3.586 \(\int \frac{e^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=13 \[ \frac{x}{c (1-a x)^2} \]

[Out]

x/(c*(1 - a*x)^2)

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Rubi [A]  time = 0.0625637, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6167, 6140, 34} \[ \frac{x}{c (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcCoth[a*x])/(c - a^2*c*x^2),x]

[Out]

x/(c*(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 6140

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 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^{4 \coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx &=\int \frac{e^{4 \tanh ^{-1}(a x)}}{c-a^2 c x^2} \, dx\\ &=\frac{\int \frac{1+a x}{(1-a x)^3} \, dx}{c}\\ &=\frac{x}{c (1-a x)^2}\\ \end{align*}

Mathematica [A]  time = 0.0096785, size = 25, normalized size = 1.92 \[ \frac{(a x+1)^2}{4 a c (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcCoth[a*x])/(c - a^2*c*x^2),x]

[Out]

(1 + a*x)^2/(4*a*c*(1 - a*x)^2)

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Maple [B]  time = 0.046, size = 28, normalized size = 2.2 \begin{align*}{\frac{1}{c} \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(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c),x)

[Out]

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

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Maxima [A]  time = 1.09274, size = 26, normalized size = 2. \begin{align*} \frac{x}{a^{2} c x^{2} - 2 \, a c x + c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(a^2*c*x^2 - 2*a*c*x + c)

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Fricas [A]  time = 1.48406, size = 39, normalized size = 3. \begin{align*} \frac{x}{a^{2} c x^{2} - 2 \, a c x + c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(a^2*c*x^2 - 2*a*c*x + c)

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Sympy [B]  time = 0.343586, size = 17, normalized size = 1.31 \begin{align*} \frac{x}{a^{2} c x^{2} - 2 a c x + c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(a**2*c*x**2 - 2*a*c*x + c)

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Giac [B]  time = 1.11978, size = 36, normalized size = 2.77 \begin{align*} \frac{\frac{1}{{\left (a x - 1\right )} a} + \frac{1}{{\left (a x - 1\right )}^{2} a}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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