3.168 \(\int \frac {(1-a^2 x^2) \tanh ^{-1}(a x)}{x^3} \, dx\)

Optimal. Leaf size=56 \[ \frac {1}{2} a^2 \text {Li}_2(-a x)-\frac {1}{2} a^2 \text {Li}_2(a x)+\frac {1}{2} a^2 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)}{2 x^2}-\frac {a}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6014, 5916, 325, 206, 5912} \[ \frac {1}{2} a^2 \text {PolyLog}(2,-a x)-\frac {1}{2} a^2 \text {PolyLog}(2,a x)+\frac {1}{2} a^2 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)}{2 x^2}-\frac {a}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)*ArcTanh[a*x])/x^3,x]

[Out]

-a/(2*x) + (a^2*ArcTanh[a*x])/2 - ArcTanh[a*x]/(2*x^2) + (a^2*PolyLog[2, -(a*x)])/2 - (a^2*PolyLog[2, a*x])/2

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 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 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]

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 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]))

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 68, normalized size = 1.21 \[ -\frac {1}{2} a^2 (\text {Li}_2(a x)-\text {Li}_2(-a x))-\frac {1}{4} a^2 \log (1-a x)+\frac {1}{4} a^2 \log (a x+1)-\frac {\tanh ^{-1}(a x)}{2 x^2}-\frac {a}{2 x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((1 - a^2*x^2)*ArcTanh[a*x])/x^3,x]

[Out]

-1/2*a/x - ArcTanh[a*x]/(2*x^2) - (a^2*Log[1 - a*x])/4 + (a^2*Log[1 + a*x])/4 - (a^2*(-PolyLog[2, -(a*x)] + Po
lyLog[2, a*x]))/2

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^3,x, algorithm="fricas")

[Out]

integral(-(a^2*x^2 - 1)*arctanh(a*x)/x^3, x)

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giac [B]  time = 2.18, size = 330, normalized size = 5.89 \[ a^{2} {\left (\frac {\log \left (\frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}}\right )}{a} - \frac {\log \left ({\left | \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - 1 \right |}\right )}{a} + \frac {\frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - 2}{a {\left (\frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - 1\right )}} - \frac {2 \, \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{a - \frac {a {\left (\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1\right )}}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}} - 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{a - \frac {a {\left (\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1\right )}}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}} + 1}\right )}{a {\left (\frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - 1\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 87, normalized size = 1.55 \[ -a^{2} \arctanh \left (a x \right ) \ln \left (a x \right )-\frac {\arctanh \left (a x \right )}{2 x^{2}}-\frac {a}{2 x}-\frac {a^{2} \ln \left (a x -1\right )}{4}+\frac {a^{2} \ln \left (a x +1\right )}{4}+\frac {a^{2} \dilog \left (a x \right )}{2}+\frac {a^{2} \dilog \left (a x +1\right )}{2}+\frac {a^{2} \ln \left (a x \right ) \ln \left (a x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)*arctanh(a*x)/x^3,x)

[Out]

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

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maxima [A]  time = 0.32, size = 81, normalized size = 1.45 \[ \frac {1}{4} \, {\left (2 \, {\left (\log \left (a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-a x\right )\right )} a - 2 \, {\left (\log \left (-a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (a x\right )\right )} a + a \log \left (a x + 1\right ) - a \log \left (a x - 1\right ) - \frac {2}{x}\right )} a - \frac {1}{2} \, {\left (a^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)/x^3,x, algorithm="maxima")

[Out]

1/4*(2*(log(a*x + 1)*log(x) + dilog(-a*x))*a - 2*(log(-a*x + 1)*log(x) + dilog(a*x))*a + a*log(a*x + 1) - a*lo
g(a*x - 1) - 2/x)*a - 1/2*(a^2*log(x^2) + 1/x^2)*arctanh(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int \frac {\mathrm {atanh}\left (a\,x\right )\,\left (a^2\,x^2-1\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(atanh(a*x)*(a^2*x^2 - 1))/x^3,x)

[Out]

-int((atanh(a*x)*(a^2*x^2 - 1))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {\operatorname {atanh}{\left (a x \right )}}{x^{3}}\right )\, dx - \int \frac {a^{2} \operatorname {atanh}{\left (a x \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)*atanh(a*x)/x**3,x)

[Out]

-Integral(-atanh(a*x)/x**3, x) - Integral(a**2*atanh(a*x)/x, x)

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