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

Optimal. Leaf size=38 \[ -\frac{1}{a c (1-a x)}-\frac{2 \log (1-a x)}{a c}-\frac{x}{c} \]

[Out]

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

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Rubi [A]  time = 0.124496, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6157, 6150, 43} \[ -\frac{1}{a c (1-a x)}-\frac{2 \log (1-a x)}{a c}-\frac{x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6157

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

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0203159, size = 38, normalized size = 1. \[ -\frac{1}{a c (1-a x)}-\frac{2 \log (1-a x)}{a c}-\frac{x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 36, normalized size = 1. \begin{align*} -{\frac{x}{c}}+{\frac{1}{ac \left ( ax-1 \right ) }}-2\,{\frac{\ln \left ( ax-1 \right ) }{ac}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.962236, size = 46, normalized size = 1.21 \begin{align*} -\frac{x}{c} + \frac{1}{a^{2} c x - a c} - \frac{2 \, \log \left (a x - 1\right )}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.01568, size = 88, normalized size = 2.32 \begin{align*} -\frac{a^{2} x^{2} - a x + 2 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 1}{a^{2} c x - a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.874792, size = 37, normalized size = 0.97 \begin{align*} - a^{2} \left (- \frac{1}{a^{4} c x - a^{3} c} + \frac{x}{a^{2} c} + \frac{2 \log{\left (a x - 1 \right )}}{a^{3} c}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*(-1/(a**4*c*x - a**3*c) + x/(a**2*c) + 2*log(a*x - 1)/(a**3*c))

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Giac [A]  time = 1.17655, size = 49, normalized size = 1.29 \begin{align*} -\frac{x}{c} - \frac{2 \, \log \left ({\left | a x - 1 \right |}\right )}{a c} + \frac{1}{{\left (a x - 1\right )} a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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