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

Optimal. Leaf size=20 \[ \frac{\log (a x+1)}{a c}-\frac{x}{c} \]

[Out]

-(x/c) + Log[1 + a*x]/(a*c)

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Rubi [A]  time = 0.0875015, antiderivative size = 20, 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 = {6131, 6129, 43} \[ \frac{\log (a x+1)}{a c}-\frac{x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/c) + Log[1 + a*x]/(a*c)

Rule 6131

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

Rule 6129

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[(u*(1 + (d*x)/c)
^p*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 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 x}} \, dx &=-\frac{a \int \frac{e^{-2 \tanh ^{-1}(a x)} x}{1-a x} \, dx}{c}\\ &=-\frac{a \int \frac{x}{1+a x} \, dx}{c}\\ &=-\frac{a \int \left (\frac{1}{a}-\frac{1}{a (1+a x)}\right ) \, dx}{c}\\ &=-\frac{x}{c}+\frac{\log (1+a x)}{a c}\\ \end{align*}

Mathematica [A]  time = 0.0152997, size = 18, normalized size = 0.9 \[ \frac{\log (a x+1)-a x}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*x) + Log[1 + a*x])/(a*c)

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Maple [A]  time = 0.028, size = 21, normalized size = 1.1 \begin{align*} -{\frac{x}{c}}+{\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/x),x)

[Out]

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

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Maxima [A]  time = 0.940878, size = 27, normalized size = 1.35 \begin{align*} -\frac{x}{c} + \frac{\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/x),x, algorithm="maxima")

[Out]

-x/c + log(a*x + 1)/(a*c)

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Fricas [A]  time = 2.20964, size = 39, normalized size = 1.95 \begin{align*} -\frac{a x - \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/x),x, algorithm="fricas")

[Out]

-(a*x - log(a*x + 1))/(a*c)

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Sympy [A]  time = 0.30708, size = 19, normalized size = 0.95 \begin{align*} - a \left (\frac{x}{a c} - \frac{\log{\left (a x + 1 \right )}}{a^{2} 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/x),x)

[Out]

-a*(x/(a*c) - log(a*x + 1)/(a**2*c))

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Giac [B]  time = 1.25834, size = 55, normalized size = 2.75 \begin{align*} -\frac{a x + 1}{a c} - \frac{\log \left (\frac{{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2}{\left | a \right |}}\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/x),x, algorithm="giac")

[Out]

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