3.16 \(\int \frac {e^{2 \coth ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=33 \[ -2 a^2 \log (x)+2 a^2 \log (1-a x)+\frac {2 a}{x}+\frac {1}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6167, 6126, 77} \[ -2 a^2 \log (x)+2 a^2 \log (1-a x)+\frac {2 a}{x}+\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[a*x])/x^3,x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {e^{2 \coth ^{-1}(a x)}}{x^3} \, dx &=-\int \frac {e^{2 \tanh ^{-1}(a x)}}{x^3} \, dx\\ &=-\int \frac {1+a x}{x^3 (1-a x)} \, dx\\ &=-\int \left (\frac {1}{x^3}+\frac {2 a}{x^2}+\frac {2 a^2}{x}-\frac {2 a^3}{-1+a x}\right ) \, dx\\ &=\frac {1}{2 x^2}+\frac {2 a}{x}-2 a^2 \log (x)+2 a^2 \log (1-a x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.00 \[ -2 a^2 \log (x)+2 a^2 \log (1-a x)+\frac {2 a}{x}+\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcCoth[a*x])/x^3,x]

[Out]

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

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fricas [A]  time = 0.52, size = 35, normalized size = 1.06 \[ \frac {4 \, a^{2} x^{2} \log \left (a x - 1\right ) - 4 \, a^{2} x^{2} \log \relax (x) + 4 \, a x + 1}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 32, normalized size = 0.97 \[ 2 \, a^{2} \log \left ({\left | a x - 1 \right |}\right ) - 2 \, a^{2} \log \left ({\left | x \right |}\right ) + \frac {4 \, a x + 1}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^2*log(abs(a*x - 1)) - 2*a^2*log(abs(x)) + 1/2*(4*a*x + 1)/x^2

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maple [A]  time = 0.04, size = 31, normalized size = 0.94 \[ \frac {1}{2 x^{2}}+\frac {2 a}{x}-2 a^{2} \ln \relax (x )+2 a^{2} \ln \left (a x -1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 30, normalized size = 0.91 \[ 2 \, a^{2} \log \left (a x - 1\right ) - 2 \, a^{2} \log \relax (x) + \frac {4 \, a x + 1}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 23, normalized size = 0.70 \[ \frac {2\,a\,x+\frac {1}{2}}{x^2}-4\,a^2\,\mathrm {atanh}\left (2\,a\,x-1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)/(x^3*(a*x - 1)),x)

[Out]

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

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sympy [A]  time = 0.15, size = 26, normalized size = 0.79 \[ 2 a^{2} \left (- \log {\relax (x )} + \log {\left (x - \frac {1}{a} \right )}\right ) + \frac {4 a x + 1}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**2*(-log(x) + log(x - 1/a)) + (4*a*x + 1)/(2*x**2)

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