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

Optimal. Leaf size=123 \[ -\frac {7 a}{16 \left (1-a^2 x^2\right )}-\frac {a}{16 \left (1-a^2 x^2\right )^2}-\frac {1}{2} a \log \left (1-a^2 x^2\right )+\frac {7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac {a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+a \log (x)+\frac {15}{16} a \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)}{x} \]

[Out]

-1/16*a/(-a^2*x^2+1)^2-7/16*a/(-a^2*x^2+1)-arctanh(a*x)/x+1/4*a^2*x*arctanh(a*x)/(-a^2*x^2+1)^2+7/8*a^2*x*arct
anh(a*x)/(-a^2*x^2+1)+15/16*a*arctanh(a*x)^2+a*ln(x)-1/2*a*ln(-a^2*x^2+1)

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Rubi [A]  time = 0.24, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {6030, 5982, 5916, 266, 36, 29, 31, 5948, 5956, 261, 5960} \[ -\frac {7 a}{16 \left (1-a^2 x^2\right )}-\frac {a}{16 \left (1-a^2 x^2\right )^2}-\frac {1}{2} a \log \left (1-a^2 x^2\right )+\frac {7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac {a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+a \log (x)+\frac {15}{16} a \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(16*(1 - a^2*x^2)^2) - (7*a)/(16*(1 - a^2*x^2)) - ArcTanh[a*x]/x + (a^2*x*ArcTanh[a*x])/(4*(1 - a^2*x^2)^2)
+ (7*a^2*x*ArcTanh[a*x])/(8*(1 - a^2*x^2)) + (15*a*ArcTanh[a*x]^2)/16 + a*Log[x] - (a*Log[1 - a^2*x^2])/2

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]

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -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 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 5948

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

Rule 5956

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

Rule 5960

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q + 1))
/(4*c*d*(q + 1)^2), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x], x] -
Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]))/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*
d + e, 0] && LtQ[q, -1] && NeQ[q, -3/2]

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 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^2 \left (1-a^2 x^2\right )^3} \, dx &=a^2 \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx+\int \frac {\tanh ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac {a}{16 \left (1-a^2 x^2\right )^2}+\frac {a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {1}{4} \left (3 a^2\right ) \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx+a^2 \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac {\tanh ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {a}{16 \left (1-a^2 x^2\right )^2}+\frac {a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac {7}{16} a \tanh ^{-1}(a x)^2+a^2 \int \frac {\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx-\frac {1}{8} \left (3 a^3\right ) \int \frac {x}{\left (1-a^2 x^2\right )^2} \, dx-\frac {1}{2} a^3 \int \frac {x}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac {\tanh ^{-1}(a x)}{x^2} \, dx\\ &=-\frac {a}{16 \left (1-a^2 x^2\right )^2}-\frac {7 a}{16 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)}{x}+\frac {a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac {15}{16} a \tanh ^{-1}(a x)^2+a \int \frac {1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {a}{16 \left (1-a^2 x^2\right )^2}-\frac {7 a}{16 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)}{x}+\frac {a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac {15}{16} a \tanh ^{-1}(a x)^2+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {a}{16 \left (1-a^2 x^2\right )^2}-\frac {7 a}{16 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)}{x}+\frac {a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac {15}{16} a \tanh ^{-1}(a x)^2+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} a^3 \operatorname {Subst}\left (\int \frac {1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac {a}{16 \left (1-a^2 x^2\right )^2}-\frac {7 a}{16 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)}{x}+\frac {a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac {15}{16} a \tanh ^{-1}(a x)^2+a \log (x)-\frac {1}{2} a \log \left (1-a^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 94, normalized size = 0.76 \[ \frac {1}{16} \left (a \left (\frac {7 a^2 x^2-8}{\left (a^2 x^2-1\right )^2}-8 \log \left (1-a^2 x^2\right )+16 \log (x)\right )-\frac {2 \left (15 a^4 x^4-25 a^2 x^2+8\right ) \tanh ^{-1}(a x)}{x \left (a^2 x^2-1\right )^2}+15 a \tanh ^{-1}(a x)^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(8 - 25*a^2*x^2 + 15*a^4*x^4)*ArcTanh[a*x])/(x*(-1 + a^2*x^2)^2) + 15*a*ArcTanh[a*x]^2 + a*((-8 + 7*a^2*x
^2)/(-1 + a^2*x^2)^2 + 16*Log[x] - 8*Log[1 - a^2*x^2]))/16

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fricas [A]  time = 0.87, size = 161, normalized size = 1.31 \[ \frac {28 \, a^{3} x^{3} + 15 \, {\left (a^{5} x^{5} - 2 \, a^{3} x^{3} + a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 32 \, a x - 32 \, {\left (a^{5} x^{5} - 2 \, a^{3} x^{3} + a x\right )} \log \left (a^{2} x^{2} - 1\right ) + 64 \, {\left (a^{5} x^{5} - 2 \, a^{3} x^{3} + a x\right )} \log \relax (x) - 4 \, {\left (15 \, a^{4} x^{4} - 25 \, a^{2} x^{2} + 8\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{64 \, {\left (a^{4} x^{5} - 2 \, a^{2} x^{3} + x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(28*a^3*x^3 + 15*(a^5*x^5 - 2*a^3*x^3 + a*x)*log(-(a*x + 1)/(a*x - 1))^2 - 32*a*x - 32*(a^5*x^5 - 2*a^3*x
^3 + a*x)*log(a^2*x^2 - 1) + 64*(a^5*x^5 - 2*a^3*x^3 + a*x)*log(x) - 4*(15*a^4*x^4 - 25*a^2*x^2 + 8)*log(-(a*x
+ 1)/(a*x - 1)))/(a^4*x^5 - 2*a^2*x^3 + x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 228, normalized size = 1.85 \[ -\frac {\arctanh \left (a x \right )}{x}+\frac {a \arctanh \left (a x \right )}{16 \left (a x -1\right )^{2}}-\frac {7 a \arctanh \left (a x \right )}{16 \left (a x -1\right )}-\frac {15 a \arctanh \left (a x \right ) \ln \left (a x -1\right )}{16}-\frac {a \arctanh \left (a x \right )}{16 \left (a x +1\right )^{2}}-\frac {7 a \arctanh \left (a x \right )}{16 \left (a x +1\right )}+\frac {15 a \arctanh \left (a x \right ) \ln \left (a x +1\right )}{16}-\frac {15 a \ln \left (a x -1\right )^{2}}{64}+\frac {15 a \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{32}-\frac {15 a \ln \left (a x +1\right )^{2}}{64}+\frac {15 a \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{32}-\frac {15 a \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{32}+a \ln \left (a x \right )-\frac {a}{64 \left (a x -1\right )^{2}}+\frac {15 a}{64 \left (a x -1\right )}-\frac {a \ln \left (a x -1\right )}{2}-\frac {a}{64 \left (a x +1\right )^{2}}-\frac {15 a}{64 \left (a x +1\right )}-\frac {a \ln \left (a x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctanh(a*x)/x+1/16*a*arctanh(a*x)/(a*x-1)^2-7/16*a*arctanh(a*x)/(a*x-1)-15/16*a*arctanh(a*x)*ln(a*x-1)-1/16*
a*arctanh(a*x)/(a*x+1)^2-7/16*a*arctanh(a*x)/(a*x+1)+15/16*a*arctanh(a*x)*ln(a*x+1)-15/64*a*ln(a*x-1)^2+15/32*
a*ln(a*x-1)*ln(1/2+1/2*a*x)-15/64*a*ln(a*x+1)^2+15/32*a*ln(-1/2*a*x+1/2)*ln(a*x+1)-15/32*a*ln(-1/2*a*x+1/2)*ln
(1/2+1/2*a*x)+a*ln(a*x)-1/64*a/(a*x-1)^2+15/64*a/(a*x-1)-1/2*a*ln(a*x-1)-1/64*a/(a*x+1)^2-15/64*a/(a*x+1)-1/2*
a*ln(a*x+1)

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maxima [A]  time = 0.33, size = 204, normalized size = 1.66 \[ \frac {1}{64} \, a {\left (\frac {28 \, a^{2} x^{2} - 15 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 30 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 15 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 32}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - 32 \, \log \left (a x + 1\right ) - 32 \, \log \left (a x - 1\right ) + 64 \, \log \relax (x)\right )} + \frac {1}{16} \, {\left (15 \, a \log \left (a x + 1\right ) - 15 \, a \log \left (a x - 1\right ) - \frac {2 \, {\left (15 \, a^{4} x^{4} - 25 \, a^{2} x^{2} + 8\right )}}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*a*((28*a^2*x^2 - 15*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 30*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)*
log(a*x - 1) - 15*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2 - 32)/(a^4*x^4 - 2*a^2*x^2 + 1) - 32*log(a*x + 1) -
32*log(a*x - 1) + 64*log(x)) + 1/16*(15*a*log(a*x + 1) - 15*a*log(a*x - 1) - 2*(15*a^4*x^4 - 25*a^2*x^2 + 8)/
(a^4*x^5 - 2*a^2*x^3 + x))*arctanh(a*x)

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mupad [B]  time = 1.39, size = 183, normalized size = 1.49 \[ \frac {15\,a\,{\ln \left (a\,x+1\right )}^2}{64}-\frac {4\,a-\frac {7\,a^3\,x^2}{2}}{8\,a^4\,x^4-16\,a^2\,x^2+8}+\frac {15\,a\,{\ln \left (1-a\,x\right )}^2}{64}-\frac {a\,\ln \left (a^2\,x^2-1\right )}{2}+a\,\ln \relax (x)+\ln \left (1-a\,x\right )\,\left (\frac {\frac {15\,a^4\,x^4}{8}-\frac {25\,a^2\,x^2}{8}+1}{2\,a^4\,x^5-4\,a^2\,x^3+2\,x}-\frac {15\,a\,\ln \left (a\,x+1\right )}{32}\right )-\frac {\ln \left (a\,x+1\right )\,\left (\frac {1}{2\,a}-\frac {25\,a\,x^2}{16}+\frac {15\,a^3\,x^4}{16}\right )}{\frac {x}{a}-2\,a\,x^3+a^3\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*a*log(a*x + 1)^2)/64 - (4*a - (7*a^3*x^2)/2)/(8*a^4*x^4 - 16*a^2*x^2 + 8) + (15*a*log(1 - a*x)^2)/64 - (a*
log(a^2*x^2 - 1))/2 + a*log(x) + log(1 - a*x)*(((15*a^4*x^4)/8 - (25*a^2*x^2)/8 + 1)/(2*x - 4*a^2*x^3 + 2*a^4*
x^5) - (15*a*log(a*x + 1))/32) - (log(a*x + 1)*(1/(2*a) - (25*a*x^2)/16 + (15*a^3*x^4)/16))/(x/a - 2*a*x^3 + a
^3*x^5)

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sympy [A]  time = 4.90, size = 549, normalized size = 4.46 \[ \begin {cases} \frac {16 a^{5} x^{5} \log {\relax (x )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} - \frac {16 a^{5} x^{5} \log {\left (x - \frac {1}{a} \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} + \frac {15 a^{5} x^{5} \operatorname {atanh}^{2}{\left (a x \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} - \frac {16 a^{5} x^{5} \operatorname {atanh}{\left (a x \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} - \frac {30 a^{4} x^{4} \operatorname {atanh}{\left (a x \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} - \frac {32 a^{3} x^{3} \log {\relax (x )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} + \frac {32 a^{3} x^{3} \log {\left (x - \frac {1}{a} \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} - \frac {30 a^{3} x^{3} \operatorname {atanh}^{2}{\left (a x \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} + \frac {32 a^{3} x^{3} \operatorname {atanh}{\left (a x \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} + \frac {7 a^{3} x^{3}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} + \frac {50 a^{2} x^{2} \operatorname {atanh}{\left (a x \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} + \frac {16 a x \log {\relax (x )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} - \frac {16 a x \log {\left (x - \frac {1}{a} \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} + \frac {15 a x \operatorname {atanh}^{2}{\left (a x \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} - \frac {16 a x \operatorname {atanh}{\left (a x \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} - \frac {8 a x}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} - \frac {16 \operatorname {atanh}{\left (a x \right )}}{16 a^{4} x^{5} - 32 a^{2} x^{3} + 16 x} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((16*a**5*x**5*log(x)/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) - 16*a**5*x**5*log(x - 1/a)/(16*a**4*x**5
- 32*a**2*x**3 + 16*x) + 15*a**5*x**5*atanh(a*x)**2/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) - 16*a**5*x**5*atanh(
a*x)/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) - 30*a**4*x**4*atanh(a*x)/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) - 32*
a**3*x**3*log(x)/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) + 32*a**3*x**3*log(x - 1/a)/(16*a**4*x**5 - 32*a**2*x**3
+ 16*x) - 30*a**3*x**3*atanh(a*x)**2/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) + 32*a**3*x**3*atanh(a*x)/(16*a**4*
x**5 - 32*a**2*x**3 + 16*x) + 7*a**3*x**3/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) + 50*a**2*x**2*atanh(a*x)/(16*a
**4*x**5 - 32*a**2*x**3 + 16*x) + 16*a*x*log(x)/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) - 16*a*x*log(x - 1/a)/(16
*a**4*x**5 - 32*a**2*x**3 + 16*x) + 15*a*x*atanh(a*x)**2/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) - 16*a*x*atanh(a
*x)/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) - 8*a*x/(16*a**4*x**5 - 32*a**2*x**3 + 16*x) - 16*atanh(a*x)/(16*a**4
*x**5 - 32*a**2*x**3 + 16*x), Ne(a, 0)), (0, True))

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