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

Optimal. Leaf size=64 \[ \frac{a^3 x^2}{6}-\frac{4}{3} a \log \left (1-a^2 x^2\right )+\frac{1}{3} a^4 x^3 \tanh ^{-1}(a x)-2 a^2 x \tanh ^{-1}(a x)+a \log (x)-\frac{\tanh ^{-1}(a x)}{x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.111458, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {6012, 5910, 260, 5916, 266, 36, 29, 31, 43} \[ \frac{a^3 x^2}{6}-\frac{4}{3} a \log \left (1-a^2 x^2\right )+\frac{1}{3} a^4 x^3 \tanh ^{-1}(a x)-2 a^2 x \tanh ^{-1}(a x)+a \log (x)-\frac{\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6012

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

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]

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0162693, size = 64, normalized size = 1. \[ \frac{a^3 x^2}{6}-\frac{4}{3} a \log \left (1-a^2 x^2\right )+\frac{1}{3} a^4 x^3 \tanh ^{-1}(a x)-2 a^2 x \tanh ^{-1}(a x)+a \log (x)-\frac{\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 65, normalized size = 1. \begin{align*}{\frac{{a}^{4}{x}^{3}{\it Artanh} \left ( ax \right ) }{3}}-2\,{a}^{2}x{\it Artanh} \left ( ax \right ) -{\frac{{\it Artanh} \left ( ax \right ) }{x}}+{\frac{{x}^{2}{a}^{3}}{6}}-{\frac{4\,a\ln \left ( ax-1 \right ) }{3}}+a\ln \left ( ax \right ) -{\frac{4\,a\ln \left ( ax+1 \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.949371, size = 77, normalized size = 1.2 \begin{align*} \frac{1}{6} \,{\left (a^{2} x^{2} - 8 \, \log \left (a x + 1\right ) - 8 \, \log \left (a x - 1\right ) + 6 \, \log \left (x\right )\right )} a + \frac{1}{3} \,{\left (a^{4} x^{3} - 6 \, a^{2} x - \frac{3}{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)^2*arctanh(a*x)/x^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.99978, size = 150, normalized size = 2.34 \begin{align*} \frac{a^{3} x^{3} - 8 \, a x \log \left (a^{2} x^{2} - 1\right ) + 6 \, a x \log \left (x\right ) +{\left (a^{4} x^{4} - 6 \, a^{2} x^{2} - 3\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{6 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 2.69881, size = 68, normalized size = 1.06 \begin{align*} \begin{cases} \frac{a^{4} x^{3} \operatorname{atanh}{\left (a x \right )}}{3} + \frac{a^{3} x^{2}}{6} - 2 a^{2} x \operatorname{atanh}{\left (a x \right )} + a \log{\left (x \right )} - \frac{8 a \log{\left (x - \frac{1}{a} \right )}}{3} - \frac{8 a \operatorname{atanh}{\left (a x \right )}}{3} - \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)**2*atanh(a*x)/x**2,x)

[Out]

Piecewise((a**4*x**3*atanh(a*x)/3 + a**3*x**2/6 - 2*a**2*x*atanh(a*x) + a*log(x) - 8*a*log(x - 1/a)/3 - 8*a*at
anh(a*x)/3 - atanh(a*x)/x, Ne(a, 0)), (0, True))

________________________________________________________________________________________

Giac [A]  time = 1.21595, size = 89, normalized size = 1.39 \begin{align*} \frac{1}{6} \, a^{3} x^{2} + \frac{1}{2} \, a \log \left (x^{2}\right ) + \frac{1}{6} \,{\left (a^{4} x^{3} - 6 \, a^{2} x - \frac{3}{x}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - \frac{4}{3} \, 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)^2*arctanh(a*x)/x^2,x, algorithm="giac")

[Out]

1/6*a^3*x^2 + 1/2*a*log(x^2) + 1/6*(a^4*x^3 - 6*a^2*x - 3/x)*log(-(a*x + 1)/(a*x - 1)) - 4/3*a*log(abs(a^2*x^2
- 1))