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

Optimal. Leaf size=43 \[ \frac{1}{2 a^3 (1-a x)}+\frac{3 \log (1-a x)}{4 a^3}+\frac{\log (a x+1)}{4 a^3} \]

[Out]

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

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Rubi [A]  time = 0.113338, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {6150, 88} \[ \frac{1}{2 a^3 (1-a x)}+\frac{3 \log (1-a x)}{4 a^3}+\frac{\log (a x+1)}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0243129, size = 33, normalized size = 0.77 \[ \frac{\frac{2}{1-a x}+3 \log (1-a x)+\log (a x+1)}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 36, normalized size = 0.8 \begin{align*}{\frac{\ln \left ( ax+1 \right ) }{4\,{a}^{3}}}-{\frac{1}{2\,{a}^{3} \left ( ax-1 \right ) }}+{\frac{3\,\ln \left ( ax-1 \right ) }{4\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.969361, size = 51, normalized size = 1.19 \begin{align*} -\frac{1}{2 \,{\left (a^{4} x - a^{3}\right )}} + \frac{\log \left (a x + 1\right )}{4 \, a^{3}} + \frac{3 \, \log \left (a x - 1\right )}{4 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.74665, size = 101, normalized size = 2.35 \begin{align*} \frac{{\left (a x - 1\right )} \log \left (a x + 1\right ) + 3 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{4 \,{\left (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)^2*x^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.385481, size = 34, normalized size = 0.79 \begin{align*} - \frac{1}{2 a^{4} x - 2 a^{3}} + \frac{\frac{3 \log{\left (x - \frac{1}{a} \right )}}{4} + \frac{\log{\left (x + \frac{1}{a} \right )}}{4}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*a**4*x - 2*a**3) + (3*log(x - 1/a)/4 + log(x + 1/a)/4)/a**3

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Giac [A]  time = 1.19562, size = 50, normalized size = 1.16 \begin{align*} \frac{\log \left ({\left | a x + 1 \right |}\right )}{4 \, a^{3}} + \frac{3 \, \log \left ({\left | a x - 1 \right |}\right )}{4 \, a^{3}} - \frac{1}{2 \,{\left (a x - 1\right )} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(abs(a*x + 1))/a^3 + 3/4*log(abs(a*x - 1))/a^3 - 1/2/((a*x - 1)*a^3)