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

Optimal. Leaf size=95 \[ \frac{1-a^2 x^2}{12 a^2}+\frac{\log \left (1-a^2 x^2\right )}{6 a^2}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac{x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}+\frac{x \tanh ^{-1}(a x)}{3 a} \]

[Out]

(1 - a^2*x^2)/(12*a^2) + (x*ArcTanh[a*x])/(3*a) + (x*(1 - a^2*x^2)*ArcTanh[a*x])/(6*a) - ((1 - a^2*x^2)^2*ArcT
anh[a*x]^2)/(4*a^2) + Log[1 - a^2*x^2]/(6*a^2)

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Rubi [A]  time = 0.0486081, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {5994, 5942, 5910, 260} \[ \frac{1-a^2 x^2}{12 a^2}+\frac{\log \left (1-a^2 x^2\right )}{6 a^2}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac{x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}+\frac{x \tanh ^{-1}(a x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[x*(1 - a^2*x^2)*ArcTanh[a*x]^2,x]

[Out]

(1 - a^2*x^2)/(12*a^2) + (x*ArcTanh[a*x])/(3*a) + (x*(1 - a^2*x^2)*ArcTanh[a*x])/(6*a) - ((1 - a^2*x^2)^2*ArcT
anh[a*x]^2)/(4*a^2) + Log[1 - a^2*x^2]/(6*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 ) \tanh ^{-1}(a x)^2 \, dx &=-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac{\int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx}{2 a}\\ &=\frac{1-a^2 x^2}{12 a^2}+\frac{x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac{\int \tanh ^{-1}(a x) \, dx}{3 a}\\ &=\frac{1-a^2 x^2}{12 a^2}+\frac{x \tanh ^{-1}(a x)}{3 a}+\frac{x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}-\frac{1}{3} \int \frac{x}{1-a^2 x^2} \, dx\\ &=\frac{1-a^2 x^2}{12 a^2}+\frac{x \tanh ^{-1}(a x)}{3 a}+\frac{x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{6 a}-\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{4 a^2}+\frac{\log \left (1-a^2 x^2\right )}{6 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0295572, size = 66, normalized size = 0.69 \[ \frac{-a^2 x^2+2 \log \left (1-a^2 x^2\right )-3 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2+\left (6 a x-2 a^3 x^3\right ) \tanh ^{-1}(a x)}{12 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^2*x^2) + (6*a*x - 2*a^3*x^3)*ArcTanh[a*x] - 3*(-1 + a^2*x^2)^2*ArcTanh[a*x]^2 + 2*Log[1 - a^2*x^2])/(12*a
^2)

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Maple [B]  time = 0.045, size = 185, normalized size = 2. \begin{align*} -{\frac{{a}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{x}^{4}}{4}}+{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{x}^{2}}{2}}-{\frac{a{\it Artanh} \left ( ax \right ){x}^{3}}{6}}+{\frac{x{\it Artanh} \left ( ax \right ) }{2\,a}}+{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{4\,{a}^{2}}}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{4\,{a}^{2}}}+{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{16\,{a}^{2}}}-{\frac{\ln \left ( ax-1 \right ) }{8\,{a}^{2}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{1}{8\,{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 ) }{8\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{16\,{a}^{2}}}-{\frac{{x}^{2}}{12}}+{\frac{\ln \left ( ax-1 \right ) }{6\,{a}^{2}}}+{\frac{\ln \left ( ax+1 \right ) }{6\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*arctanh(a*x)^2*x^4+1/2*arctanh(a*x)^2*x^2-1/6*a*arctanh(a*x)*x^3+1/2*x*arctanh(a*x)/a+1/4/a^2*arctanh
(a*x)*ln(a*x-1)-1/4/a^2*arctanh(a*x)*ln(a*x+1)+1/16/a^2*ln(a*x-1)^2-1/8/a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+1/8/a^2*
ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)-1/8/a^2*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/16/a^2*ln(a*x+1)^2-1/12*x^2+1/6/a^2*ln(a
*x-1)+1/6/a^2*ln(a*x+1)

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Maxima [A]  time = 0.965173, size = 100, normalized size = 1.05 \begin{align*} -\frac{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname{artanh}\left (a x\right )^{2}}{4 \, a^{2}} - \frac{{\left (x^{2} - \frac{2 \, \log \left (a x + 1\right )}{a^{2}} - \frac{2 \, \log \left (a x - 1\right )}{a^{2}}\right )} a + 2 \,{\left (a^{2} x^{3} - 3 \, x\right )} \operatorname{artanh}\left (a x\right )}{12 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2*x^2 - 1)^2*arctanh(a*x)^2/a^2 - 1/12*((x^2 - 2*log(a*x + 1)/a^2 - 2*log(a*x - 1)/a^2)*a + 2*(a^2*x^3
- 3*x)*arctanh(a*x))/a

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Fricas [A]  time = 2.19842, size = 203, normalized size = 2.14 \begin{align*} -\frac{4 \, a^{2} x^{2} + 3 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} + 4 \,{\left (a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - 8 \, \log \left (a^{2} x^{2} - 1\right )}{48 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(4*a^2*x^2 + 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 + 4*(a^3*x^3 - 3*a*x)*log(-(a*x + 1
)/(a*x - 1)) - 8*log(a^2*x^2 - 1))/a^2

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Sympy [A]  time = 1.82278, size = 88, normalized size = 0.93 \begin{align*} \begin{cases} - \frac{a^{2} x^{4} \operatorname{atanh}^{2}{\left (a x \right )}}{4} - \frac{a x^{3} \operatorname{atanh}{\left (a x \right )}}{6} + \frac{x^{2} \operatorname{atanh}^{2}{\left (a x \right )}}{2} - \frac{x^{2}}{12} + \frac{x \operatorname{atanh}{\left (a x \right )}}{2 a} + \frac{\log{\left (x - \frac{1}{a} \right )}}{3 a^{2}} - \frac{\operatorname{atanh}^{2}{\left (a x \right )}}{4 a^{2}} + \frac{\operatorname{atanh}{\left (a x \right )}}{3 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)*atanh(a*x)**2,x)

[Out]

Piecewise((-a**2*x**4*atanh(a*x)**2/4 - a*x**3*atanh(a*x)/6 + x**2*atanh(a*x)**2/2 - x**2/12 + x*atanh(a*x)/(2
*a) + log(x - 1/a)/(3*a**2) - atanh(a*x)**2/(4*a**2) + atanh(a*x)/(3*a**2), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.16943, size = 115, normalized size = 1.21 \begin{align*} -\frac{1}{16} \,{\left (a^{2} x^{4} - 2 \, x^{2} + \frac{1}{a^{2}}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - \frac{1}{12} \, x^{2} - \frac{1}{12} \,{\left (a x^{3} - \frac{3 \, x}{a}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + \frac{\log \left (a^{2} x^{2} - 1\right )}{6 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(a^2*x^4 - 2*x^2 + 1/a^2)*log(-(a*x + 1)/(a*x - 1))^2 - 1/12*x^2 - 1/12*(a*x^3 - 3*x/a)*log(-(a*x + 1)/(
a*x - 1)) + 1/6*log(a^2*x^2 - 1)/a^2