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

Optimal. Leaf size=138 \[ \frac{\left (1-a^2 x^2\right )^2}{60 a^2}+\frac{2 \left (1-a^2 x^2\right )}{45 a^2}+\frac{4 \log \left (1-a^2 x^2\right )}{45 a^2}-\frac{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac{x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}+\frac{4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac{8 x \tanh ^{-1}(a x)}{45 a} \]

[Out]

(2*(1 - a^2*x^2))/(45*a^2) + (1 - a^2*x^2)^2/(60*a^2) + (8*x*ArcTanh[a*x])/(45*a) + (4*x*(1 - a^2*x^2)*ArcTanh
[a*x])/(45*a) + (x*(1 - a^2*x^2)^2*ArcTanh[a*x])/(15*a) - ((1 - a^2*x^2)^3*ArcTanh[a*x]^2)/(6*a^2) + (4*Log[1
- a^2*x^2])/(45*a^2)

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Rubi [A]  time = 0.088121, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5994, 5942, 5910, 260} \[ \frac{\left (1-a^2 x^2\right )^2}{60 a^2}+\frac{2 \left (1-a^2 x^2\right )}{45 a^2}+\frac{4 \log \left (1-a^2 x^2\right )}{45 a^2}-\frac{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac{x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}+\frac{4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac{8 x \tanh ^{-1}(a x)}{45 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(1 - a^2*x^2))/(45*a^2) + (1 - a^2*x^2)^2/(60*a^2) + (8*x*ArcTanh[a*x])/(45*a) + (4*x*(1 - a^2*x^2)*ArcTanh
[a*x])/(45*a) + (x*(1 - a^2*x^2)^2*ArcTanh[a*x])/(15*a) - ((1 - a^2*x^2)^3*ArcTanh[a*x]^2)/(6*a^2) + (4*Log[1
- a^2*x^2])/(45*a^2)

Rule 5994

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

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 x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2 \, dx &=-\frac{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac{\int \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx}{3 a}\\ &=\frac{\left (1-a^2 x^2\right )^2}{60 a^2}+\frac{x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}-\frac{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac{4 \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx}{15 a}\\ &=\frac{2 \left (1-a^2 x^2\right )}{45 a^2}+\frac{\left (1-a^2 x^2\right )^2}{60 a^2}+\frac{4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac{x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}-\frac{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac{8 \int \tanh ^{-1}(a x) \, dx}{45 a}\\ &=\frac{2 \left (1-a^2 x^2\right )}{45 a^2}+\frac{\left (1-a^2 x^2\right )^2}{60 a^2}+\frac{8 x \tanh ^{-1}(a x)}{45 a}+\frac{4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac{x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}-\frac{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}-\frac{8}{45} \int \frac{x}{1-a^2 x^2} \, dx\\ &=\frac{2 \left (1-a^2 x^2\right )}{45 a^2}+\frac{\left (1-a^2 x^2\right )^2}{60 a^2}+\frac{8 x \tanh ^{-1}(a x)}{45 a}+\frac{4 x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{45 a}+\frac{x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{15 a}-\frac{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}{6 a^2}+\frac{4 \log \left (1-a^2 x^2\right )}{45 a^2}\\ \end{align*}

Mathematica [A]  time = 0.053743, size = 82, normalized size = 0.59 \[ \frac{3 a^4 x^4-14 a^2 x^2+16 \log \left (1-a^2 x^2\right )+4 a x \left (3 a^4 x^4-10 a^2 x^2+15\right ) \tanh ^{-1}(a x)+30 \left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2}{180 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*a^2*x^2 + 3*a^4*x^4 + 4*a*x*(15 - 10*a^2*x^2 + 3*a^4*x^4)*ArcTanh[a*x] + 30*(-1 + a^2*x^2)^3*ArcTanh[a*x]
^2 + 16*Log[1 - a^2*x^2])/(180*a^2)

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Maple [A]  time = 0.049, size = 219, normalized size = 1.6 \begin{align*}{\frac{{a}^{4} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{x}^{6}}{6}}-{\frac{{a}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{x}^{4}}{2}}+{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{x}^{2}}{2}}+{\frac{{a}^{3}{\it Artanh} \left ( ax \right ){x}^{5}}{15}}-{\frac{2\,a{\it Artanh} \left ( ax \right ){x}^{3}}{9}}+{\frac{x{\it Artanh} \left ( ax \right ) }{3\,a}}+{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{6\,{a}^{2}}}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{6\,{a}^{2}}}+{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{24\,{a}^{2}}}-{\frac{\ln \left ( ax-1 \right ) }{12\,{a}^{2}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{1}{12\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{\ln \left ( ax+1 \right ) }{12\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{24\,{a}^{2}}}+{\frac{{a}^{2}{x}^{4}}{60}}-{\frac{7\,{x}^{2}}{90}}+{\frac{4\,\ln \left ( ax-1 \right ) }{45\,{a}^{2}}}+{\frac{4\,\ln \left ( ax+1 \right ) }{45\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^4*arctanh(a*x)^2*x^6-1/2*a^2*arctanh(a*x)^2*x^4+1/2*arctanh(a*x)^2*x^2+1/15*a^3*arctanh(a*x)*x^5-2/9*a*a
rctanh(a*x)*x^3+1/3*x*arctanh(a*x)/a+1/6/a^2*arctanh(a*x)*ln(a*x-1)-1/6/a^2*arctanh(a*x)*ln(a*x+1)+1/24/a^2*ln
(a*x-1)^2-1/12/a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+1/12/a^2*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/12/a^2*ln(-1/2*a*x+1/
2)*ln(a*x+1)+1/24/a^2*ln(a*x+1)^2+1/60*a^2*x^4-7/90*x^2+4/45/a^2*ln(a*x-1)+4/45/a^2*ln(a*x+1)

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Maxima [A]  time = 0.964607, size = 126, normalized size = 0.91 \begin{align*} \frac{{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname{artanh}\left (a x\right )^{2}}{6 \, a^{2}} + \frac{{\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 + 4 \,{\left (3 \, a^{4} x^{5} - 10 \, a^{2} x^{3} + 15 \, x\right )} \operatorname{artanh}\left (a x\right )}{180 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a^2*x^2 - 1)^3*arctanh(a*x)^2/a^2 + 1/180*((3*a^2*x^4 - 14*x^2 + 16*log(a*x + 1)/a^2 + 16*log(a*x - 1)/a^
2)*a + 4*(3*a^4*x^5 - 10*a^2*x^3 + 15*x)*arctanh(a*x))/a

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Fricas [A]  time = 2.00645, size = 261, normalized size = 1.89 \begin{align*} \frac{6 \, a^{4} x^{4} - 28 \, a^{2} x^{2} + 15 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} + 4 \,{\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 ) + 32 \, \log \left (a^{2} x^{2} - 1\right )}{360 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/360*(6*a^4*x^4 - 28*a^2*x^2 + 15*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^2 + 4*(3*a^
5*x^5 - 10*a^3*x^3 + 15*a*x)*log(-(a*x + 1)/(a*x - 1)) + 32*log(a^2*x^2 - 1))/a^2

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Sympy [A]  time = 3.57061, size = 133, normalized size = 0.96 \begin{align*} \begin{cases} \frac{a^{4} x^{6} \operatorname{atanh}^{2}{\left (a x \right )}}{6} + \frac{a^{3} x^{5} \operatorname{atanh}{\left (a x \right )}}{15} - \frac{a^{2} x^{4} \operatorname{atanh}^{2}{\left (a x \right )}}{2} + \frac{a^{2} x^{4}}{60} - \frac{2 a x^{3} \operatorname{atanh}{\left (a x \right )}}{9} + \frac{x^{2} \operatorname{atanh}^{2}{\left (a x \right )}}{2} - \frac{7 x^{2}}{90} + \frac{x \operatorname{atanh}{\left (a x \right )}}{3 a} + \frac{8 \log{\left (x - \frac{1}{a} \right )}}{45 a^{2}} - \frac{\operatorname{atanh}^{2}{\left (a x \right )}}{6 a^{2}} + \frac{8 \operatorname{atanh}{\left (a x \right )}}{45 a^{2}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x**6*atanh(a*x)**2/6 + a**3*x**5*atanh(a*x)/15 - a**2*x**4*atanh(a*x)**2/2 + a**2*x**4/60 - 2*
a*x**3*atanh(a*x)/9 + x**2*atanh(a*x)**2/2 - 7*x**2/90 + x*atanh(a*x)/(3*a) + 8*log(x - 1/a)/(45*a**2) - atanh
(a*x)**2/(6*a**2) + 8*atanh(a*x)/(45*a**2), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.20939, size = 151, normalized size = 1.09 \begin{align*} \frac{1}{60} \, a^{2} x^{4} + \frac{1}{24} \,{\left (a^{4} x^{6} - 3 \, a^{2} x^{4} + 3 \, x^{2} - \frac{1}{a^{2}}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - \frac{7}{90} \, x^{2} + \frac{1}{90} \,{\left (3 \, a^{3} x^{5} - 10 \, a x^{3} + \frac{15 \, x}{a}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + \frac{4 \, \log \left (a^{2} x^{2} - 1\right )}{45 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*a^2*x^4 + 1/24*(a^4*x^6 - 3*a^2*x^4 + 3*x^2 - 1/a^2)*log(-(a*x + 1)/(a*x - 1))^2 - 7/90*x^2 + 1/90*(3*a^3
*x^5 - 10*a*x^3 + 15*x/a)*log(-(a*x + 1)/(a*x - 1)) + 4/45*log(a^2*x^2 - 1)/a^2