3.940 \(\int \frac{e^{\tanh ^{-1}(a x)} x^4}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=72 \[ -\frac{3}{4 a^5 (1-a x)}-\frac{1}{8 a^5 (a x+1)}+\frac{1}{8 a^5 (1-a x)^2}-\frac{11 \log (1-a x)}{16 a^5}-\frac{5 \log (a x+1)}{16 a^5} \]

[Out]

1/(8*a^5*(1 - a*x)^2) - 3/(4*a^5*(1 - a*x)) - 1/(8*a^5*(1 + a*x)) - (11*Log[1 - a*x])/(16*a^5) - (5*Log[1 + a*
x])/(16*a^5)

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Rubi [A]  time = 0.124757, antiderivative size = 72, 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} \[ -\frac{3}{4 a^5 (1-a x)}-\frac{1}{8 a^5 (a x+1)}+\frac{1}{8 a^5 (1-a x)^2}-\frac{11 \log (1-a x)}{16 a^5}-\frac{5 \log (a x+1)}{16 a^5} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(5/2),x]

[Out]

1/(8*a^5*(1 - a*x)^2) - 3/(4*a^5*(1 - a*x)) - 1/(8*a^5*(1 + a*x)) - (11*Log[1 - a*x])/(16*a^5) - (5*Log[1 + a*
x])/(16*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 )^{5/2}} \, dx &=\int \frac{x^4}{(1-a x)^3 (1+a x)^2} \, dx\\ &=\int \left (-\frac{1}{4 a^4 (-1+a x)^3}-\frac{3}{4 a^4 (-1+a x)^2}-\frac{11}{16 a^4 (-1+a x)}+\frac{1}{8 a^4 (1+a x)^2}-\frac{5}{16 a^4 (1+a x)}\right ) \, dx\\ &=\frac{1}{8 a^5 (1-a x)^2}-\frac{3}{4 a^5 (1-a x)}-\frac{1}{8 a^5 (1+a x)}-\frac{11 \log (1-a x)}{16 a^5}-\frac{5 \log (1+a x)}{16 a^5}\\ \end{align*}

Mathematica [A]  time = 0.057497, size = 55, normalized size = 0.76 \[ \frac{\frac{2 \left (5 a^2 x^2+3 a x-6\right )}{(a x-1)^2 (a x+1)}-11 \log (1-a x)-5 \log (a x+1)}{16 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(5/2),x]

[Out]

((2*(-6 + 3*a*x + 5*a^2*x^2))/((-1 + a*x)^2*(1 + a*x)) - 11*Log[1 - a*x] - 5*Log[1 + a*x])/(16*a^5)

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Maple [A]  time = 0.036, size = 60, normalized size = 0.8 \begin{align*} -{\frac{1}{8\,{a}^{5} \left ( ax+1 \right ) }}-{\frac{5\,\ln \left ( ax+1 \right ) }{16\,{a}^{5}}}+{\frac{1}{8\,{a}^{5} \left ( ax-1 \right ) ^{2}}}+{\frac{3}{4\,{a}^{5} \left ( ax-1 \right ) }}-{\frac{11\,\ln \left ( ax-1 \right ) }{16\,{a}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/a^5/(a*x+1)-5/16*ln(a*x+1)/a^5+1/8/a^5/(a*x-1)^2+3/4/a^5/(a*x-1)-11/16/a^5*ln(a*x-1)

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Maxima [A]  time = 0.949897, size = 89, normalized size = 1.24 \begin{align*} \frac{5 \, a^{2} x^{2} + 3 \, a x - 6}{8 \,{\left (a^{8} x^{3} - a^{7} x^{2} - a^{6} x + a^{5}\right )}} - \frac{5 \, \log \left (a x + 1\right )}{16 \, a^{5}} - \frac{11 \, \log \left (a x - 1\right )}{16 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^3*x^4,x, algorithm="maxima")

[Out]

1/8*(5*a^2*x^2 + 3*a*x - 6)/(a^8*x^3 - a^7*x^2 - a^6*x + a^5) - 5/16*log(a*x + 1)/a^5 - 11/16*log(a*x - 1)/a^5

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Fricas [A]  time = 2.06296, size = 217, normalized size = 3.01 \begin{align*} \frac{10 \, a^{2} x^{2} + 6 \, a x - 5 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) - 11 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 12}{16 \,{\left (a^{8} x^{3} - a^{7} x^{2} - 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)^3*x^4,x, algorithm="fricas")

[Out]

1/16*(10*a^2*x^2 + 6*a*x - 5*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) - 11*(a^3*x^3 - a^2*x^2 - a*x + 1)*log
(a*x - 1) - 12)/(a^8*x^3 - a^7*x^2 - a^6*x + a^5)

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Sympy [A]  time = 0.548471, size = 66, normalized size = 0.92 \begin{align*} \frac{5 a^{2} x^{2} + 3 a x - 6}{8 a^{8} x^{3} - 8 a^{7} x^{2} - 8 a^{6} x + 8 a^{5}} - \frac{\frac{11 \log{\left (x - \frac{1}{a} \right )}}{16} + \frac{5 \log{\left (x + \frac{1}{a} \right )}}{16}}{a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a**2*x**2 + 3*a*x - 6)/(8*a**8*x**3 - 8*a**7*x**2 - 8*a**6*x + 8*a**5) - (11*log(x - 1/a)/16 + 5*log(x + 1/
a)/16)/a**5

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Giac [A]  time = 1.20655, size = 78, normalized size = 1.08 \begin{align*} -\frac{5 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{5}} - \frac{11 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{5}} + \frac{5 \, a^{2} x^{2} + 3 \, a x - 6}{8 \,{\left (a x + 1\right )}{\left (a x - 1\right )}^{2} a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^3*x^4,x, algorithm="giac")

[Out]

-5/16*log(abs(a*x + 1))/a^5 - 11/16*log(abs(a*x - 1))/a^5 + 1/8*(5*a^2*x^2 + 3*a*x - 6)/((a*x + 1)*(a*x - 1)^2
*a^5)