3.929 \(\int \frac{e^{\tanh ^{-1}(a x)} x^4}{(1-a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=58 \[ \frac{x^2}{2 a^3}+\frac{x}{a^4}+\frac{1}{2 a^5 (1-a x)}+\frac{7 \log (1-a x)}{4 a^5}+\frac{\log (a x+1)}{4 a^5} \]

[Out]

x/a^4 + x^2/(2*a^3) + 1/(2*a^5*(1 - a*x)) + (7*Log[1 - a*x])/(4*a^5) + Log[1 + a*x]/(4*a^5)

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Rubi [A]  time = 0.115073, antiderivative size = 58, 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} \[ \frac{x^2}{2 a^3}+\frac{x}{a^4}+\frac{1}{2 a^5 (1-a x)}+\frac{7 \log (1-a x)}{4 a^5}+\frac{\log (a x+1)}{4 a^5} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(3/2),x]

[Out]

x/a^4 + x^2/(2*a^3) + 1/(2*a^5*(1 - a*x)) + (7*Log[1 - a*x])/(4*a^5) + Log[1 + a*x]/(4*a^5)

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^4}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\int \frac{x^4}{(1-a x)^2 (1+a x)} \, dx\\ &=\int \left (\frac{1}{a^4}+\frac{x}{a^3}+\frac{1}{2 a^4 (-1+a x)^2}+\frac{7}{4 a^4 (-1+a x)}+\frac{1}{4 a^4 (1+a x)}\right ) \, dx\\ &=\frac{x}{a^4}+\frac{x^2}{2 a^3}+\frac{1}{2 a^5 (1-a x)}+\frac{7 \log (1-a x)}{4 a^5}+\frac{\log (1+a x)}{4 a^5}\\ \end{align*}

Mathematica [A]  time = 0.060876, size = 45, normalized size = 0.78 \[ \frac{2 \left (a^2 x^2+2 a x+\frac{1}{1-a x}\right )+7 \log (1-a x)+\log (a x+1)}{4 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(3/2),x]

[Out]

(2*(2*a*x + a^2*x^2 + (1 - a*x)^(-1)) + 7*Log[1 - a*x] + Log[1 + a*x])/(4*a^5)

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Maple [A]  time = 0.035, size = 49, normalized size = 0.8 \begin{align*}{\frac{{x}^{2}}{2\,{a}^{3}}}+{\frac{x}{{a}^{4}}}+{\frac{\ln \left ( ax+1 \right ) }{4\,{a}^{5}}}-{\frac{1}{2\,{a}^{5} \left ( ax-1 \right ) }}+{\frac{7\,\ln \left ( ax-1 \right ) }{4\,{a}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^2*x^4,x)

[Out]

1/2*x^2/a^3+x/a^4+1/4*ln(a*x+1)/a^5-1/2/a^5/(a*x-1)+7/4/a^5*ln(a*x-1)

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Maxima [A]  time = 0.94287, size = 70, normalized size = 1.21 \begin{align*} -\frac{1}{2 \,{\left (a^{6} x - a^{5}\right )}} + \frac{a x^{2} + 2 \, x}{2 \, a^{4}} + \frac{\log \left (a x + 1\right )}{4 \, a^{5}} + \frac{7 \, \log \left (a x - 1\right )}{4 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^2*x^4,x, algorithm="maxima")

[Out]

-1/2/(a^6*x - a^5) + 1/2*(a*x^2 + 2*x)/a^4 + 1/4*log(a*x + 1)/a^5 + 7/4*log(a*x - 1)/a^5

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Fricas [A]  time = 1.77871, size = 144, normalized size = 2.48 \begin{align*} \frac{2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 4 \, a x +{\left (a x - 1\right )} \log \left (a x + 1\right ) + 7 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{4 \,{\left (a^{6} x - a^{5}\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^4,x, algorithm="fricas")

[Out]

1/4*(2*a^3*x^3 + 2*a^2*x^2 - 4*a*x + (a*x - 1)*log(a*x + 1) + 7*(a*x - 1)*log(a*x - 1) - 2)/(a^6*x - a^5)

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Sympy [A]  time = 0.426931, size = 48, normalized size = 0.83 \begin{align*} - \frac{1}{2 a^{6} x - 2 a^{5}} + \frac{x^{2}}{2 a^{3}} + \frac{x}{a^{4}} + \frac{\frac{7 \log{\left (x - \frac{1}{a} \right )}}{4} + \frac{\log{\left (x + \frac{1}{a} \right )}}{4}}{a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**2*x**4,x)

[Out]

-1/(2*a**6*x - 2*a**5) + x**2/(2*a**3) + x/a**4 + (7*log(x - 1/a)/4 + log(x + 1/a)/4)/a**5

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Giac [A]  time = 1.16013, size = 76, normalized size = 1.31 \begin{align*} \frac{\log \left ({\left | a x + 1 \right |}\right )}{4 \, a^{5}} + \frac{7 \, \log \left ({\left | a x - 1 \right |}\right )}{4 \, a^{5}} + \frac{a^{3} x^{2} + 2 \, a^{2} x}{2 \, a^{6}} - \frac{1}{2 \,{\left (a x - 1\right )} a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^2*x^4,x, algorithm="giac")

[Out]

1/4*log(abs(a*x + 1))/a^5 + 7/4*log(abs(a*x - 1))/a^5 + 1/2*(a^3*x^2 + 2*a^2*x)/a^6 - 1/2/((a*x - 1)*a^5)