3.1055 \(\int \frac{e^{2 \tanh ^{-1}(a x)}}{x^2 (c-a^2 c x^2)} \, dx\)

Optimal. Leaf size=43 \[ \frac{a}{c (1-a x)}+\frac{2 a \log (x)}{c}-\frac{2 a \log (1-a x)}{c}-\frac{1}{c x} \]

[Out]

-(1/(c*x)) + a/(c*(1 - a*x)) + (2*a*Log[x])/c - (2*a*Log[1 - a*x])/c

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Rubi [A]  time = 0.0938533, antiderivative size = 43, 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} \[ \frac{a}{c (1-a x)}+\frac{2 a \log (x)}{c}-\frac{2 a \log (1-a x)}{c}-\frac{1}{c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0301129, size = 35, normalized size = 0.81 \[ \frac{\frac{a}{1-a x}+2 a \log (x)-2 a \log (1-a x)-\frac{1}{x}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-x^(-1) + a/(1 - a*x) + 2*a*Log[x] - 2*a*Log[1 - a*x])/c

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Maple [A]  time = 0.033, size = 43, normalized size = 1. \begin{align*} -{\frac{1}{cx}}+2\,{\frac{a\ln \left ( x \right ) }{c}}-{\frac{a}{c \left ( ax-1 \right ) }}-2\,{\frac{a\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^2/(-a^2*c*x^2+c),x)

[Out]

-1/c/x+2*a*ln(x)/c-1/c*a/(a*x-1)-2/c*a*ln(a*x-1)

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Maxima [A]  time = 0.966991, size = 57, normalized size = 1.33 \begin{align*} -\frac{2 \, a \log \left (a x - 1\right )}{c} + \frac{2 \, a \log \left (x\right )}{c} - \frac{2 \, a x - 1}{a c x^{2} - c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.00823, size = 122, normalized size = 2.84 \begin{align*} -\frac{2 \, a x + 2 \,{\left (a^{2} x^{2} - a x\right )} \log \left (a x - 1\right ) - 2 \,{\left (a^{2} x^{2} - a x\right )} \log \left (x\right ) - 1}{a c x^{2} - c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.411368, size = 31, normalized size = 0.72 \begin{align*} \frac{2 a \left (\log{\left (x \right )} - \log{\left (x - \frac{1}{a} \right )}\right )}{c} - \frac{2 a x - 1}{a c x^{2} - c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*(log(x) - log(x - 1/a))/c - (2*a*x - 1)/(a*c*x**2 - c*x)

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Giac [A]  time = 1.13968, size = 61, normalized size = 1.42 \begin{align*} -\frac{2 \, a \log \left ({\left | a x - 1 \right |}\right )}{c} + \frac{2 \, a \log \left ({\left | x \right |}\right )}{c} - \frac{2 \, a x - 1}{{\left (a x^{2} - x\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^2/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

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