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

Optimal. Leaf size=38 \[ -a \log \left (1-a^2 x^2\right )+a^2 (-x) \tanh ^{-1}(a x)+a \log (x)-\frac{\tanh ^{-1}(a x)}{x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0511531, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {6014, 5916, 266, 36, 29, 31, 5910, 260} \[ -a \log \left (1-a^2 x^2\right )+a^2 (-x) \tanh ^{-1}(a x)+a \log (x)-\frac{\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

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

Rubi steps

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

Mathematica [A]  time = 0.0097317, size = 38, normalized size = 1. \[ -a \log \left (1-a^2 x^2\right )+a^2 (-x) \tanh ^{-1}(a x)+a \log (x)-\frac{\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.033, size = 45, normalized size = 1.2 \begin{align*} -{a}^{2}x{\it Artanh} \left ( ax \right ) -{\frac{{\it Artanh} \left ( ax \right ) }{x}}-a\ln \left ( ax-1 \right ) +a\ln \left ( ax \right ) -a\ln \left ( ax+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.94342, size = 49, normalized size = 1.29 \begin{align*} -a{\left (\log \left (a x + 1\right ) + \log \left (a x - 1\right ) - \log \left (x\right )\right )} -{\left (a^{2} x + \frac{1}{x}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.23007, size = 122, normalized size = 3.21 \begin{align*} -\frac{2 \, a x \log \left (a^{2} x^{2} - 1\right ) - 2 \, a x \log \left (x\right ) +{\left (a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.23114, size = 41, normalized size = 1.08 \begin{align*} \begin{cases} - a^{2} x \operatorname{atanh}{\left (a x \right )} + a \log{\left (x \right )} - 2 a \log{\left (x - \frac{1}{a} \right )} - 2 a \operatorname{atanh}{\left (a x \right )} - \frac{\operatorname{atanh}{\left (a x \right )}}{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((-a**2*x**2+1)*atanh(a*x)/x**2,x)

[Out]

Piecewise((-a**2*x*atanh(a*x) + a*log(x) - 2*a*log(x - 1/a) - 2*a*atanh(a*x) - atanh(a*x)/x, Ne(a, 0)), (0, Tr
ue))

________________________________________________________________________________________

Giac [A]  time = 1.15541, size = 65, normalized size = 1.71 \begin{align*} \frac{1}{2} \, a \log \left (x^{2}\right ) - \frac{1}{2} \,{\left (a^{2} x + \frac{1}{x}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - a \log \left ({\left | a^{2} x^{2} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*log(x^2) - 1/2*(a^2*x + 1/x)*log(-(a*x + 1)/(a*x - 1)) - a*log(abs(a^2*x^2 - 1))