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

Optimal. Leaf size=59 \[ \frac{1}{2 (1-a x)}+\frac{1}{8 (a x+1)}+\frac{1}{8 (1-a x)^2}-\frac{11}{16} \log (1-a x)-\frac{5}{16} \log (a x+1)+\log (x) \]

[Out]

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

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Rubi [A]  time = 0.117907, antiderivative size = 59, 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} \[ \frac{1}{2 (1-a x)}+\frac{1}{8 (a x+1)}+\frac{1}{8 (1-a x)^2}-\frac{11}{16} \log (1-a x)-\frac{5}{16} \log (a x+1)+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0452678, size = 54, normalized size = 0.92 \[ \frac{1}{16} \left (\frac{8}{1-a x}+\frac{2}{a x+1}+\frac{2}{(a x-1)^2}-11 \log (1-a x)-5 \log (a x+1)+16 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.038, size = 47, normalized size = 0.8 \begin{align*} \ln \left ( x \right ) +{\frac{1}{8\,ax+8}}-{\frac{5\,\ln \left ( ax+1 \right ) }{16}}+{\frac{1}{8\, \left ( ax-1 \right ) ^{2}}}-{\frac{1}{2\,ax-2}}-{\frac{11\,\ln \left ( ax-1 \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.958549, size = 77, normalized size = 1.31 \begin{align*} -\frac{3 \, a^{2} x^{2} + a x - 6}{8 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )}} - \frac{5}{16} \, \log \left (a x + 1\right ) - \frac{11}{16} \, \log \left (a x - 1\right ) + \log \left (x\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,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.13587, size = 269, normalized size = 4.56 \begin{align*} -\frac{6 \, a^{2} x^{2} + 2 \, 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 ) - 16 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (x\right ) - 12}{16 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\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,x, algorithm="fricas")

[Out]

-1/16*(6*a^2*x^2 + 2*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) - 16*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(x) - 12)/(a^3*x^3 - a^2*x^2 - a*x + 1)

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Sympy [A]  time = 0.590731, size = 60, normalized size = 1.02 \begin{align*} - \frac{3 a^{2} x^{2} + a x - 6}{8 a^{3} x^{3} - 8 a^{2} x^{2} - 8 a x + 8} + \log{\left (x \right )} - \frac{11 \log{\left (x - \frac{1}{a} \right )}}{16} - \frac{5 \log{\left (x + \frac{1}{a} \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16475, size = 69, normalized size = 1.17 \begin{align*} -\frac{3 \, a^{2} x^{2} + a x - 6}{8 \,{\left (a x + 1\right )}{\left (a x - 1\right )}^{2}} - \frac{5}{16} \, \log \left ({\left | a x + 1 \right |}\right ) - \frac{11}{16} \, \log \left ({\left | a x - 1 \right |}\right ) + \log \left ({\left | x \right |}\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,x, algorithm="giac")

[Out]

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