3.28 \(\int e^{4 \coth ^{-1}(a x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {4}{a (1-a x)}+\frac {4 \log (1-a x)}{a}+x \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6167, 6125, 43} \[ \frac {4}{a (1-a x)}+\frac {4 \log (1-a x)}{a}+x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6125

Int[E^(ArcTanh[(a_.)*(x_)]*(n_)), x_Symbol] :> Int[(1 + a*x)^(n/2)/(1 - a*x)^(n/2), x] /; FreeQ[{a, 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 e^{4 \coth ^{-1}(a x)} \, dx &=\int e^{4 \tanh ^{-1}(a x)} \, dx\\ &=\int \frac {(1+a x)^2}{(1-a x)^2} \, dx\\ &=\int \left (1+\frac {4}{(-1+a x)^2}+\frac {4}{-1+a x}\right ) \, dx\\ &=x+\frac {4}{a (1-a x)}+\frac {4 \log (1-a x)}{a}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 0.96 \[ -\frac {4}{a (a x-1)}+\frac {4 \log (1-a x)}{a}+x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 38, normalized size = 1.41 \[ \frac {a^{2} x^{2} - a x + 4 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 4}{a^{2} x - a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 46, normalized size = 1.70 \[ \frac {a x - 1}{a} - \frac {4 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {4}{{\left (a x - 1\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 26, normalized size = 0.96 \[ x +\frac {4 \ln \left (a x -1\right )}{a}-\frac {4}{a \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 26, normalized size = 0.96 \[ x + \frac {4 \, \log \left (a x - 1\right )}{a} - \frac {4}{a^{2} x - a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 4*log(a*x - 1)/a - 4/(a^2*x - a)

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mupad [B]  time = 0.04, size = 25, normalized size = 0.93 \[ x-\frac {4}{a\,\left (a\,x-1\right )}+\frac {4\,\ln \left (a\,x-1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)^2/(a*x - 1)^2,x)

[Out]

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

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sympy [A]  time = 0.12, size = 19, normalized size = 0.70 \[ x - \frac {4}{a^{2} x - a} + \frac {4 \log {\left (a x - 1 \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 4/(a**2*x - a) + 4*log(a*x - 1)/a

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