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

Optimal. Leaf size=77 \[ -\frac{1}{2} a^4 \text{PolyLog}(2,-a x)+\frac{1}{2} a^4 \text{PolyLog}(2,a x)+\frac{a^2 \tanh ^{-1}(a x)}{x^2}+\frac{3 a^3}{4 x}-\frac{3}{4} a^4 \tanh ^{-1}(a x)-\frac{a}{12 x^3}-\frac{\tanh ^{-1}(a x)}{4 x^4} \]

[Out]

-a/(12*x^3) + (3*a^3)/(4*x) - (3*a^4*ArcTanh[a*x])/4 - ArcTanh[a*x]/(4*x^4) + (a^2*ArcTanh[a*x])/x^2 - (a^4*Po
lyLog[2, -(a*x)])/2 + (a^4*PolyLog[2, a*x])/2

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Rubi [A]  time = 0.100928, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6012, 5916, 325, 206, 5912} \[ -\frac{1}{2} a^4 \text{PolyLog}(2,-a x)+\frac{1}{2} a^4 \text{PolyLog}(2,a x)+\frac{a^2 \tanh ^{-1}(a x)}{x^2}+\frac{3 a^3}{4 x}-\frac{3}{4} a^4 \tanh ^{-1}(a x)-\frac{a}{12 x^3}-\frac{\tanh ^{-1}(a x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(12*x^3) + (3*a^3)/(4*x) - (3*a^4*ArcTanh[a*x])/4 - ArcTanh[a*x]/(4*x^4) + (a^2*ArcTanh[a*x])/x^2 - (a^4*Po
lyLog[2, -(a*x)])/2 + (a^4*PolyLog[2, a*x])/2

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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])

Rule 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, x]

Rubi steps

\begin{align*} \int \frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{x^5} \, dx &=\int \left (\frac{\tanh ^{-1}(a x)}{x^5}-\frac{2 a^2 \tanh ^{-1}(a x)}{x^3}+\frac{a^4 \tanh ^{-1}(a x)}{x}\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int \frac{\tanh ^{-1}(a x)}{x^3} \, dx\right )+a^4 \int \frac{\tanh ^{-1}(a x)}{x} \, dx+\int \frac{\tanh ^{-1}(a x)}{x^5} \, dx\\ &=-\frac{\tanh ^{-1}(a x)}{4 x^4}+\frac{a^2 \tanh ^{-1}(a x)}{x^2}-\frac{1}{2} a^4 \text{Li}_2(-a x)+\frac{1}{2} a^4 \text{Li}_2(a x)+\frac{1}{4} a \int \frac{1}{x^4 \left (1-a^2 x^2\right )} \, dx-a^3 \int \frac{1}{x^2 \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{a}{12 x^3}+\frac{a^3}{x}-\frac{\tanh ^{-1}(a x)}{4 x^4}+\frac{a^2 \tanh ^{-1}(a x)}{x^2}-\frac{1}{2} a^4 \text{Li}_2(-a x)+\frac{1}{2} a^4 \text{Li}_2(a x)+\frac{1}{4} a^3 \int \frac{1}{x^2 \left (1-a^2 x^2\right )} \, dx-a^5 \int \frac{1}{1-a^2 x^2} \, dx\\ &=-\frac{a}{12 x^3}+\frac{3 a^3}{4 x}-a^4 \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)}{4 x^4}+\frac{a^2 \tanh ^{-1}(a x)}{x^2}-\frac{1}{2} a^4 \text{Li}_2(-a x)+\frac{1}{2} a^4 \text{Li}_2(a x)+\frac{1}{4} a^5 \int \frac{1}{1-a^2 x^2} \, dx\\ &=-\frac{a}{12 x^3}+\frac{3 a^3}{4 x}-\frac{3}{4} a^4 \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)}{4 x^4}+\frac{a^2 \tanh ^{-1}(a x)}{x^2}-\frac{1}{2} a^4 \text{Li}_2(-a x)+\frac{1}{2} a^4 \text{Li}_2(a x)\\ \end{align*}

Mathematica [A]  time = 0.0815758, size = 84, normalized size = 1.09 \[ \frac{1}{24} \left (-12 a^4 \text{PolyLog}(2,-a x)+12 a^4 \text{PolyLog}(2,a x)+\frac{24 a^2 \tanh ^{-1}(a x)}{x^2}+\frac{18 a^3}{x}+9 a^4 \log (1-a x)-9 a^4 \log (a x+1)-\frac{2 a}{x^3}-\frac{6 \tanh ^{-1}(a x)}{x^4}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*a)/x^3 + (18*a^3)/x - (6*ArcTanh[a*x])/x^4 + (24*a^2*ArcTanh[a*x])/x^2 + 9*a^4*Log[1 - a*x] - 9*a^4*Log[1
+ a*x] - 12*a^4*PolyLog[2, -(a*x)] + 12*a^4*PolyLog[2, a*x])/24

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Maple [A]  time = 0.047, size = 105, normalized size = 1.4 \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) }{4\,{x}^{4}}}+{a}^{4}{\it Artanh} \left ( ax \right ) \ln \left ( ax \right ) +{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) }{{x}^{2}}}-{\frac{{a}^{4}{\it dilog} \left ( ax \right ) }{2}}-{\frac{{a}^{4}{\it dilog} \left ( ax+1 \right ) }{2}}-{\frac{{a}^{4}\ln \left ( ax \right ) \ln \left ( ax+1 \right ) }{2}}+{\frac{3\,{a}^{4}\ln \left ( ax-1 \right ) }{8}}-{\frac{a}{12\,{x}^{3}}}+{\frac{3\,{a}^{3}}{4\,x}}-{\frac{3\,{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)^2*arctanh(a*x)/x^5,x)

[Out]

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

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Maxima [A]  time = 0.964769, size = 151, normalized size = 1.96 \begin{align*} -\frac{1}{24} \,{\left (12 \,{\left (\log \left (a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-a x\right )\right )} a^{3} - 12 \,{\left (\log \left (-a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (a x\right )\right )} a^{3} + 9 \, a^{3} \log \left (a x + 1\right ) - 9 \, a^{3} \log \left (a x - 1\right ) - \frac{2 \,{\left (9 \, a^{2} x^{2} - 1\right )}}{x^{3}}\right )} a + \frac{1}{4} \,{\left (2 \, a^{4} \log \left (x^{2}\right ) + \frac{4 \, a^{2} x^{2} - 1}{x^{4}}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(12*(log(a*x + 1)*log(x) + dilog(-a*x))*a^3 - 12*(log(-a*x + 1)*log(x) + dilog(a*x))*a^3 + 9*a^3*log(a*x
+ 1) - 9*a^3*log(a*x - 1) - 2*(9*a^2*x^2 - 1)/x^3)*a + 1/4*(2*a^4*log(x^2) + (4*a^2*x^2 - 1)/x^4)*arctanh(a*x
)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname{artanh}\left (a x\right )}{x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)/x^5, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname{atanh}{\left (a x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname{artanh}\left (a x\right )}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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