### 3.29 $$\int \frac{e^{4 \coth ^{-1}(a x)}}{x} \, dx$$

Optimal. Leaf size=13 $\frac{4}{1-a x}+\log (x)$

[Out]

4/(1 - a*x) + Log[x]

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Rubi [A]  time = 0.0392385, antiderivative size = 13, 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} $\frac{4}{1-a x}+\log (x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0097751, size = 13, normalized size = 1. $\frac{4}{1-a x}+\log (x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

4/(1 - a*x) + Log[x]

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Maple [A]  time = 0.046, size = 13, normalized size = 1. \begin{align*} -4\, \left ( ax-1 \right ) ^{-1}+\ln \left ( x \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,x)

[Out]

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

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Maxima [A]  time = 1.01998, size = 16, normalized size = 1.23 \begin{align*} -\frac{4}{a x - 1} + \log \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(a*x - 1) + log(x)

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Fricas [A]  time = 1.78683, size = 46, normalized size = 3.54 \begin{align*} \frac{{\left (a x - 1\right )} \log \left (x\right ) - 4}{a x - 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,x, algorithm="fricas")

[Out]

((a*x - 1)*log(x) - 4)/(a*x - 1)

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Sympy [A]  time = 0.355664, size = 8, normalized size = 0.62 \begin{align*} \log{\left (x \right )} - \frac{4}{a x - 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,x)

[Out]

log(x) - 4/(a*x - 1)

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Giac [B]  time = 1.12011, size = 77, normalized size = 5.92 \begin{align*} -a{\left (\frac{\log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | a \right |}}\right )}{a} - \frac{\log \left ({\left | -\frac{1}{a x - 1} - 1 \right |}\right )}{a} + \frac{4}{{\left (a x - 1\right )} a}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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