3.1054 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{x (c-a^2 c x^2)} \, dx\)

Optimal. Leaf size=31 \[ \frac{1}{c (1-a x)}-\frac{\log (1-a x)}{c}+\frac{\log (x)}{c} \]

[Out]

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

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Rubi [A]  time = 0.0866635, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {6150, 44} \[ \frac{1}{c (1-a x)}-\frac{\log (1-a x)}{c}+\frac{\log (x)}{c} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])/(x*(c - a^2*c*x^2)),x]

[Out]

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

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0198567, size = 24, normalized size = 0.77 \[ \frac{\frac{1}{1-a x}-\log (1-a x)+\log (x)}{c} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])/(x*(c - a^2*c*x^2)),x]

[Out]

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

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Maple [A]  time = 0.031, size = 31, normalized size = 1. \begin{align*}{\frac{\ln \left ( x \right ) }{c}}-{\frac{1}{c \left ( ax-1 \right ) }}-{\frac{\ln \left ( ax-1 \right ) }{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.958175, size = 41, normalized size = 1.32 \begin{align*} -\frac{\log \left (a x - 1\right )}{c} + \frac{\log \left (x\right )}{c} - \frac{1}{a c x - c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.03213, size = 84, normalized size = 2.71 \begin{align*} -\frac{{\left (a x - 1\right )} \log \left (a x - 1\right ) -{\left (a x - 1\right )} \log \left (x\right ) + 1}{a c x - c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.364284, size = 19, normalized size = 0.61 \begin{align*} - \frac{1}{a c x - c} + \frac{\log{\left (x \right )} - \log{\left (x - \frac{1}{a} \right )}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20644, size = 43, normalized size = 1.39 \begin{align*} -\frac{\log \left ({\left | a x - 1 \right |}\right )}{c} + \frac{\log \left ({\left | x \right |}\right )}{c} - \frac{1}{{\left (a x - 1\right )} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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