \left (1-a^2 x^2\right )-\frac{2}{3} a^3 \log (x)+\frac{a^2 \tanh ^{-1}(a x)}{x}-\frac{a}{6 x^2}-\frac{\tanh ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(6*x^2) - ArcTanh[a*x]/(3*x^3) + (a^2*ArcTanh[a*x])/x - (2*a^3*Log[x])/3 + (a^3*Log[1 - a^2*x^2])/3

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])

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

Mathematica [A]  time = 0.0154878, size = 58, normalized size = 1. \[ \frac{1}{3} a^3 \log \left (1-a^2 x^2\right )-\frac{2}{3} a^3 \log (x)+\frac{a^2 \tanh ^{-1}(a x)}{x}-\frac{a}{6 x^2}-\frac{\tanh ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(6*x^2) - ArcTanh[a*x]/(3*x^3) + (a^2*ArcTanh[a*x])/x - (2*a^3*Log[x])/3 + (a^3*Log[1 - a^2*x^2])/3

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Maple [A]  time = 0.038, size = 59, normalized size = 1. \begin{align*}{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) }{x}}-{\frac{{\it Artanh} \left ( ax \right ) }{3\,{x}^{3}}}+{\frac{{a}^{3}\ln \left ( ax-1 \right ) }{3}}-{\frac{a}{6\,{x}^{2}}}-{\frac{2\,{a}^{3}\ln \left ( ax \right ) }{3}}+{\frac{{a}^{3}\ln \left ( ax+1 \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)*arctanh(a*x)/x^4,x)

[Out]

a^2*arctanh(a*x)/x-1/3*arctanh(a*x)/x^3+1/3*a^3*ln(a*x-1)-1/6*a/x^2-2/3*a^3*ln(a*x)+1/3*a^3*ln(a*x+1)

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Maxima [A]  time = 0.945411, size = 72, normalized size = 1.24 \begin{align*} \frac{1}{6} \,{\left (2 \, a^{2} \log \left (a^{2} x^{2} - 1\right ) - 2 \, a^{2} \log \left (x^{2}\right ) - \frac{1}{x^{2}}\right )} a + \frac{{\left (3 \, a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.24186, size = 144, normalized size = 2.48 \begin{align*} \frac{2 \, a^{3} x^{3} \log \left (a^{2} x^{2} - 1\right ) - 4 \, a^{3} x^{3} \log \left (x\right ) - a x +{\left (3 \, a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.7318, size = 63, normalized size = 1.09 \begin{align*} \begin{cases} - \frac{2 a^{3} \log{\left (x \right )}}{3} + \frac{2 a^{3} \log{\left (x - \frac{1}{a} \right )}}{3} + \frac{2 a^{3} \operatorname{atanh}{\left (a x \right )}}{3} + \frac{a^{2} \operatorname{atanh}{\left (a x \right )}}{x} - \frac{a}{6 x^{2}} - \frac{\operatorname{atanh}{\left (a x \right )}}{3 x^{3}} & \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**4,x)

[Out]

Piecewise((-2*a**3*log(x)/3 + 2*a**3*log(x - 1/a)/3 + 2*a**3*atanh(a*x)/3 + a**2*atanh(a*x)/x - a/(6*x**2) - a
tanh(a*x)/(3*x**3), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.17024, size = 99, normalized size = 1.71 \begin{align*} -\frac{1}{3} \, a^{3} \log \left (x^{2}\right ) + \frac{1}{3} \, a^{3} \log \left ({\left | a^{2} x^{2} - 1 \right |}\right ) + \frac{2 \, a^{3} x^{2} - a}{6 \, x^{2}} + \frac{{\left (3 \, a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^4,x, algorithm="giac")

[Out]

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