3.30 $$\int \frac{e^{4 \coth ^{-1}(a x)}}{x^2} \, dx$$

Optimal. Leaf size=32 $\frac{4 a}{1-a x}+4 a \log (x)-4 a \log (1-a x)-\frac{1}{x}$

[Out]

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

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Rubi [A]  time = 0.0505939, antiderivative size = 32, 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, 88} $\frac{4 a}{1-a x}+4 a \log (x)-4 a \log (1-a x)-\frac{1}{x}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(4*ArcCoth[a*x])/x^2,x]

[Out]

-x^(-1) + (4*a)/(1 - a*x) + 4*a*Log[x] - 4*a*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 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^{4 \coth ^{-1}(a x)}}{x^2} \, dx &=\int \frac{e^{4 \tanh ^{-1}(a x)}}{x^2} \, dx\\ &=\int \frac{(1+a x)^2}{x^2 (1-a x)^2} \, dx\\ &=\int \left (\frac{1}{x^2}+\frac{4 a}{x}+\frac{4 a^2}{(-1+a x)^2}-\frac{4 a^2}{-1+a x}\right ) \, dx\\ &=-\frac{1}{x}+\frac{4 a}{1-a x}+4 a \log (x)-4 a \log (1-a x)\\ \end{align*}

Mathematica [A]  time = 0.0235628, size = 32, normalized size = 1. $\frac{4 a}{1-a x}+4 a \log (x)-4 a \log (1-a x)-\frac{1}{x}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(4*ArcCoth[a*x])/x^2,x]

[Out]

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

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Maple [A]  time = 0.047, size = 31, normalized size = 1. \begin{align*} -{x}^{-1}+4\,a\ln \left ( x \right ) -4\,{\frac{a}{ax-1}}-4\,a\ln \left ( ax-1 \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x+4*a*ln(x)-4*a/(a*x-1)-4*a*ln(a*x-1)

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Maxima [A]  time = 0.969968, size = 46, normalized size = 1.44 \begin{align*} -4 \, a \log \left (a x - 1\right ) + 4 \, a \log \left (x\right ) - \frac{5 \, a x - 1}{a x^{2} - x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.78587, size = 116, normalized size = 3.62 \begin{align*} -\frac{5 \, a x + 4 \,{\left (a^{2} x^{2} - a x\right )} \log \left (a x - 1\right ) - 4 \,{\left (a^{2} x^{2} - a x\right )} \log \left (x\right ) - 1}{a x^{2} - x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.412016, size = 26, normalized size = 0.81 \begin{align*} 4 a \left (\log{\left (x \right )} - \log{\left (x - \frac{1}{a} \right )}\right ) - \frac{5 a x - 1}{a x^{2} - x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a*(log(x) - log(x - 1/a)) - (5*a*x - 1)/(a*x**2 - x)

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Giac [A]  time = 1.13818, size = 54, normalized size = 1.69 \begin{align*} 4 \, a \log \left ({\left | -\frac{1}{a x - 1} - 1 \right |}\right ) - \frac{4 \, a}{a x - 1} + \frac{a}{\frac{1}{a x - 1} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/x^2,x, algorithm="giac")

[Out]

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