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

Optimal. Leaf size=84 \[ -\frac {1}{2} a^2 \text {Li}_2\left (\frac {2}{a x+1}-1\right )+\frac {1}{2} a^2 \tanh ^{-1}(a x)^2+\frac {1}{2} a^2 \tanh ^{-1}(a x)+a^2 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)}{2 x^2}-\frac {a}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {5982, 5916, 325, 206, 5988, 5932, 2447} \[ -\frac {1}{2} a^2 \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )+\frac {1}{2} a^2 \tanh ^{-1}(a x)^2+\frac {1}{2} a^2 \tanh ^{-1}(a x)+a^2 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)}{2 x^2}-\frac {a}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(2*x) + (a^2*ArcTanh[a*x])/2 - ArcTanh[a*x]/(2*x^2) + (a^2*ArcTanh[a*x]^2)/2 + a^2*ArcTanh[a*x]*Log[2 - 2/(
1 + a*x)] - (a^2*PolyLog[2, -1 + 2/(1 + 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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, 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 5932

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

Rule 5982

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

Rule 5988

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 60, normalized size = 0.71 \[ -\frac {1}{2} a^2 \left (-\tanh ^{-1}(a x) \left (-\frac {1}{a^2 x^2}+\tanh ^{-1}(a x)+2 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )+1\right )+\text {Li}_2\left (e^{-2 \tanh ^{-1}(a x)}\right )+\frac {1}{a x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (a x\right )}{a^{2} x^{5} - x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-arctanh(a*x)/(a^2*x^5 - x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\operatorname {artanh}\left (a x\right )}{{\left (a^{2} x^{2} - 1\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 209, normalized size = 2.49 \[ -\frac {\arctanh \left (a x \right )}{2 x^{2}}+a^{2} \arctanh \left (a x \right ) \ln \left (a x \right )-\frac {a^{2} \arctanh \left (a x \right ) \ln \left (a x -1\right )}{2}-\frac {a^{2} \arctanh \left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {a}{2 x}-\frac {a^{2} \ln \left (a x -1\right )}{4}+\frac {a^{2} \ln \left (a x +1\right )}{4}-\frac {a^{2} \ln \left (a x -1\right )^{2}}{8}+\frac {a^{2} \dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{2}+\frac {a^{2} \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4}+\frac {a^{2} \ln \left (a x +1\right )^{2}}{8}+\frac {a^{2} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4}-\frac {a^{2} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{4}-\frac {a^{2} \dilog \left (a x \right )}{2}-\frac {a^{2} \dilog \left (a x +1\right )}{2}-\frac {a^{2} \ln \left (a x \right ) \ln \left (a x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctanh(a*x)/x^2+a^2*arctanh(a*x)*ln(a*x)-1/2*a^2*arctanh(a*x)*ln(a*x-1)-1/2*a^2*arctanh(a*x)*ln(a*x+1)-1
/2*a/x-1/4*a^2*ln(a*x-1)+1/4*a^2*ln(a*x+1)-1/8*a^2*ln(a*x-1)^2+1/2*a^2*dilog(1/2+1/2*a*x)+1/4*a^2*ln(a*x-1)*ln
(1/2+1/2*a*x)+1/8*a^2*ln(a*x+1)^2+1/4*a^2*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/4*a^2*ln(-1/2*a*x+1/2)*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 [B]  time = 0.33, size = 162, normalized size = 1.93 \[ \frac {1}{8} \, {\left (4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )} a - 4 \, {\left (\log \left (a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-a x\right )\right )} a + 4 \, {\left (\log \left (-a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (a x\right )\right )} a + 2 \, a \log \left (a x + 1\right ) - 2 \, a \log \left (a x - 1\right ) + \frac {a x \log \left (a x + 1\right )^{2} - 2 \, a x \log \left (a x + 1\right ) \log \left (a x - 1\right ) - a x \log \left (a x - 1\right )^{2} - 4}{x}\right )} a - \frac {1}{2} \, {\left (a^{2} \log \left (a^{2} x^{2} - 1\right ) - a^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))*a - 4*(log(a*x + 1)*log(x) + dilog(-a*x))*a +
4*(log(-a*x + 1)*log(x) + dilog(a*x))*a + 2*a*log(a*x + 1) - 2*a*log(a*x - 1) + (a*x*log(a*x + 1)^2 - 2*a*x*l
og(a*x + 1)*log(a*x - 1) - a*x*log(a*x - 1)^2 - 4)/x)*a - 1/2*(a^2*log(a^2*x^2 - 1) - a^2*log(x^2) + 1/x^2)*ar
ctanh(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {\mathrm {atanh}\left (a\,x\right )}{x^3\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\operatorname {atanh}{\left (a x \right )}}{a^{2} x^{5} - x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(atanh(a*x)/(a**2*x**5 - x**3), x)

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