3.1.29 \(\int \frac {e^{4 \coth ^{-1}(a x)}}{x} \, dx\) [29]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 13, 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 = {6302, 6261, 90} \begin {gather*} \frac {4}{1-a x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4/(1 - a*x) + Log[x]

Rule 90

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 6261

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 6302

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^{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*}

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Mathematica [A]
time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} \frac {4}{1-a x}+\log (x) \end {gather*}

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 = 0.13, size = 13, normalized size = 1.00

method	result	size
default	\(-\frac {4}{a x -1}+\ln \left (x \right )\)	\(13\)
norman	\(-\frac {4 a x}{a x -1}+\ln \left (x \right )\)	\(15\)
risch	\(-\frac {4}{a x -1}+\ln \left (-x \right )\)	\(15\)
meijerg	\(\frac {3 a x}{-a x +1}+\frac {2 a x}{-2 a x +2}+1+\ln \left (x \right )+\ln \left (-a \right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 12, normalized size = 0.92 \begin {gather*} -\frac {4}{a x - 1} + \log \left (x\right ) \end {gather*}

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 = 0.34, size = 18, normalized size = 1.38 \begin {gather*} \frac {{\left (a x - 1\right )} \log \left (x\right ) - 4}{a x - 1} \end {gather*}

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 = 0.09, size = 8, normalized size = 0.62 \begin {gather*} \log {\left (x \right )} - \frac {4}{a x - 1} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (12) = 24\).
time = 0.40, size = 57, normalized size = 4.38 \begin {gather*} -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 {gather*}

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))

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Mupad [B]
time = 0.03, size = 12, normalized size = 0.92 \begin {gather*} \ln \left (x\right )-\frac {4}{a\,x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)^2/(x*(a*x - 1)^2),x)

[Out]

log(x) - 4/(a*x - 1)

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