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*Log[x] + 2*a^2*Log[1 - a*x]

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Rubi [A]  time = 0.0474221, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, 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 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]

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 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]))))

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*}

Mathematica [A]  time = 0.0136995, size = 33, normalized size = 1. $-2 a^2 \log (x)+2 a^2 \log (1-a x)+\frac{2 a}{x}+\frac{1}{2 x^2}$

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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Maple [A]  time = 0.047, size = 31, normalized size = 0.9 \begin{align*}{\frac{1}{2\,{x}^{2}}}+2\,{\frac{a}{x}}-2\,{a}^{2}\ln \left ( x \right ) +2\,{a}^{2}\ln \left ( ax-1 \right ) \end{align*}

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 = 0.982735, size = 41, normalized size = 1.24 \begin{align*} 2 \, a^{2} \log \left (a x - 1\right ) - 2 \, a^{2} \log \left (x\right ) + \frac{4 \, a x + 1}{2 \, x^{2}} \end{align*}

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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Fricas [A]  time = 1.83743, size = 88, normalized size = 2.67 \begin{align*} \frac{4 \, a^{2} x^{2} \log \left (a x - 1\right ) - 4 \, a^{2} x^{2} \log \left (x\right ) + 4 \, a x + 1}{2 \, x^{2}} \end{align*}

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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Sympy [A]  time = 2.12677, size = 26, normalized size = 0.79 \begin{align*} 2 a^{2} \left (- \log{\left (x \right )} + \log{\left (x - \frac{1}{a} \right )}\right ) + \frac{4 a x + 1}{2 x^{2}} \end{align*}

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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Giac [A]  time = 1.1481, size = 43, normalized size = 1.3 \begin{align*} 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}} \end{align*}

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