3.224 \(\int (1-a^2 x^2)^3 \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=144 \[ \frac{\left (1-a^2 x^2\right )^3}{42 a}+\frac{3 \left (1-a^2 x^2\right )^2}{70 a}+\frac{4 \left (1-a^2 x^2\right )}{35 a}+\frac{8 \log \left (1-a^2 x^2\right )}{35 a}+\frac{1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac{6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{16}{35} x \tanh ^{-1}(a x) \]

[Out]

(4*(1 - a^2*x^2))/(35*a) + (3*(1 - a^2*x^2)^2)/(70*a) + (1 - a^2*x^2)^3/(42*a) + (16*x*ArcTanh[a*x])/35 + (8*x
*(1 - a^2*x^2)*ArcTanh[a*x])/35 + (6*x*(1 - a^2*x^2)^2*ArcTanh[a*x])/35 + (x*(1 - a^2*x^2)^3*ArcTanh[a*x])/7 +
(8*Log[1 - a^2*x^2])/(35*a)

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Rubi [A]  time = 0.0674981, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {5942, 5910, 260} \[ \frac{\left (1-a^2 x^2\right )^3}{42 a}+\frac{3 \left (1-a^2 x^2\right )^2}{70 a}+\frac{4 \left (1-a^2 x^2\right )}{35 a}+\frac{8 \log \left (1-a^2 x^2\right )}{35 a}+\frac{1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac{6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{16}{35} x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^3*ArcTanh[a*x],x]

[Out]

(4*(1 - a^2*x^2))/(35*a) + (3*(1 - a^2*x^2)^2)/(70*a) + (1 - a^2*x^2)^3/(42*a) + (16*x*ArcTanh[a*x])/35 + (8*x
*(1 - a^2*x^2)*ArcTanh[a*x])/35 + (6*x*(1 - a^2*x^2)^2*ArcTanh[a*x])/35 + (x*(1 - a^2*x^2)^3*ArcTanh[a*x])/7 +
(8*Log[1 - a^2*x^2])/(35*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 )^3 \tanh ^{-1}(a x) \, dx &=\frac{\left (1-a^2 x^2\right )^3}{42 a}+\frac{1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac{6}{7} \int \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx\\ &=\frac{3 \left (1-a^2 x^2\right )^2}{70 a}+\frac{\left (1-a^2 x^2\right )^3}{42 a}+\frac{6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac{24}{35} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx\\ &=\frac{4 \left (1-a^2 x^2\right )}{35 a}+\frac{3 \left (1-a^2 x^2\right )^2}{70 a}+\frac{\left (1-a^2 x^2\right )^3}{42 a}+\frac{8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac{16}{35} \int \tanh ^{-1}(a x) \, dx\\ &=\frac{4 \left (1-a^2 x^2\right )}{35 a}+\frac{3 \left (1-a^2 x^2\right )^2}{70 a}+\frac{\left (1-a^2 x^2\right )^3}{42 a}+\frac{16}{35} x \tanh ^{-1}(a x)+\frac{8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)-\frac{1}{35} (16 a) \int \frac{x}{1-a^2 x^2} \, dx\\ &=\frac{4 \left (1-a^2 x^2\right )}{35 a}+\frac{3 \left (1-a^2 x^2\right )^2}{70 a}+\frac{\left (1-a^2 x^2\right )^3}{42 a}+\frac{16}{35} x \tanh ^{-1}(a x)+\frac{8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac{1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac{8 \log \left (1-a^2 x^2\right )}{35 a}\\ \end{align*}

Mathematica [A]  time = 0.0458864, size = 79, normalized size = 0.55 \[ \frac{-5 a^6 x^6+24 a^4 x^4-57 a^2 x^2+48 \log \left (1-a^2 x^2\right )-6 a x \left (5 a^6 x^6-21 a^4 x^4+35 a^2 x^2-35\right ) \tanh ^{-1}(a x)}{210 a} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - a^2*x^2)^3*ArcTanh[a*x],x]

[Out]

(-57*a^2*x^2 + 24*a^4*x^4 - 5*a^6*x^6 - 6*a*x*(-35 + 35*a^2*x^2 - 21*a^4*x^4 + 5*a^6*x^6)*ArcTanh[a*x] + 48*Lo
g[1 - a^2*x^2])/(210*a)

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Maple [A]  time = 0.029, size = 88, normalized size = 0.6 \begin{align*} -{\frac{{a}^{6}{\it Artanh} \left ( ax \right ){x}^{7}}{7}}+{\frac{3\,{a}^{4}{\it Artanh} \left ( ax \right ){x}^{5}}{5}}-{a}^{2}{\it Artanh} \left ( ax \right ){x}^{3}+x{\it Artanh} \left ( ax \right ) -{\frac{{a}^{5}{x}^{6}}{42}}+{\frac{4\,{x}^{4}{a}^{3}}{35}}-{\frac{19\,a{x}^{2}}{70}}+{\frac{8\,\ln \left ( ax-1 \right ) }{35\,a}}+{\frac{8\,\ln \left ( ax+1 \right ) }{35\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)^3*arctanh(a*x),x)

[Out]

-1/7*a^6*arctanh(a*x)*x^7+3/5*a^4*arctanh(a*x)*x^5-a^2*arctanh(a*x)*x^3+x*arctanh(a*x)-1/42*a^5*x^6+4/35*x^4*a
^3-19/70*a*x^2+8/35/a*ln(a*x-1)+8/35/a*ln(a*x+1)

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Maxima [A]  time = 0.966892, size = 111, normalized size = 0.77 \begin{align*} -\frac{1}{210} \,{\left (5 \, a^{4} x^{6} - 24 \, a^{2} x^{4} + 57 \, x^{2} - \frac{48 \, \log \left (a x + 1\right )}{a^{2}} - \frac{48 \, \log \left (a x - 1\right )}{a^{2}}\right )} a - \frac{1}{35} \,{\left (5 \, a^{6} x^{7} - 21 \, a^{4} x^{5} + 35 \, a^{2} x^{3} - 35 \, 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)^3*arctanh(a*x),x, algorithm="maxima")

[Out]

-1/210*(5*a^4*x^6 - 24*a^2*x^4 + 57*x^2 - 48*log(a*x + 1)/a^2 - 48*log(a*x - 1)/a^2)*a - 1/35*(5*a^6*x^7 - 21*
a^4*x^5 + 35*a^2*x^3 - 35*x)*arctanh(a*x)

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Fricas [A]  time = 2.39057, size = 198, normalized size = 1.38 \begin{align*} -\frac{5 \, a^{6} x^{6} - 24 \, a^{4} x^{4} + 57 \, a^{2} x^{2} + 3 \,{\left (5 \, a^{7} x^{7} - 21 \, a^{5} x^{5} + 35 \, a^{3} x^{3} - 35 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - 48 \, \log \left (a^{2} x^{2} - 1\right )}{210 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^3*arctanh(a*x),x, algorithm="fricas")

[Out]

-1/210*(5*a^6*x^6 - 24*a^4*x^4 + 57*a^2*x^2 + 3*(5*a^7*x^7 - 21*a^5*x^5 + 35*a^3*x^3 - 35*a*x)*log(-(a*x + 1)/
(a*x - 1)) - 48*log(a^2*x^2 - 1))/a

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Sympy [A]  time = 4.30289, size = 97, normalized size = 0.67 \begin{align*} \begin{cases} - \frac{a^{6} x^{7} \operatorname{atanh}{\left (a x \right )}}{7} - \frac{a^{5} x^{6}}{42} + \frac{3 a^{4} x^{5} \operatorname{atanh}{\left (a x \right )}}{5} + \frac{4 a^{3} x^{4}}{35} - a^{2} x^{3} \operatorname{atanh}{\left (a x \right )} - \frac{19 a x^{2}}{70} + x \operatorname{atanh}{\left (a x \right )} + \frac{16 \log{\left (x - \frac{1}{a} \right )}}{35 a} + \frac{16 \operatorname{atanh}{\left (a x \right )}}{35 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)**3*atanh(a*x),x)

[Out]

Piecewise((-a**6*x**7*atanh(a*x)/7 - a**5*x**6/42 + 3*a**4*x**5*atanh(a*x)/5 + 4*a**3*x**4/35 - a**2*x**3*atan
h(a*x) - 19*a*x**2/70 + x*atanh(a*x) + 16*log(x - 1/a)/(35*a) + 16*atanh(a*x)/(35*a), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.20726, size = 124, normalized size = 0.86 \begin{align*} -\frac{1}{70} \,{\left (5 \, a^{6} x^{7} - 21 \, a^{4} x^{5} + 35 \, a^{2} x^{3} - 35 \, x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + \frac{8 \, \log \left ({\left | a^{2} x^{2} - 1 \right |}\right )}{35 \, a} - \frac{5 \, a^{11} x^{6} - 24 \, a^{9} x^{4} + 57 \, a^{7} x^{2}}{210 \, a^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^3*arctanh(a*x),x, algorithm="giac")

[Out]

-1/70*(5*a^6*x^7 - 21*a^4*x^5 + 35*a^2*x^3 - 35*x)*log(-(a*x + 1)/(a*x - 1)) + 8/35*log(abs(a^2*x^2 - 1))/a -
1/210*(5*a^11*x^6 - 24*a^9*x^4 + 57*a^7*x^2)/a^6