3.941 \(\int \frac{e^{\tanh ^{-1}(a x)} x^3}{(1-a^2 x^2)^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.123667, 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} \[ -\frac{1}{2 a^4 (1-a x)}+\frac{1}{8 a^4 (a x+1)}+\frac{1}{8 a^4 (1-a x)^2}+\frac{3 \tanh ^{-1}(a x)}{8 a^4} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^3)/(1 - a^2*x^2)^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0319314, size = 53, normalized size = 0.95 \[ \frac{5 a^2 x^2-a x+3 (a x-1)^2 (a x+1) \tanh ^{-1}(a x)-2}{8 a^4 (a x-1)^2 (a x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^3)/(1 - a^2*x^2)^(5/2),x]

[Out]

(-2 - a*x + 5*a^2*x^2 + 3*(-1 + a*x)^2*(1 + a*x)*ArcTanh[a*x])/(8*a^4*(-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}^{4} \left ( ax+1 \right ) }}+{\frac{3\,\ln \left ( ax+1 \right ) }{16\,{a}^{4}}}+{\frac{1}{8\,{a}^{4} \left ( ax-1 \right ) ^{2}}}+{\frac{1}{2\,{a}^{4} \left ( ax-1 \right ) }}-{\frac{3\,\ln \left ( ax-1 \right ) }{16\,{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.950193, size = 89, normalized size = 1.59 \begin{align*} \frac{5 \, a^{2} x^{2} - a x - 2}{8 \,{\left (a^{7} x^{3} - a^{6} x^{2} - a^{5} x + a^{4}\right )}} + \frac{3 \, \log \left (a x + 1\right )}{16 \, a^{4}} - \frac{3 \, \log \left (a x - 1\right )}{16 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(5*a^2*x^2 - a*x - 2)/(a^7*x^3 - a^6*x^2 - a^5*x + a^4) + 3/16*log(a*x + 1)/a^4 - 3/16*log(a*x - 1)/a^4

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Fricas [B]  time = 2.3696, size = 215, normalized size = 3.84 \begin{align*} \frac{10 \, a^{2} x^{2} - 2 \, a x + 3 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) - 3 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 4}{16 \,{\left (a^{7} x^{3} - a^{6} x^{2} - a^{5} x + a^{4}\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^3,x, algorithm="fricas")

[Out]

1/16*(10*a^2*x^2 - 2*a*x + 3*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) - 3*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(
a*x - 1) - 4)/(a^7*x^3 - a^6*x^2 - a^5*x + a^4)

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Sympy [A]  time = 0.488704, size = 65, normalized size = 1.16 \begin{align*} \frac{5 a^{2} x^{2} - a x - 2}{8 a^{7} x^{3} - 8 a^{6} x^{2} - 8 a^{5} x + 8 a^{4}} - \frac{\frac{3 \log{\left (x - \frac{1}{a} \right )}}{16} - \frac{3 \log{\left (x + \frac{1}{a} \right )}}{16}}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a**2*x**2 - a*x - 2)/(8*a**7*x**3 - 8*a**6*x**2 - 8*a**5*x + 8*a**4) - (3*log(x - 1/a)/16 - 3*log(x + 1/a)/
16)/a**4

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Giac [A]  time = 1.22821, size = 78, normalized size = 1.39 \begin{align*} \frac{3 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{4}} - \frac{3 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{4}} + \frac{5 \, a^{2} x^{2} - a x - 2}{8 \,{\left (a x + 1\right )}{\left (a x - 1\right )}^{2} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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