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

Optimal. Leaf size=11 \[ \frac{\tanh ^{-1}(a x)}{a c^2} \]

[Out]

ArcTanh[a*x]/(a*c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0268221, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6129, 35, 206} \[ \frac{\tanh ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[a*x]/(a*c^2)

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 35

Int[1/(((a_) + (b_.)*(x_))*((c_) + (d_.)*(x_))), x_Symbol] :> Int[1/(a*c + b*d*x^2), x] /; FreeQ[{a, b, c, d},
x] && EqQ[b*c + a*d, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.007319, size = 11, normalized size = 1. \[ \frac{\tanh ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[a*x]/(a*c^2)

________________________________________________________________________________________

Maple [B]  time = 0.032, size = 30, normalized size = 2.7 \begin{align*}{\frac{\ln \left ( ax+1 \right ) }{2\,a{c}^{2}}}-{\frac{\ln \left ( ax-1 \right ) }{2\,a{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 0.945176, size = 39, normalized size = 3.55 \begin{align*} \frac{\log \left (a x + 1\right )}{2 \, a c^{2}} - \frac{\log \left (a x - 1\right )}{2 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.57252, size = 58, normalized size = 5.27 \begin{align*} \frac{\log \left (a x + 1\right ) - \log \left (a x - 1\right )}{2 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B]  time = 0.158413, size = 22, normalized size = 2. \begin{align*} - \frac{\frac{\log{\left (x - \frac{1}{a} \right )}}{2} - \frac{\log{\left (x + \frac{1}{a} \right )}}{2}}{a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.27475, size = 34, normalized size = 3.09 \begin{align*} \frac{\log \left ({\left | -\frac{2 \, c}{a c x - c} - 1 \right |}\right )}{2 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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