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]

-1/4*a/(-a^2*x^2+1)-arctanh(a*x)/x+1/2*a^2*x*arctanh(a*x)/(-a^2*x^2+1)+3/4*a*arctanh(a*x)^2+a*ln(x)-1/2*a*ln(-
a^2*x^2+1)

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Rubi [A]  time = 0.15, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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 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 )^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*}

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Mathematica [A]  time = 0.14, 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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fricas [A]  time = 0.90, size = 118, normalized size = 1.44 \[ \frac {3 \, {\left (a^{3} x^{3} - a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, a x - 8 \, {\left (a^{3} x^{3} - a x\right )} \log \left (a^{2} x^{2} - 1\right ) + 16 \, {\left (a^{3} x^{3} - a x\right )} \log \relax (x) - 4 \, {\left (3 \, a^{2} x^{2} - 2\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{16 \, {\left (a^{2} x^{3} - x\right )}} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 180, normalized size = 2.20 \[ -\frac {\arctanh \left (a x \right )}{x}-\frac {a \arctanh \left (a x \right )}{4 \left (a x -1\right )}-\frac {3 a \arctanh \left (a x \right ) \ln \left (a x -1\right )}{4}-\frac {a \arctanh \left (a x \right )}{4 \left (a x +1\right )}+\frac {3 a \arctanh \left (a x \right ) \ln \left (a x +1\right )}{4}-\frac {3 a \ln \left (a x -1\right )^{2}}{16}+\frac {3 a \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8}-\frac {3 a \ln \left (a x +1\right )^{2}}{16}-\frac {3 a \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{8}+\frac {3 a \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{8}+a \ln \left (a x \right )-\frac {a \ln \left (a x -1\right )}{2}+\frac {a}{8 a x -8}-\frac {a \ln \left (a x +1\right )}{2}-\frac {a}{8 \left (a x +1\right )} \]

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]

-arctanh(a*x)/x-1/4*a*arctanh(a*x)/(a*x-1)-3/4*a*arctanh(a*x)*ln(a*x-1)-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)+a*ln(a*x)-1/2*a*ln(a*x-1)+1/8*a/(a*x-1)-1/2*a*ln(a*x+1)-1/8*
a/(a*x+1)

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maxima [B]  time = 0.33, size = 150, normalized size = 1.83 \[ -\frac {1}{16} \, a {\left (\frac {3 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 6 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + 3 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4}{a^{2} x^{2} - 1} + 8 \, \log \left (a x + 1\right ) + 8 \, \log \left (a x - 1\right ) - 16 \, \log \relax (x)\right )} + \frac {1}{4} \, {\left (3 \, a \log \left (a x + 1\right ) - 3 \, a \log \left (a x - 1\right ) - \frac {2 \, {\left (3 \, a^{2} x^{2} - 2\right )}}{a^{2} x^{3} - x}\right )} \operatorname {artanh}\left (a x\right ) \]

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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mupad [B]  time = 1.15, size = 132, normalized size = 1.61 \[ \frac {3\,a\,{\ln \left (a\,x+1\right )}^2}{16}+\frac {3\,a\,{\ln \left (1-a\,x\right )}^2}{16}+\frac {a}{2\,\left (2\,a^2\,x^2-2\right )}-\frac {a\,\ln \left (a^2\,x^2-1\right )}{2}+a\,\ln \relax (x)-\ln \left (1-a\,x\right )\,\left (\frac {\frac {3\,a^2\,x^2}{2}-1}{2\,x-2\,a^2\,x^3}+\frac {3\,a\,\ln \left (a\,x+1\right )}{8}\right )+\frac {\ln \left (a\,x+1\right )\,\left (\frac {3\,a\,x^2}{4}-\frac {1}{2\,a}\right )}{\frac {x}{a}-a\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.72, size = 253, normalized size = 3.09 \[ \begin {cases} \frac {4 a^{3} x^{3} \log {\relax (x )}}{4 a^{2} x^{3} - 4 x} - \frac {4 a^{3} x^{3} \log {\left (x - \frac {1}{a} \right )}}{4 a^{2} x^{3} - 4 x} + \frac {3 a^{3} x^{3} \operatorname {atanh}^{2}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} - \frac {4 a^{3} x^{3} \operatorname {atanh}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} - \frac {6 a^{2} x^{2} \operatorname {atanh}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} - \frac {4 a x \log {\relax (x )}}{4 a^{2} x^{3} - 4 x} + \frac {4 a x \log {\left (x - \frac {1}{a} \right )}}{4 a^{2} x^{3} - 4 x} - \frac {3 a x \operatorname {atanh}^{2}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} + \frac {4 a x \operatorname {atanh}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} + \frac {a x}{4 a^{2} x^{3} - 4 x} + \frac {4 \operatorname {atanh}{\left (a x \right )}}{4 a^{2} x^{3} - 4 x} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

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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