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

Optimal. Leaf size=82 \[ -\frac{a}{4 \left (1-a^2 x^2\right )}-\frac{1}{2} a \log \left (1-a^2 x^2\right )+\frac{a^2 x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+a \log (x)+\frac{3}{4} a \tanh ^{-1}(a x)^2-\frac{\tanh ^{-1}(a x)}{x} \]

[Out]

-a/(4*(1 - a^2*x^2)) - ArcTanh[a*x]/x + (a^2*x*ArcTanh[a*x])/(2*(1 - a^2*x^2)) + (3*a*ArcTanh[a*x]^2)/4 + a*Lo
g[x] - (a*Log[1 - a^2*x^2])/2

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Rubi [A]  time = 0.145359, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6030, 5982, 5916, 266, 36, 29, 31, 5948, 5956, 261} \[ -\frac{a}{4 \left (1-a^2 x^2\right )}-\frac{1}{2} a \log \left (1-a^2 x^2\right )+\frac{a^2 x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+a \log (x)+\frac{3}{4} a \tanh ^{-1}(a x)^2-\frac{\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(4*(1 - a^2*x^2)) - ArcTanh[a*x]/x + (a^2*x*ArcTanh[a*x])/(2*(1 - a^2*x^2)) + (3*a*ArcTanh[a*x]^2)/4 + a*Lo
g[x] - (a*Log[1 - a^2*x^2])/2

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]

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 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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.141376, size = 77, normalized size = 0.94 \[ \frac{1}{4} \left (a \left (\frac{1}{a^2 x^2-1}-2 \log \left (1-a^2 x^2\right )+4 \log (a x)\right )-\frac{2 \left (3 a^2 x^2-2\right ) \tanh ^{-1}(a x)}{x \left (a^2 x^2-1\right )}+3 a \tanh ^{-1}(a x)^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(-2 + 3*a^2*x^2)*ArcTanh[a*x])/(x*(-1 + a^2*x^2)) + 3*a*ArcTanh[a*x]^2 + a*((-1 + a^2*x^2)^(-1) + 4*Log[a
*x] - 2*Log[1 - a^2*x^2]))/4

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Maple [B]  time = 0.062, size = 180, normalized size = 2.2 \begin{align*} -{\frac{a{\it Artanh} \left ( ax \right ) }{4\,ax-4}}-{\frac{3\,a{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{4}}-{\frac{{\it Artanh} \left ( ax \right ) }{x}}-{\frac{a{\it Artanh} \left ( ax \right ) }{4\,ax+4}}+{\frac{3\,a{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{4}}-{\frac{3\,a \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{16}}+{\frac{3\,a\ln \left ( ax-1 \right ) }{8}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{3\,a \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{16}}-{\frac{3\,a}{8}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{3\,a\ln \left ( ax+1 \right ) }{8}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }-{\frac{a\ln \left ( ax-1 \right ) }{2}}+{\frac{a}{8\,ax-8}}+a\ln \left ( ax \right ) -{\frac{a\ln \left ( ax+1 \right ) }{2}}-{\frac{a}{8\,ax+8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*arctanh(a*x)/(a*x-1)-3/4*a*arctanh(a*x)*ln(a*x-1)-arctanh(a*x)/x-1/4*a*arctanh(a*x)/(a*x+1)+3/4*a*arcta
nh(a*x)*ln(a*x+1)-3/16*a*ln(a*x-1)^2+3/8*a*ln(a*x-1)*ln(1/2+1/2*a*x)-3/16*a*ln(a*x+1)^2-3/8*a*ln(-1/2*a*x+1/2)
*ln(1/2+1/2*a*x)+3/8*a*ln(-1/2*a*x+1/2)*ln(a*x+1)-1/2*a*ln(a*x-1)+1/8*a/(a*x-1)+a*ln(a*x)-1/2*a*ln(a*x+1)-1/8*
a/(a*x+1)

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Maxima [B]  time = 0.98279, size = 203, normalized size = 2.48 \begin{align*} -\frac{1}{16} \, a{\left (\frac{3 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 6 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + 3 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4}{a^{2} x^{2} - 1} + 8 \, \log \left (a x + 1\right ) + 8 \, \log \left (a x - 1\right ) - 16 \, \log \left (x\right )\right )} + \frac{1}{4} \,{\left (3 \, a \log \left (a x + 1\right ) - 3 \, a \log \left (a x - 1\right ) - \frac{2 \,{\left (3 \, a^{2} x^{2} - 2\right )}}{a^{2} x^{3} - x}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*a*((3*(a^2*x^2 - 1)*log(a*x + 1)^2 - 6*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) + 3*(a^2*x^2 - 1)*log(a*x
- 1)^2 - 4)/(a^2*x^2 - 1) + 8*log(a*x + 1) + 8*log(a*x - 1) - 16*log(x)) + 1/4*(3*a*log(a*x + 1) - 3*a*log(a*
x - 1) - 2*(3*a^2*x^2 - 2)/(a^2*x^3 - x))*arctanh(a*x)

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Fricas [A]  time = 2.12923, size = 252, normalized size = 3.07 \begin{align*} \frac{3 \,{\left (a^{3} x^{3} - a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} + 4 \, a x - 8 \,{\left (a^{3} x^{3} - a x\right )} \log \left (a^{2} x^{2} - 1\right ) + 16 \,{\left (a^{3} x^{3} - a x\right )} \log \left (x\right ) - 4 \,{\left (3 \, a^{2} x^{2} - 2\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{16 \,{\left (a^{2} x^{3} - x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*(a^3*x^3 - a*x)*log(-(a*x + 1)/(a*x - 1))^2 + 4*a*x - 8*(a^3*x^3 - a*x)*log(a^2*x^2 - 1) + 16*(a^3*x^3
- a*x)*log(x) - 4*(3*a^2*x^2 - 2)*log(-(a*x + 1)/(a*x - 1)))/(a^2*x^3 - x)

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Sympy [A]  time = 4.4774, size = 253, normalized size = 3.09 \begin{align*} \begin{cases} \frac{4 a^{3} x^{3} \log{\left (x \right )}}{4 a^{2} x^{3} - 4 x} - \frac{4 a^{3} x^{3} \log{\left (x - \frac{1}{a} \right )}}{4 a^{2} x^{3} - 4 x} + \frac{3 a^{3} x^{3} \operatorname{atanh}^{2}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} - \frac{4 a^{3} x^{3} \operatorname{atanh}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} - \frac{6 a^{2} x^{2} \operatorname{atanh}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} - \frac{4 a x \log{\left (x \right )}}{4 a^{2} x^{3} - 4 x} + \frac{4 a x \log{\left (x - \frac{1}{a} \right )}}{4 a^{2} x^{3} - 4 x} - \frac{3 a x \operatorname{atanh}^{2}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} + \frac{4 a x \operatorname{atanh}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} + \frac{a x}{4 a^{2} x^{3} - 4 x} + \frac{4 \operatorname{atanh}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((4*a**3*x**3*log(x)/(4*a**2*x**3 - 4*x) - 4*a**3*x**3*log(x - 1/a)/(4*a**2*x**3 - 4*x) + 3*a**3*x**3
*atanh(a*x)**2/(4*a**2*x**3 - 4*x) - 4*a**3*x**3*atanh(a*x)/(4*a**2*x**3 - 4*x) - 6*a**2*x**2*atanh(a*x)/(4*a*
*2*x**3 - 4*x) - 4*a*x*log(x)/(4*a**2*x**3 - 4*x) + 4*a*x*log(x - 1/a)/(4*a**2*x**3 - 4*x) - 3*a*x*atanh(a*x)*
*2/(4*a**2*x**3 - 4*x) + 4*a*x*atanh(a*x)/(4*a**2*x**3 - 4*x) + a*x/(4*a**2*x**3 - 4*x) + 4*atanh(a*x)/(4*a**2
*x**3 - 4*x), Ne(a, 0)), (0, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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