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

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 = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6126, 88} \[ \frac {4}{1-a x}+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

4/(1 - a*x) + Log[x]

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]))

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]

Rubi steps

\begin {align*} \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 \[ \frac {4}{1-a x}+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

4/(1 - a*x) + Log[x]

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fricas [A]  time = 0.44, size = 18, normalized size = 1.38 \[ \frac {{\left (a x - 1\right )} \log \relax (x) - 4}{a x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/x,x, algorithm="fricas")

[Out]

((a*x - 1)*log(x) - 4)/(a*x - 1)

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giac [A]  time = 0.16, size = 13, normalized size = 1.00 \[ -\frac {4}{a x - 1} + \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/x,x, algorithm="giac")

[Out]

-4/(a*x - 1) + log(abs(x))

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maple [A]  time = 0.03, size = 13, normalized size = 1.00 \[ \ln \relax (x )-\frac {4}{a x -1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 12, normalized size = 0.92 \[ -\frac {4}{a x - 1} + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/x,x, algorithm="maxima")

[Out]

-4/(a*x - 1) + log(x)

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mupad [B]  time = 0.04, size = 12, normalized size = 0.92 \[ \ln \relax (x)-\frac {4}{a\,x-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - 4/(a*x - 1)

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sympy [A]  time = 0.16, size = 8, normalized size = 0.62 \[ \log {\relax (x )} - \frac {4}{a x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2/x,x)

[Out]

log(x) - 4/(a*x - 1)

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