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

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Rubi [A]  time = 0.12, antiderivative size = 59, normalized size of antiderivative = 1.00, 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 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 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])

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*}

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Mathematica [A]  time = 0.05, 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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fricas [B]  time = 0.59, size = 122, normalized size = 2.07 \[ -\frac {6 \, a^{2} x^{2} + 2 \, a x + 5 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 11 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \relax (x) - 12}{16 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )}} \]

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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giac [A]  time = 0.18, size = 51, normalized size = 0.86 \[ -\frac {3 \, a^{2} x^{2} + a x - 6}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2}} - \frac {5}{16} \, \log \left ({\left | a x + 1 \right |}\right ) - \frac {11}{16} \, \log \left ({\left | a x - 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.04, size = 47, normalized size = 0.80 \[ \ln \relax (x )+\frac {1}{8 \left (a x -1\right )^{2}}-\frac {1}{2 \left (a x -1\right )}-\frac {11 \ln \left (a x -1\right )}{16}+\frac {1}{8 a x +8}-\frac {5 \ln \left (a x +1\right )}{16} \]

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

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maxima [A]  time = 0.31, size = 57, normalized size = 0.97 \[ -\frac {3 \, a^{2} x^{2} + a x - 6}{8 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )}} - \frac {5}{16} \, \log \left (a x + 1\right ) - \frac {11}{16} \, \log \left (a x - 1\right ) + \log \relax (x) \]

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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mupad [B]  time = 0.07, size = 57, normalized size = 0.97 \[ \ln \relax (x)-\frac {11\,\ln \left (1-a\,x\right )}{16}-\frac {5\,\ln \left (a\,x+1\right )}{16}+\frac {\frac {3\,a^2\,x^2}{8}+\frac {a\,x}{8}-\frac {3}{4}}{-a^3\,x^3+a^2\,x^2+a\,x-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a*x + 1)/(x*(a^2*x^2 - 1)^3),x)

[Out]

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

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sympy [A]  time = 0.39, size = 60, normalized size = 1.02 \[ - \frac {3 a^{2} x^{2} + a x - 6}{8 a^{3} x^{3} - 8 a^{2} x^{2} - 8 a x + 8} + \log {\relax (x )} - \frac {11 \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {5 \log {\left (x + \frac {1}{a} \right )}}{16} \]

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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