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

Optimal. Leaf size=125 \[ \frac{3}{32 a^2 \left (1-a^2 x^2\right )}+\frac{1}{32 a^2 \left (1-a^2 x^2\right )^2}+\frac{\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{3 x \tanh ^{-1}(a x)}{16 a \left (1-a^2 x^2\right )}-\frac{x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)^2}{32 a^2} \]

[Out]

1/(32*a^2*(1 - a^2*x^2)^2) + 3/(32*a^2*(1 - a^2*x^2)) - (x*ArcTanh[a*x])/(8*a*(1 - a^2*x^2)^2) - (3*x*ArcTanh[
a*x])/(16*a*(1 - a^2*x^2)) - (3*ArcTanh[a*x]^2)/(32*a^2) + ArcTanh[a*x]^2/(4*a^2*(1 - a^2*x^2)^2)

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Rubi [A]  time = 0.093837, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5994, 5960, 5956, 261} \[ \frac{3}{32 a^2 \left (1-a^2 x^2\right )}+\frac{1}{32 a^2 \left (1-a^2 x^2\right )^2}+\frac{\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{3 x \tanh ^{-1}(a x)}{16 a \left (1-a^2 x^2\right )}-\frac{x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)^2}{32 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(32*a^2*(1 - a^2*x^2)^2) + 3/(32*a^2*(1 - a^2*x^2)) - (x*ArcTanh[a*x])/(8*a*(1 - a^2*x^2)^2) - (3*x*ArcTanh[
a*x])/(16*a*(1 - a^2*x^2)) - (3*ArcTanh[a*x]^2)/(32*a^2) + ArcTanh[a*x]^2/(4*a^2*(1 - a^2*x^2)^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 5960

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q + 1))
/(4*c*d*(q + 1)^2), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x], x] -
Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]))/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*
d + e, 0] && LtQ[q, -1] && NeQ[q, -3/2]

Rule 5956

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{x \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx &=\frac{\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{\int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx}{2 a}\\ &=\frac{1}{32 a^2 \left (1-a^2 x^2\right )^2}-\frac{x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac{\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac{3 \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx}{8 a}\\ &=\frac{1}{32 a^2 \left (1-a^2 x^2\right )^2}-\frac{x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac{3 x \tanh ^{-1}(a x)}{16 a \left (1-a^2 x^2\right )}-\frac{3 \tanh ^{-1}(a x)^2}{32 a^2}+\frac{\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}+\frac{3}{16} \int \frac{x}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac{1}{32 a^2 \left (1-a^2 x^2\right )^2}+\frac{3}{32 a^2 \left (1-a^2 x^2\right )}-\frac{x \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac{3 x \tanh ^{-1}(a x)}{16 a \left (1-a^2 x^2\right )}-\frac{3 \tanh ^{-1}(a x)^2}{32 a^2}+\frac{\tanh ^{-1}(a x)^2}{4 a^2 \left (1-a^2 x^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.0698054, size = 71, normalized size = 0.57 \[ \frac{-3 a^2 x^2+2 a x \left (3 a^2 x^2-5\right ) \tanh ^{-1}(a x)+\left (-3 a^4 x^4+6 a^2 x^2+5\right ) \tanh ^{-1}(a x)^2+4}{32 a^2 \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4 - 3*a^2*x^2 + 2*a*x*(-5 + 3*a^2*x^2)*ArcTanh[a*x] + (5 + 6*a^2*x^2 - 3*a^4*x^4)*ArcTanh[a*x]^2)/(32*a^2*(-1
+ a^2*x^2)^2)

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Maple [B]  time = 0.058, size = 247, normalized size = 2. \begin{align*}{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{4\,{a}^{2} \left ({a}^{2}{x}^{2}-1 \right ) ^{2}}}-{\frac{{\it Artanh} \left ( ax \right ) }{32\,{a}^{2} \left ( ax-1 \right ) ^{2}}}+{\frac{3\,{\it Artanh} \left ( ax \right ) }{32\,{a}^{2} \left ( ax-1 \right ) }}+{\frac{3\,{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{32\,{a}^{2}}}+{\frac{{\it Artanh} \left ( ax \right ) }{32\,{a}^{2} \left ( ax+1 \right ) ^{2}}}+{\frac{3\,{\it Artanh} \left ( ax \right ) }{32\,{a}^{2} \left ( ax+1 \right ) }}-{\frac{3\,{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{32\,{a}^{2}}}+{\frac{3\, \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{128\,{a}^{2}}}-{\frac{3\,\ln \left ( ax-1 \right ) }{64\,{a}^{2}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{3\, \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{128\,{a}^{2}}}-{\frac{3\,\ln \left ( ax+1 \right ) }{64\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{3}{64\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{1}{128\,{a}^{2} \left ( ax-1 \right ) ^{2}}}-{\frac{7}{128\,{a}^{2} \left ( ax-1 \right ) }}+{\frac{1}{128\,{a}^{2} \left ( ax+1 \right ) ^{2}}}+{\frac{7}{128\,{a}^{2} \left ( ax+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a^2/(a^2*x^2-1)^2*arctanh(a*x)^2-1/32/a^2*arctanh(a*x)/(a*x-1)^2+3/32/a^2*arctanh(a*x)/(a*x-1)+3/32/a^2*ar
ctanh(a*x)*ln(a*x-1)+1/32/a^2*arctanh(a*x)/(a*x+1)^2+3/32/a^2*arctanh(a*x)/(a*x+1)-3/32/a^2*arctanh(a*x)*ln(a*
x+1)+3/128/a^2*ln(a*x-1)^2-3/64/a^2*ln(a*x-1)*ln(1/2+1/2*a*x)+3/128/a^2*ln(a*x+1)^2-3/64/a^2*ln(-1/2*a*x+1/2)*
ln(a*x+1)+3/64/a^2*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)+1/128/a^2/(a*x-1)^2-7/128/a^2/(a*x-1)+1/128/a^2/(a*x+1)^2+
7/128/a^2/(a*x+1)

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Maxima [A]  time = 0.982664, size = 278, normalized size = 2.22 \begin{align*} \frac{{\left (\frac{2 \,{\left (3 \, a^{2} x^{3} - 5 \, x\right )}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - \frac{3 \, \log \left (a x + 1\right )}{a} + \frac{3 \, \log \left (a x - 1\right )}{a}\right )} \operatorname{artanh}\left (a x\right )}{32 \, a} - \frac{12 \, a^{2} x^{2} - 3 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 6 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 3 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 16}{128 \,{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )}} + \frac{\operatorname{artanh}\left (a x\right )^{2}}{4 \,{\left (a^{2} x^{2} - 1\right )}^{2} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(2*(3*a^2*x^3 - 5*x)/(a^4*x^4 - 2*a^2*x^2 + 1) - 3*log(a*x + 1)/a + 3*log(a*x - 1)/a)*arctanh(a*x)/a - 1/
128*(12*a^2*x^2 - 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 6*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)*log(a*
x - 1) - 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2 - 16)/(a^6*x^4 - 2*a^4*x^2 + a^2) + 1/4*arctanh(a*x)^2/((a
^2*x^2 - 1)^2*a^2)

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Fricas [A]  time = 2.0753, size = 219, normalized size = 1.75 \begin{align*} -\frac{12 \, a^{2} x^{2} +{\left (3 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 5\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 4 \,{\left (3 \, a^{3} x^{3} - 5 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - 16}{128 \,{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(12*a^2*x^2 + (3*a^4*x^4 - 6*a^2*x^2 - 5)*log(-(a*x + 1)/(a*x - 1))^2 - 4*(3*a^3*x^3 - 5*a*x)*log(-(a*x
+ 1)/(a*x - 1)) - 16)/(a^6*x^4 - 2*a^4*x^2 + a^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x \operatorname{atanh}^{2}{\left (a x \right )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(a*x)**2/(-a**2*x**2+1)**3,x)

[Out]

-Integral(x*atanh(a*x)**2/(a**6*x**6 - 3*a**4*x**4 + 3*a**2*x**2 - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x \operatorname{artanh}\left (a x\right )^{2}}{{\left (a^{2} x^{2} - 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-x*arctanh(a*x)^2/(a^2*x^2 - 1)^3, x)