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

Optimal. Leaf size=115 \[ -\frac{2 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{3 a}+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^2+\frac{2 \tanh ^{-1}(a x)^2}{3 a}-\frac{4 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)}{3 a}-\frac{x}{3} \]

[Out]

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

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Rubi [A]  time = 0.10326, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {5944, 5910, 5984, 5918, 2402, 2315, 8} \[ -\frac{2 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{3 a}+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^2+\frac{2 \tanh ^{-1}(a x)^2}{3 a}-\frac{4 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)}{3 a}-\frac{x}{3} \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)*ArcTanh[a*x]^2,x]

[Out]

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

Rule 5944

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(b*p*(d + e*x^2)^q
*(a + b*ArcTanh[c*x])^(p - 1))/(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])^p, x], x] - Dist[(b^2*d*p*(p - 1))/(2*q*(2*q + 1)), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x]
)^(p - 2), x], x] + Simp[(x*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^p)/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x]
&& EqQ[c^2*d + e, 0] && GtQ[q, 0] && GtQ[p, 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 5984

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

Rule 5918

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

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2 \, dx &=\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac{\int 1 \, dx}{3}+\frac{2}{3} \int \tanh ^{-1}(a x)^2 \, dx\\ &=-\frac{x}{3}+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^2+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac{1}{3} (4 a) \int \frac{x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=-\frac{x}{3}+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac{2 \tanh ^{-1}(a x)^2}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^2+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac{4}{3} \int \frac{\tanh ^{-1}(a x)}{1-a x} \, dx\\ &=-\frac{x}{3}+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac{2 \tanh ^{-1}(a x)^2}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^2+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac{4 \tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{3 a}+\frac{4}{3} \int \frac{\log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac{x}{3}+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac{2 \tanh ^{-1}(a x)^2}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^2+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac{4 \tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{3 a}-\frac{4 \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-a x}\right )}{3 a}\\ &=-\frac{x}{3}+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac{2 \tanh ^{-1}(a x)^2}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^2+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac{4 \tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{3 a}-\frac{2 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{3 a}\\ \end{align*}

Mathematica [A]  time = 0.0498392, size = 71, normalized size = 0.62 \[ -\frac{-2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x) \left (a^2 x^2+4 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-1\right )+a x+(a x-1)^2 (a x+2) \tanh ^{-1}(a x)^2}{3 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 182, normalized size = 1.6 \begin{align*} -{\frac{{a}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{x}^{3}}{3}}+x \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}-{\frac{a{\it Artanh} \left ( ax \right ){x}^{2}}{3}}+{\frac{2\,{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{3\,a}}+{\frac{2\,{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{3\,a}}-{\frac{x}{3}}-{\frac{\ln \left ( ax-1 \right ) }{6\,a}}+{\frac{\ln \left ( ax+1 \right ) }{6\,a}}+{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{6\,a}}-{\frac{2}{3\,a}{\it dilog} \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{\ln \left ( ax-1 \right ) }{3\,a}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{1}{3\,a}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{\ln \left ( ax+1 \right ) }{3\,a}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }-{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{6\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.98019, size = 194, normalized size = 1.69 \begin{align*} -\frac{1}{6} \, a^{2}{\left (\frac{2 \, a x + \log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) - \log \left (a x - 1\right )^{2} + \log \left (a x - 1\right )}{a^{3}} + \frac{4 \,{\left (\log \left (a x - 1\right ) \log \left (\frac{1}{2} \, a x + \frac{1}{2}\right ) +{\rm Li}_2\left (-\frac{1}{2} \, a x + \frac{1}{2}\right )\right )}}{a^{3}} - \frac{\log \left (a x + 1\right )}{a^{3}}\right )} - \frac{1}{3} \,{\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 \operatorname{artanh}\left (a x\right ) - \frac{1}{3} \,{\left (a^{2} x^{3} - 3 \, x\right )} \operatorname{artanh}\left (a x\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a^2*x^2 - 1)*arctanh(a*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int a^{2} x^{2} \operatorname{atanh}^{2}{\left (a x \right )}\, dx - \int - \operatorname{atanh}^{2}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a**2*x**2*atanh(a*x)**2, x) - Integral(-atanh(a*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -{\left (a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(a^2*x^2 - 1)*arctanh(a*x)^2, x)