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

Optimal. Leaf size=35 \[ \frac{1}{a c (a x+1)}+\frac{2 \log (a x+1)}{a c}-\frac{x}{c} \]

[Out]

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

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Rubi [A]  time = 0.132358, antiderivative size = 35, 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} \[ \frac{1}{a c (a x+1)}+\frac{2 \log (a x+1)}{a c}-\frac{x}{c} \]

Antiderivative was successfully verified.

[In]

Int[1/(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 = 0.0268276, size = 28, normalized size = 0.8 \[ \frac{\frac{1}{a^2 x+a}+\frac{2 \log (a x+1)}{a}-x}{c} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - c/(a^2*x^2))),x]

[Out]

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

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Maple [A]  time = 0.035, 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(1/(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*ln(a*x+1)/a/c

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Maxima [A]  time = 0.963086, size = 45, normalized size = 1.29 \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(1/(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 = 1.79469, size = 88, normalized size = 2.51 \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(1/(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 = 0.733326, size = 37, normalized size = 1.06 \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(1/(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 = 1.31822, size = 74, normalized size = 2.11 \begin{align*} -\frac{a x + 1}{a c} - \frac{2 \, \log \left (\frac{{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2}{\left | a \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(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2),x, algorithm="giac")

[Out]

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