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

Optimal. Leaf size=127 \[ \frac{x^4}{32 \left (1-a^2 x^2\right )^2}-\frac{3}{32 a^4 \left (1-a^2 x^2\right )}+\frac{x^4 \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac{x^3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)}{16 a^3 \left (1-a^2 x^2\right )}-\frac{3 \tanh ^{-1}(a x)^2}{32 a^4} \]

[Out]

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

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Rubi [A]  time = 0.183527, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6008, 6002, 5998, 5948} \[ \frac{x^4}{32 \left (1-a^2 x^2\right )^2}-\frac{3}{32 a^4 \left (1-a^2 x^2\right )}+\frac{x^4 \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac{x^3 \tanh ^{-1}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac{3 x \tanh ^{-1}(a x)}{16 a^3 \left (1-a^2 x^2\right )}-\frac{3 \tanh ^{-1}(a x)^2}{32 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 6002

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

Rule 5998

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(
q + 1))/(4*c^3*d*(q + 1)^2), x] + (Dist[1/(2*c^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*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q, -5/2]

Rule 5948

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

Rubi steps

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

Mathematica [A]  time = 0.0927196, size = 71, normalized size = 0.56 \[ \frac{5 a^2 x^2+\left (5 a^4 x^4+6 a^2 x^2-3\right ) \tanh ^{-1}(a x)^2+\left (6 a x-10 a^3 x^3\right ) \tanh ^{-1}(a x)-4}{32 a^4 \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4 + 5*a^2*x^2 + (6*a*x - 10*a^3*x^3)*ArcTanh[a*x] + (-3 + 6*a^2*x^2 + 5*a^4*x^4)*ArcTanh[a*x]^2)/(32*a^4*(-1
+ a^2*x^2)^2)

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Maple [B]  time = 0.07, size = 297, normalized size = 2.3 \begin{align*}{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{16\,{a}^{4} \left ( ax-1 \right ) ^{2}}}+{\frac{3\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{16\,{a}^{4} \left ( ax-1 \right ) }}+{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{16\,{a}^{4} \left ( ax+1 \right ) ^{2}}}-{\frac{3\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{16\,{a}^{4} \left ( ax+1 \right ) }}-{\frac{{\it Artanh} \left ( ax \right ) }{32\,{a}^{4} \left ( ax-1 \right ) ^{2}}}-{\frac{5\,{\it Artanh} \left ( ax \right ) }{32\,{a}^{4} \left ( ax-1 \right ) }}-{\frac{5\,{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{32\,{a}^{4}}}+{\frac{{\it Artanh} \left ( ax \right ) }{32\,{a}^{4} \left ( ax+1 \right ) ^{2}}}-{\frac{5\,{\it Artanh} \left ( ax \right ) }{32\,{a}^{4} \left ( ax+1 \right ) }}+{\frac{5\,{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{32\,{a}^{4}}}-{\frac{5\, \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{128\,{a}^{4}}}+{\frac{5\,\ln \left ( ax-1 \right ) }{64\,{a}^{4}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{5\, \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{128\,{a}^{4}}}-{\frac{5}{64\,{a}^{4}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{5\,\ln \left ( ax+1 \right ) }{64\,{a}^{4}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{1}{128\,{a}^{4} \left ( ax-1 \right ) ^{2}}}+{\frac{9}{128\,{a}^{4} \left ( ax-1 \right ) }}+{\frac{1}{128\,{a}^{4} \left ( ax+1 \right ) ^{2}}}-{\frac{9}{128\,{a}^{4} \left ( ax+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a^4*arctanh(a*x)^2/(a*x-1)^2+3/16/a^4*arctanh(a*x)^2/(a*x-1)+1/16/a^4*arctanh(a*x)^2/(a*x+1)^2-3/16/a^4*a
rctanh(a*x)^2/(a*x+1)-1/32/a^4*arctanh(a*x)/(a*x-1)^2-5/32/a^4*arctanh(a*x)/(a*x-1)-5/32/a^4*arctanh(a*x)*ln(a
*x-1)+1/32/a^4*arctanh(a*x)/(a*x+1)^2-5/32/a^4*arctanh(a*x)/(a*x+1)+5/32/a^4*arctanh(a*x)*ln(a*x+1)-5/128/a^4*
ln(a*x-1)^2+5/64/a^4*ln(a*x-1)*ln(1/2+1/2*a*x)-5/128/a^4*ln(a*x+1)^2-5/64/a^4*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)
+5/64/a^4*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/128/a^4/(a*x-1)^2+9/128/a^4/(a*x-1)+1/128/a^4/(a*x+1)^2-9/128/a^4/(a*x+
1)

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Maxima [B]  time = 0.985992, size = 305, normalized size = 2.4 \begin{align*} -\frac{1}{32} \, a{\left (\frac{2 \,{\left (5 \, a^{2} x^{3} - 3 \, x\right )}}{a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}} - \frac{5 \, \log \left (a x + 1\right )}{a^{5}} + \frac{5 \, \log \left (a x - 1\right )}{a^{5}}\right )} \operatorname{artanh}\left (a x\right ) + \frac{{\left (20 \, a^{2} x^{2} - 5 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 10 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 5 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 16\right )} a^{2}}{128 \,{\left (a^{10} x^{4} - 2 \, a^{8} x^{2} + a^{6}\right )}} + \frac{{\left (2 \, a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )^{2}}{4 \,{\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*a*(2*(5*a^2*x^3 - 3*x)/(a^8*x^4 - 2*a^6*x^2 + a^4) - 5*log(a*x + 1)/a^5 + 5*log(a*x - 1)/a^5)*arctanh(a*
x) + 1/128*(20*a^2*x^2 - 5*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 10*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1
)*log(a*x - 1) - 5*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2 - 16)*a^2/(a^10*x^4 - 2*a^8*x^2 + a^6) + 1/4*(2*a^
2*x^2 - 1)*arctanh(a*x)^2/(a^8*x^4 - 2*a^6*x^2 + a^4)

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Fricas [A]  time = 1.9317, size = 217, normalized size = 1.71 \begin{align*} \frac{20 \, a^{2} x^{2} +{\left (5 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 3\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 4 \,{\left (5 \, a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - 16}{128 \,{\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(20*a^2*x^2 + (5*a^4*x^4 + 6*a^2*x^2 - 3)*log(-(a*x + 1)/(a*x - 1))^2 - 4*(5*a^3*x^3 - 3*a*x)*log(-(a*x
+ 1)/(a*x - 1)) - 16)/(a^8*x^4 - 2*a^6*x^2 + a^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{3} \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**3*atanh(a*x)**2/(-a**2*x**2+1)**3,x)

[Out]

-Integral(x**3*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^{3} \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^3*arctanh(a*x)^2/(-a^2*x^2+1)^3,x, algorithm="giac")

[Out]

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