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

Optimal. Leaf size=14 \[ -\frac{x}{c^2 (1-a x)^2} \]

[Out]

-(x/(c^2*(1 - a*x)^2))

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Rubi [A]  time = 0.0510269, antiderivative size = 14, 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 = {6167, 6129, 34} \[ -\frac{x}{c^2 (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/(c^2*(1 - a*x)^2))

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/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 34

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

Rubi steps

\begin{align*} \int \frac{e^{2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)}}{(c-a c x)^2} \, dx\\ &=-\frac{\int \frac{1+a x}{(1-a x)^3} \, dx}{c^2}\\ &=-\frac{x}{c^2 (1-a x)^2}\\ \end{align*}

Mathematica [A]  time = 0.0081405, size = 25, normalized size = 1.79 \[ -\frac{(a x+1)^2}{4 a c^2 (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 + a*x)^2/(4*a*c^2*(1 - a*x)^2)

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Maple [B]  time = 0.047, size = 30, normalized size = 2.1 \begin{align*}{\frac{1}{{c}^{2}} \left ( -{\frac{1}{a \left ( ax-1 \right ) ^{2}}}-{\frac{1}{a \left ( ax-1 \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00542, size = 35, normalized size = 2.5 \begin{align*} -\frac{x}{a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(a^2*c^2*x^2 - 2*a*c^2*x + c^2)

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Fricas [A]  time = 1.35504, size = 49, normalized size = 3.5 \begin{align*} -\frac{x}{a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(a^2*c^2*x^2 - 2*a*c^2*x + c^2)

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Sympy [A]  time = 0.370919, size = 24, normalized size = 1.71 \begin{align*} - \frac{x}{a^{2} c^{2} x^{2} - 2 a c^{2} x + c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.11755, size = 46, normalized size = 3.29 \begin{align*} -\frac{1}{{\left (a c x - c\right )}^{2} a} - \frac{1}{{\left (a c x - c\right )} a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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