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

Optimal. Leaf size=77 \[ -\frac{x^3}{16 a \left (1-a^2 x^2\right )^2}+\frac{3 x}{32 a^3 \left (1-a^2 x^2\right )}+\frac{x^4 \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)}{32 a^4} \]

[Out]

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

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Rubi [A]  time = 0.0627185, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6008, 288, 206} \[ -\frac{x^3}{16 a \left (1-a^2 x^2\right )^2}+\frac{3 x}{32 a^3 \left (1-a^2 x^2\right )}+\frac{x^4 \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}-\frac{3 \tanh ^{-1}(a x)}{32 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^3/(16*a*(1 - a^2*x^2)^2) + (3*x)/(32*a^3*(1 - a^2*x^2)) - (3*ArcTanh[a*x])/(32*a^4) + (x^4*ArcTanh[a*x])/(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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0830387, size = 98, normalized size = 1.27 \[ -\frac{5 x}{32 a^3 \left (a^2 x^2-1\right )}-\frac{x}{16 a^3 \left (a^2 x^2-1\right )^2}+\frac{\left (2 a^2 x^2-1\right ) \tanh ^{-1}(a x)}{4 a^4 \left (a^2 x^2-1\right )^2}-\frac{5 \log (1-a x)}{64 a^4}+\frac{5 \log (a x+1)}{64 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(16*a^3*(-1 + a^2*x^2)^2) - (5*x)/(32*a^3*(-1 + a^2*x^2)) + ((-1 + 2*a^2*x^2)*ArcTanh[a*x])/(4*a^4*(-1 + a^
2*x^2)^2) - (5*Log[1 - a*x])/(64*a^4) + (5*Log[1 + a*x])/(64*a^4)

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Maple [A]  time = 0.043, size = 136, normalized size = 1.8 \begin{align*}{\frac{{\it Artanh} \left ( ax \right ) }{16\,{a}^{4} \left ( ax-1 \right ) ^{2}}}+{\frac{3\,{\it Artanh} \left ( ax \right ) }{16\,{a}^{4} \left ( ax-1 \right ) }}+{\frac{{\it Artanh} \left ( ax \right ) }{16\,{a}^{4} \left ( ax+1 \right ) ^{2}}}-{\frac{3\,{\it Artanh} \left ( ax \right ) }{16\,{a}^{4} \left ( ax+1 \right ) }}-{\frac{1}{64\,{a}^{4} \left ( ax-1 \right ) ^{2}}}-{\frac{5}{64\,{a}^{4} \left ( ax-1 \right ) }}-{\frac{5\,\ln \left ( ax-1 \right ) }{64\,{a}^{4}}}+{\frac{1}{64\,{a}^{4} \left ( ax+1 \right ) ^{2}}}-{\frac{5}{64\,{a}^{4} \left ( ax+1 \right ) }}+{\frac{5\,\ln \left ( ax+1 \right ) }{64\,{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.96068, size = 134, normalized size = 1.74 \begin{align*} -\frac{1}{64} \, 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 )} + \frac{{\left (2 \, a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )}{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)/(-a^2*x^2+1)^3,x, algorithm="maxima")

[Out]

-1/64*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) + 1/4*(2*a
^2*x^2 - 1)*arctanh(a*x)/(a^8*x^4 - 2*a^6*x^2 + a^4)

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Fricas [A]  time = 2.02289, size = 151, normalized size = 1.96 \begin{align*} -\frac{10 \, a^{3} x^{3} - 6 \, a x -{\left (5 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 3\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{64 \,{\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)/(-a^2*x^2+1)^3,x, algorithm="fricas")

[Out]

-1/64*(10*a^3*x^3 - 6*a*x - (5*a^4*x^4 + 6*a^2*x^2 - 3)*log(-(a*x + 1)/(a*x - 1)))/(a^8*x^4 - 2*a^6*x^2 + a^4)

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Sympy [A]  time = 4.0009, size = 158, normalized size = 2.05 \begin{align*} \begin{cases} \frac{5 a^{4} x^{4} \operatorname{atanh}{\left (a x \right )}}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} - \frac{5 a^{3} x^{3}}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} + \frac{6 a^{2} x^{2} \operatorname{atanh}{\left (a x \right )}}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} + \frac{3 a x}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} - \frac{3 \operatorname{atanh}{\left (a x \right )}}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} & \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**3*atanh(a*x)/(-a**2*x**2+1)**3,x)

[Out]

Piecewise((5*a**4*x**4*atanh(a*x)/(32*a**8*x**4 - 64*a**6*x**2 + 32*a**4) - 5*a**3*x**3/(32*a**8*x**4 - 64*a**
6*x**2 + 32*a**4) + 6*a**2*x**2*atanh(a*x)/(32*a**8*x**4 - 64*a**6*x**2 + 32*a**4) + 3*a*x/(32*a**8*x**4 - 64*
a**6*x**2 + 32*a**4) - 3*atanh(a*x)/(32*a**8*x**4 - 64*a**6*x**2 + 32*a**4), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.23055, size = 127, normalized size = 1.65 \begin{align*} \frac{5 \, \log \left ({\left | a x + 1 \right |}\right )}{64 \, a^{4}} - \frac{5 \, \log \left ({\left | a x - 1 \right |}\right )}{64 \, a^{4}} - \frac{5 \, a^{2} x^{3} - 3 \, x}{32 \,{\left (a^{2} x^{2} - 1\right )}^{2} a^{3}} + \frac{{\left (2 \, a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{8 \,{\left (a^{2} x^{2} - 1\right )}^{2} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/64*log(abs(a*x + 1))/a^4 - 5/64*log(abs(a*x - 1))/a^4 - 1/32*(5*a^2*x^3 - 3*x)/((a^2*x^2 - 1)^2*a^3) + 1/8*(
2*a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))/((a^2*x^2 - 1)^2*a^4)