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

Optimal. Leaf size=91 \[ -\frac {a x}{4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}-\frac {1}{2} \text {Li}_2\left (\frac {2}{a x+1}-1\right )+\frac {1}{2} \tanh ^{-1}(a x)^2-\frac {1}{4} \tanh ^{-1}(a x)+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 91, 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 = {6030, 5988, 5932, 2447, 5994, 199, 206} \[ -\frac {1}{2} \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )-\frac {a x}{4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {1}{2} \tanh ^{-1}(a x)^2-\frac {1}{4} \tanh ^{-1}(a x)+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x)/(4*(1 - a^2*x^2)) - ArcTanh[a*x]/4 + ArcTanh[a*x]/(2*(1 - a^2*x^2)) + ArcTanh[a*x]^2/2 + ArcTanh[a*x]*L
og[2 - 2/(1 + a*x)] - PolyLog[2, -1 + 2/(1 + a*x)]/2

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 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 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 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]

Rule 5994

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

Rule 6030

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

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 63, normalized size = 0.69 \[ \frac {1}{8} \left (-4 \text {Li}_2\left (e^{-2 \tanh ^{-1}(a x)}\right )+4 \tanh ^{-1}(a x)^2-\sinh \left (2 \tanh ^{-1}(a x)\right )+2 \tanh ^{-1}(a x) \left (4 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )+\cosh \left (2 \tanh ^{-1}(a x)\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*ArcTanh[a*x]^2 + 2*ArcTanh[a*x]*(Cosh[2*ArcTanh[a*x]] + 4*Log[1 - E^(-2*ArcTanh[a*x])]) - 4*PolyLog[2, E^(-
2*ArcTanh[a*x])] - Sinh[2*ArcTanh[a*x]])/8

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fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (a x\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctanh(a*x)/(a^4*x^5 - 2*a^2*x^3 + x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 190, normalized size = 2.09 \[ \arctanh \left (a x \right ) \ln \left (a x \right )-\frac {\arctanh \left (a x \right )}{4 \left (a x -1\right )}-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2}+\frac {\arctanh \left (a x \right )}{4 a x +4}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {\dilog \left (a x \right )}{2}-\frac {\dilog \left (a x +1\right )}{2}-\frac {\ln \left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {\ln \left (a x -1\right )^{2}}{8}+\frac {\dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{2}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{4}+\frac {\ln \left (a x +1\right )^{2}}{8}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {1}{2}+\frac {a x}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}+\frac {1}{8 a x -8}+\frac {\ln \left (a x -1\right )}{8}+\frac {1}{8 a x +8}-\frac {\ln \left (a x +1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 206, normalized size = 2.26 \[ \frac {1}{8} \, a {\left (\frac {{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} + 2 \, a x}{a^{3} x^{2} - a} + \frac {4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a} - \frac {4 \, {\left (\log \left (a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-a x\right )\right )}}{a} + \frac {4 \, {\left (\log \left (-a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (a x\right )\right )}}{a} - \frac {\log \left (a x + 1\right )}{a} + \frac {\log \left (a x - 1\right )}{a}\right )} - \frac {1}{2} \, {\left (\frac {1}{a^{2} x^{2} - 1} + \log \left (a^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*(((a^2*x^2 - 1)*log(a*x + 1)^2 - 2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) - (a^2*x^2 - 1)*log(a*x - 1)^
2 + 2*a*x)/(a^3*x^2 - a) + 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 - log(a*x + 1)/a + log(a*x - 1)/a) - 1/2*(1/(a^
2*x^2 - 1) + log(a^2*x^2 - 1) - log(x^2))*arctanh(a*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)/(x*(a^2*x^2 - 1)^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (a x \right )}}{x \left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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