3.942 \(\int \frac{e^{\tanh ^{-1}(a x)} x^2}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=56 \[ -\frac{1}{4 a^3 (1-a x)}-\frac{1}{8 a^3 (a x+1)}+\frac{1}{8 a^3 (1-a x)^2}-\frac{\tanh ^{-1}(a x)}{8 a^3} \]

[Out]

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

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Rubi [A]  time = 0.122589, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6150, 88, 207} \[ -\frac{1}{4 a^3 (1-a x)}-\frac{1}{8 a^3 (a x+1)}+\frac{1}{8 a^3 (1-a x)^2}-\frac{\tanh ^{-1}(a x)}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^2)/(1 - a^2*x^2)^(5/2),x]

[Out]

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

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 207

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

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)} x^2}{\left (1-a^2 x^2\right )^{5/2}} \, dx &=\int \frac{x^2}{(1-a x)^3 (1+a x)^2} \, dx\\ &=\int \left (-\frac{1}{4 a^2 (-1+a x)^3}-\frac{1}{4 a^2 (-1+a x)^2}+\frac{1}{8 a^2 (1+a x)^2}+\frac{1}{8 a^2 \left (-1+a^2 x^2\right )}\right ) \, dx\\ &=\frac{1}{8 a^3 (1-a x)^2}-\frac{1}{4 a^3 (1-a x)}-\frac{1}{8 a^3 (1+a x)}+\frac{\int \frac{1}{-1+a^2 x^2} \, dx}{8 a^2}\\ &=\frac{1}{8 a^3 (1-a x)^2}-\frac{1}{4 a^3 (1-a x)}-\frac{1}{8 a^3 (1+a x)}-\frac{\tanh ^{-1}(a x)}{8 a^3}\\ \end{align*}

Mathematica [A]  time = 0.0302282, size = 52, normalized size = 0.93 \[ \frac{a^2 x^2+3 a x-(a x-1)^2 (a x+1) \tanh ^{-1}(a x)-2}{8 a^3 (a x-1)^2 (a x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^2)/(1 - a^2*x^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.035, size = 60, normalized size = 1.1 \begin{align*} -{\frac{1}{8\,{a}^{3} \left ( ax+1 \right ) }}-{\frac{\ln \left ( ax+1 \right ) }{16\,{a}^{3}}}+{\frac{1}{8\,{a}^{3} \left ( ax-1 \right ) ^{2}}}+{\frac{1}{4\,{a}^{3} \left ( ax-1 \right ) }}+{\frac{\ln \left ( ax-1 \right ) }{16\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.94536, size = 88, normalized size = 1.57 \begin{align*} \frac{a^{2} x^{2} + 3 \, a x - 2}{8 \,{\left (a^{6} x^{3} - a^{5} x^{2} - a^{4} x + a^{3}\right )}} - \frac{\log \left (a x + 1\right )}{16 \, a^{3}} + \frac{\log \left (a x - 1\right )}{16 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.33037, size = 208, normalized size = 3.71 \begin{align*} \frac{2 \, a^{2} x^{2} + 6 \, a x -{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) +{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 4}{16 \,{\left (a^{6} x^{3} - a^{5} x^{2} - a^{4} x + a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.47591, size = 61, normalized size = 1.09 \begin{align*} \frac{a^{2} x^{2} + 3 a x - 2}{8 a^{6} x^{3} - 8 a^{5} x^{2} - 8 a^{4} x + 8 a^{3}} - \frac{- \frac{\log{\left (x - \frac{1}{a} \right )}}{16} + \frac{\log{\left (x + \frac{1}{a} \right )}}{16}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18126, size = 77, normalized size = 1.38 \begin{align*} -\frac{\log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{3}} + \frac{\log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{3}} + \frac{a^{2} x^{2} + 3 \, a x - 2}{8 \,{\left (a x + 1\right )}{\left (a x - 1\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*log(abs(a*x + 1))/a^3 + 1/16*log(abs(a*x - 1))/a^3 + 1/8*(a^2*x^2 + 3*a*x - 2)/((a*x + 1)*(a*x - 1)^2*a^
3)