3.181 \(\int \frac{(1-a^2 x^2) \tanh ^{-1}(a x)^2}{x^5} \, dx\)

Optimal. Leaf size=89 \[ -\frac{a^2}{12 x^2}+\frac{1}{6} a^4 \log \left (1-a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 x^4}-\frac{1}{3} a^4 \log (x)+\frac{a^3 \tanh ^{-1}(a x)}{2 x}-\frac{a \tanh ^{-1}(a x)}{6 x^3} \]

[Out]

-a^2/(12*x^2) - (a*ArcTanh[a*x])/(6*x^3) + (a^3*ArcTanh[a*x])/(2*x) - ((1 - a^2*x^2)^2*ArcTanh[a*x]^2)/(4*x^4)
- (a^4*Log[x])/3 + (a^4*Log[1 - a^2*x^2])/6

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Rubi [A]  time = 0.107831, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6008, 6014, 5916, 266, 44, 36, 29, 31} \[ -\frac{a^2}{12 x^2}+\frac{1}{6} a^4 \log \left (1-a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 x^4}-\frac{1}{3} a^4 \log (x)+\frac{a^3 \tanh ^{-1}(a x)}{2 x}-\frac{a \tanh ^{-1}(a x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)*ArcTanh[a*x]^2)/x^5,x]

[Out]

-a^2/(12*x^2) - (a*ArcTanh[a*x])/(6*x^3) + (a^3*ArcTanh[a*x])/(2*x) - ((1 - a^2*x^2)^2*ArcTanh[a*x]^2)/(4*x^4)
- (a^4*Log[x])/3 + (a^4*Log[1 - a^2*x^2])/6

Rule 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 6014

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

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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]

Rubi steps

\begin{align*} \int \frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{x^5} \, dx &=-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 x^4}+\frac{1}{2} a \int \frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{x^4} \, dx\\ &=-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 x^4}+\frac{1}{2} a \int \frac{\tanh ^{-1}(a x)}{x^4} \, dx-\frac{1}{2} a^3 \int \frac{\tanh ^{-1}(a x)}{x^2} \, dx\\ &=-\frac{a \tanh ^{-1}(a x)}{6 x^3}+\frac{a^3 \tanh ^{-1}(a x)}{2 x}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 x^4}+\frac{1}{6} a^2 \int \frac{1}{x^3 \left (1-a^2 x^2\right )} \, dx-\frac{1}{2} a^4 \int \frac{1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{a \tanh ^{-1}(a x)}{6 x^3}+\frac{a^3 \tanh ^{-1}(a x)}{2 x}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 x^4}+\frac{1}{12} a^2 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-a^2 x\right )} \, dx,x,x^2\right )-\frac{1}{4} a^4 \operatorname{Subst}\left (\int \frac{1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{a \tanh ^{-1}(a x)}{6 x^3}+\frac{a^3 \tanh ^{-1}(a x)}{2 x}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 x^4}+\frac{1}{12} a^2 \operatorname{Subst}\left (\int \left (\frac{1}{x^2}+\frac{a^2}{x}-\frac{a^4}{-1+a^2 x}\right ) \, dx,x,x^2\right )-\frac{1}{4} a^4 \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )-\frac{1}{4} a^6 \operatorname{Subst}\left (\int \frac{1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac{a^2}{12 x^2}-\frac{a \tanh ^{-1}(a x)}{6 x^3}+\frac{a^3 \tanh ^{-1}(a x)}{2 x}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 x^4}-\frac{1}{3} a^4 \log (x)+\frac{1}{6} a^4 \log \left (1-a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0326074, size = 82, normalized size = 0.92 \[ \frac{-a^2 x^2-4 a^4 x^4 \log (x)+2 a^4 x^4 \log \left (1-a^2 x^2\right )-3 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2+\left (6 a^3 x^3-2 a x\right ) \tanh ^{-1}(a x)}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)*ArcTanh[a*x]^2)/x^5,x]

[Out]

(-(a^2*x^2) + (-2*a*x + 6*a^3*x^3)*ArcTanh[a*x] - 3*(-1 + a^2*x^2)^2*ArcTanh[a*x]^2 - 4*a^4*x^4*Log[x] + 2*a^4
*x^4*Log[1 - a^2*x^2])/(12*x^4)

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Maple [B]  time = 0.06, size = 199, normalized size = 2.2 \begin{align*} -{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{4\,{x}^{4}}}+{\frac{{a}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{2\,{x}^{2}}}+{\frac{{a}^{4}{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{4}}+{\frac{{a}^{3}{\it Artanh} \left ( ax \right ) }{2\,x}}-{\frac{a{\it Artanh} \left ( ax \right ) }{6\,{x}^{3}}}-{\frac{{a}^{4}{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{4}}+{\frac{{a}^{4} \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{16}}-{\frac{{a}^{4}\ln \left ( ax-1 \right ) }{8}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{{a}^{4}}{8}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{{a}^{4}\ln \left ( ax+1 \right ) }{8}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{{a}^{4} \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{16}}+{\frac{{a}^{4}\ln \left ( ax-1 \right ) }{6}}-{\frac{{a}^{2}}{12\,{x}^{2}}}-{\frac{{a}^{4}\ln \left ( ax \right ) }{3}}+{\frac{{a}^{4}\ln \left ( ax+1 \right ) }{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)*arctanh(a*x)^2/x^5,x)

[Out]

-1/4*arctanh(a*x)^2/x^4+1/2*a^2*arctanh(a*x)^2/x^2+1/4*a^4*arctanh(a*x)*ln(a*x-1)+1/2*a^3*arctanh(a*x)/x-1/6*a
*arctanh(a*x)/x^3-1/4*a^4*arctanh(a*x)*ln(a*x+1)+1/16*a^4*ln(a*x-1)^2-1/8*a^4*ln(a*x-1)*ln(1/2+1/2*a*x)+1/8*a^
4*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/8*a^4*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/16*a^4*ln(a*x+1)^2+1/6*a^4*ln(a*x-1)-1
/12*a^2/x^2-1/3*a^4*ln(a*x)+1/6*a^4*ln(a*x+1)

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Maxima [B]  time = 0.967811, size = 221, normalized size = 2.48 \begin{align*} -\frac{1}{48} \,{\left (16 \, a^{2} \log \left (x\right ) - \frac{3 \, a^{2} x^{2} \log \left (a x + 1\right )^{2} + 3 \, a^{2} x^{2} \log \left (a x - 1\right )^{2} + 8 \, a^{2} x^{2} \log \left (a x - 1\right ) - 2 \,{\left (3 \, a^{2} x^{2} \log \left (a x - 1\right ) - 4 \, a^{2} x^{2}\right )} \log \left (a x + 1\right ) - 4}{x^{2}}\right )} a^{2} - \frac{1}{12} \,{\left (3 \, a^{3} \log \left (a x + 1\right ) - 3 \, a^{3} \log \left (a x - 1\right ) - \frac{2 \,{\left (3 \, a^{2} x^{2} - 1\right )}}{x^{3}}\right )} a \operatorname{artanh}\left (a x\right ) + \frac{{\left (2 \, a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)^2/x^5,x, algorithm="maxima")

[Out]

-1/48*(16*a^2*log(x) - (3*a^2*x^2*log(a*x + 1)^2 + 3*a^2*x^2*log(a*x - 1)^2 + 8*a^2*x^2*log(a*x - 1) - 2*(3*a^
2*x^2*log(a*x - 1) - 4*a^2*x^2)*log(a*x + 1) - 4)/x^2)*a^2 - 1/12*(3*a^3*log(a*x + 1) - 3*a^3*log(a*x - 1) - 2
*(3*a^2*x^2 - 1)/x^3)*a*arctanh(a*x) + 1/4*(2*a^2*x^2 - 1)*arctanh(a*x)^2/x^4

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Fricas [A]  time = 2.51928, size = 239, normalized size = 2.69 \begin{align*} \frac{8 \, a^{4} x^{4} \log \left (a^{2} x^{2} - 1\right ) - 16 \, a^{4} x^{4} \log \left (x\right ) - 4 \, a^{2} x^{2} - 3 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} + 4 \,{\left (3 \, a^{3} x^{3} - a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{48 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)^2/x^5,x, algorithm="fricas")

[Out]

1/48*(8*a^4*x^4*log(a^2*x^2 - 1) - 16*a^4*x^4*log(x) - 4*a^2*x^2 - 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/
(a*x - 1))^2 + 4*(3*a^3*x^3 - a*x)*log(-(a*x + 1)/(a*x - 1)))/x^4

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Sympy [A]  time = 2.46072, size = 102, normalized size = 1.15 \begin{align*} \begin{cases} - \frac{a^{4} \log{\left (x \right )}}{3} + \frac{a^{4} \log{\left (x - \frac{1}{a} \right )}}{3} - \frac{a^{4} \operatorname{atanh}^{2}{\left (a x \right )}}{4} + \frac{a^{4} \operatorname{atanh}{\left (a x \right )}}{3} + \frac{a^{3} \operatorname{atanh}{\left (a x \right )}}{2 x} + \frac{a^{2} \operatorname{atanh}^{2}{\left (a x \right )}}{2 x^{2}} - \frac{a^{2}}{12 x^{2}} - \frac{a \operatorname{atanh}{\left (a x \right )}}{6 x^{3}} - \frac{\operatorname{atanh}^{2}{\left (a x \right )}}{4 x^{4}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)*atanh(a*x)**2/x**5,x)

[Out]

Piecewise((-a**4*log(x)/3 + a**4*log(x - 1/a)/3 - a**4*atanh(a*x)**2/4 + a**4*atanh(a*x)/3 + a**3*atanh(a*x)/(
2*x) + a**2*atanh(a*x)**2/(2*x**2) - a**2/(12*x**2) - a*atanh(a*x)/(6*x**3) - atanh(a*x)**2/(4*x**4), Ne(a, 0)
), (0, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )^{2}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)*arctanh(a*x)^2/x^5,x, algorithm="giac")

[Out]

integrate(-(a^2*x^2 - 1)*arctanh(a*x)^2/x^5, x)