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

Optimal. Leaf size=104 \[ \frac{\left (1-a^2 x^2\right )^2}{20 a}+\frac{2 \left (1-a^2 x^2\right )}{15 a}+\frac{4 \log \left (1-a^2 x^2\right )}{15 a}+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{8}{15} x \tanh ^{-1}(a x) \]

[Out]

(2*(1 - a^2*x^2))/(15*a) + (1 - a^2*x^2)^2/(20*a) + (8*x*ArcTanh[a*x])/15 + (4*x*(1 - a^2*x^2)*ArcTanh[a*x])/1
5 + (x*(1 - a^2*x^2)^2*ArcTanh[a*x])/5 + (4*Log[1 - a^2*x^2])/(15*a)

________________________________________________________________________________________

Rubi [A]  time = 0.0437777, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {5942, 5910, 260} \[ \frac{\left (1-a^2 x^2\right )^2}{20 a}+\frac{2 \left (1-a^2 x^2\right )}{15 a}+\frac{4 \log \left (1-a^2 x^2\right )}{15 a}+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{8}{15} x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^2*ArcTanh[a*x],x]

[Out]

(2*(1 - a^2*x^2))/(15*a) + (1 - a^2*x^2)^2/(20*a) + (8*x*ArcTanh[a*x])/15 + (4*x*(1 - a^2*x^2)*ArcTanh[a*x])/1
5 + (x*(1 - a^2*x^2)^2*ArcTanh[a*x])/5 + (4*Log[1 - a^2*x^2])/(15*a)

Rule 5942

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

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 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx &=\frac{\left (1-a^2 x^2\right )^2}{20 a}+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{4}{5} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx\\ &=\frac{2 \left (1-a^2 x^2\right )}{15 a}+\frac{\left (1-a^2 x^2\right )^2}{20 a}+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{8}{15} \int \tanh ^{-1}(a x) \, dx\\ &=\frac{2 \left (1-a^2 x^2\right )}{15 a}+\frac{\left (1-a^2 x^2\right )^2}{20 a}+\frac{8}{15} x \tanh ^{-1}(a x)+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)-\frac{1}{15} (8 a) \int \frac{x}{1-a^2 x^2} \, dx\\ &=\frac{2 \left (1-a^2 x^2\right )}{15 a}+\frac{\left (1-a^2 x^2\right )^2}{20 a}+\frac{8}{15} x \tanh ^{-1}(a x)+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{4 \log \left (1-a^2 x^2\right )}{15 a}\\ \end{align*}

Mathematica [A]  time = 0.0166325, size = 71, normalized size = 0.68 \[ \frac{a^3 x^4}{20}+\frac{4 \log \left (1-a^2 x^2\right )}{15 a}+\frac{1}{5} a^4 x^5 \tanh ^{-1}(a x)-\frac{2}{3} a^2 x^3 \tanh ^{-1}(a x)-\frac{7 a x^2}{30}+x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*a*x^2)/30 + (a^3*x^4)/20 + x*ArcTanh[a*x] - (2*a^2*x^3*ArcTanh[a*x])/3 + (a^4*x^5*ArcTanh[a*x])/5 + (4*Log
[1 - a^2*x^2])/(15*a)

________________________________________________________________________________________

Maple [A]  time = 0.03, size = 68, normalized size = 0.7 \begin{align*}{\frac{{a}^{4}{\it Artanh} \left ( ax \right ){x}^{5}}{5}}-{\frac{2\,{a}^{2}{\it Artanh} \left ( ax \right ){x}^{3}}{3}}+x{\it Artanh} \left ( ax \right ) +{\frac{{x}^{4}{a}^{3}}{20}}-{\frac{7\,a{x}^{2}}{30}}+{\frac{4\,\ln \left ( ax-1 \right ) }{15\,a}}+{\frac{4\,\ln \left ( ax+1 \right ) }{15\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^4*arctanh(a*x)*x^5-2/3*a^2*arctanh(a*x)*x^3+x*arctanh(a*x)+1/20*x^4*a^3-7/30*a*x^2+4/15/a*ln(a*x-1)+4/15
/a*ln(a*x+1)

________________________________________________________________________________________

Maxima [A]  time = 0.975285, size = 89, normalized size = 0.86 \begin{align*} \frac{1}{60} \,{\left (3 \, a^{2} x^{4} - 14 \, x^{2} + \frac{16 \, \log \left (a x + 1\right )}{a^{2}} + \frac{16 \, \log \left (a x - 1\right )}{a^{2}}\right )} a + \frac{1}{15} \,{\left (3 \, a^{4} x^{5} - 10 \, a^{2} x^{3} + 15 \, 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, algorithm="maxima")

[Out]

1/60*(3*a^2*x^4 - 14*x^2 + 16*log(a*x + 1)/a^2 + 16*log(a*x - 1)/a^2)*a + 1/15*(3*a^4*x^5 - 10*a^2*x^3 + 15*x)
*arctanh(a*x)

________________________________________________________________________________________

Fricas [A]  time = 1.95209, size = 161, normalized size = 1.55 \begin{align*} \frac{3 \, a^{4} x^{4} - 14 \, a^{2} x^{2} + 2 \,{\left (3 \, a^{5} x^{5} - 10 \, a^{3} x^{3} + 15 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + 16 \, \log \left (a^{2} x^{2} - 1\right )}{60 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*a^4*x^4 - 14*a^2*x^2 + 2*(3*a^5*x^5 - 10*a^3*x^3 + 15*a*x)*log(-(a*x + 1)/(a*x - 1)) + 16*log(a^2*x^2
- 1))/a

________________________________________________________________________________________

Sympy [A]  time = 1.77116, size = 75, normalized size = 0.72 \begin{align*} \begin{cases} \frac{a^{4} x^{5} \operatorname{atanh}{\left (a x \right )}}{5} + \frac{a^{3} x^{4}}{20} - \frac{2 a^{2} x^{3} \operatorname{atanh}{\left (a x \right )}}{3} - \frac{7 a x^{2}}{30} + x \operatorname{atanh}{\left (a x \right )} + \frac{8 \log{\left (x - \frac{1}{a} \right )}}{15 a} + \frac{8 \operatorname{atanh}{\left (a x \right )}}{15 a} & \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)

[Out]

Piecewise((a**4*x**5*atanh(a*x)/5 + a**3*x**4/20 - 2*a**2*x**3*atanh(a*x)/3 - 7*a*x**2/30 + x*atanh(a*x) + 8*l
og(x - 1/a)/(15*a) + 8*atanh(a*x)/(15*a), Ne(a, 0)), (0, True))

________________________________________________________________________________________

Giac [A]  time = 1.16773, size = 103, normalized size = 0.99 \begin{align*} \frac{1}{30} \,{\left (3 \, a^{4} x^{5} - 10 \, a^{2} x^{3} + 15 \, x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + \frac{4 \, \log \left ({\left | a^{2} x^{2} - 1 \right |}\right )}{15 \, a} + \frac{3 \, a^{7} x^{4} - 14 \, a^{5} x^{2}}{60 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(3*a^4*x^5 - 10*a^2*x^3 + 15*x)*log(-(a*x + 1)/(a*x - 1)) + 4/15*log(abs(a^2*x^2 - 1))/a + 1/60*(3*a^7*x^
4 - 14*a^5*x^2)/a^4