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

Optimal. Leaf size=73 \[ \frac {a^3}{2 (1-a x)}+2 a^3 \log (x)-\frac {9}{4} a^3 \log (1-a x)+\frac {1}{4} a^3 \log (a x+1)-\frac {2 a^2}{x}-\frac {a}{2 x^2}-\frac {1}{3 x^3} \]

[Out]

-1/3/x^3-1/2*a/x^2-2*a^2/x+1/2*a^3/(-a*x+1)+2*a^3*ln(x)-9/4*a^3*ln(-a*x+1)+1/4*a^3*ln(a*x+1)

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Rubi [A]  time = 0.12, antiderivative size = 73, 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} \[ \frac {a^3}{2 (1-a x)}-\frac {2 a^2}{x}+2 a^3 \log (x)-\frac {9}{4} a^3 \log (1-a x)+\frac {1}{4} a^3 \log (a x+1)-\frac {a}{2 x^2}-\frac {1}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(3*x^3) - a/(2*x^2) - (2*a^2)/x + a^3/(2*(1 - a*x)) + 2*a^3*Log[x] - (9*a^3*Log[1 - a*x])/4 + (a^3*Log[1 +
a*x])/4

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^4 \left (1-a^2 x^2\right )^{3/2}} \, dx &=\int \frac {1}{x^4 (1-a x)^2 (1+a x)} \, dx\\ &=\int \left (\frac {1}{x^4}+\frac {a}{x^3}+\frac {2 a^2}{x^2}+\frac {2 a^3}{x}+\frac {a^4}{2 (-1+a x)^2}-\frac {9 a^4}{4 (-1+a x)}+\frac {a^4}{4 (1+a x)}\right ) \, dx\\ &=-\frac {1}{3 x^3}-\frac {a}{2 x^2}-\frac {2 a^2}{x}+\frac {a^3}{2 (1-a x)}+2 a^3 \log (x)-\frac {9}{4} a^3 \log (1-a x)+\frac {1}{4} a^3 \log (1+a x)\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 67, normalized size = 0.92 \[ \frac {1}{12} \left (\frac {6 a^3}{1-a x}+24 a^3 \log (x)-27 a^3 \log (1-a x)+3 a^3 \log (a x+1)-\frac {24 a^2}{x}-\frac {6 a}{x^2}-\frac {4}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4/x^3 - (6*a)/x^2 - (24*a^2)/x + (6*a^3)/(1 - a*x) + 24*a^3*Log[x] - 27*a^3*Log[1 - a*x] + 3*a^3*Log[1 + a*x
])/12

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fricas [A]  time = 0.59, size = 105, normalized size = 1.44 \[ -\frac {30 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 2 \, a x - 3 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (a x + 1\right ) + 27 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (a x - 1\right ) - 24 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \relax (x) - 4}{12 \, {\left (a x^{4} - x^{3}\right )}} \]

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]

-1/12*(30*a^3*x^3 - 18*a^2*x^2 - 2*a*x - 3*(a^4*x^4 - a^3*x^3)*log(a*x + 1) + 27*(a^4*x^4 - a^3*x^3)*log(a*x -
1) - 24*(a^4*x^4 - a^3*x^3)*log(x) - 4)/(a*x^4 - x^3)

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giac [A]  time = 0.15, size = 67, normalized size = 0.92 \[ \frac {1}{4} \, a^{3} \log \left ({\left | a x + 1 \right |}\right ) - \frac {9}{4} \, a^{3} \log \left ({\left | a x - 1 \right |}\right ) + 2 \, a^{3} \log \left ({\left | x \right |}\right ) - \frac {15 \, a^{3} x^{3} - 9 \, a^{2} x^{2} - a x - 2}{6 \, {\left (a x - 1\right )} x^{3}} \]

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*a^3*log(abs(a*x + 1)) - 9/4*a^3*log(abs(a*x - 1)) + 2*a^3*log(abs(x)) - 1/6*(15*a^3*x^3 - 9*a^2*x^2 - a*x
- 2)/((a*x - 1)*x^3)

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maple [A]  time = 0.04, size = 62, normalized size = 0.85 \[ -\frac {1}{3 x^{3}}-\frac {a}{2 x^{2}}-\frac {2 a^{2}}{x}+2 a^{3} \ln \relax (x )-\frac {a^{3}}{2 \left (a x -1\right )}-\frac {9 a^{3} \ln \left (a x -1\right )}{4}+\frac {a^{3} \ln \left (a x +1\right )}{4} \]

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]

-1/3/x^3-1/2*a/x^2-2*a^2/x+2*a^3*ln(x)-1/2*a^3/(a*x-1)-9/4*a^3*ln(a*x-1)+1/4*a^3*ln(a*x+1)

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maxima [A]  time = 0.32, size = 67, normalized size = 0.92 \[ \frac {1}{4} \, a^{3} \log \left (a x + 1\right ) - \frac {9}{4} \, a^{3} \log \left (a x - 1\right ) + 2 \, a^{3} \log \relax (x) - \frac {15 \, a^{3} x^{3} - 9 \, a^{2} x^{2} - a x - 2}{6 \, {\left (a x^{4} - x^{3}\right )}} \]

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]

1/4*a^3*log(a*x + 1) - 9/4*a^3*log(a*x - 1) + 2*a^3*log(x) - 1/6*(15*a^3*x^3 - 9*a^2*x^2 - a*x - 2)/(a*x^4 - x
^3)

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mupad [B]  time = 0.91, size = 66, normalized size = 0.90 \[ 2\,a^3\,\ln \relax (x)-\frac {9\,a^3\,\ln \left (a\,x-1\right )}{4}+\frac {a^3\,\ln \left (a\,x+1\right )}{4}+\frac {-\frac {5\,a^3\,x^3}{2}+\frac {3\,a^2\,x^2}{2}+\frac {a\,x}{6}+\frac {1}{3}}{a\,x^4-x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^3*log(x) - (9*a^3*log(a*x - 1))/4 + (a^3*log(a*x + 1))/4 + ((a*x)/6 + (3*a^2*x^2)/2 - (5*a^3*x^3)/2 + 1/3)
/(a*x^4 - x^3)

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sympy [A]  time = 0.43, size = 66, normalized size = 0.90 \[ 2 a^{3} \log {\relax (x )} - \frac {9 a^{3} \log {\left (x - \frac {1}{a} \right )}}{4} + \frac {a^{3} \log {\left (x + \frac {1}{a} \right )}}{4} + \frac {- 15 a^{3} x^{3} + 9 a^{2} x^{2} + a x + 2}{6 a x^{4} - 6 x^{3}} \]

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]

2*a**3*log(x) - 9*a**3*log(x - 1/a)/4 + a**3*log(x + 1/a)/4 + (-15*a**3*x**3 + 9*a**2*x**2 + a*x + 2)/(6*a*x**
4 - 6*x**3)

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