### 3.587 $$\int \frac{e^{4 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^2} \, dx$$

Optimal. Leaf size=18 $\frac{1}{3 a c^2 (1-a x)^3}$

[Out]

1/(3*a*c^2*(1 - a*x)^3)

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Rubi [A]  time = 0.0597022, antiderivative size = 18, 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, 32} $\frac{1}{3 a c^2 (1-a x)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

1/(3*a*c^2*(1 - a*x)^3)

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx &=\int \frac{e^{4 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx\\ &=\frac{\int \frac{1}{(1-a x)^4} \, dx}{c^2}\\ &=\frac{1}{3 a c^2 (1-a x)^3}\\ \end{align*}

Mathematica [A]  time = 0.0180535, size = 17, normalized size = 0.94 $-\frac{1}{3 a c^2 (a x-1)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(3*a*c^2*(-1 + a*x)^3)

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Maple [A]  time = 0.039, size = 16, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,a{c}^{2} \left ( ax-1 \right ) ^{3}}} \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)^2,x)

[Out]

-1/3/c^2/a/(a*x-1)^3

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Maxima [B]  time = 1.01847, size = 55, normalized size = 3.06 \begin{align*} -\frac{1}{3 \,{\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \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)^2,x, algorithm="maxima")

[Out]

-1/3/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2)

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Fricas [B]  time = 1.39524, size = 78, normalized size = 4.33 \begin{align*} -\frac{1}{3 \,{\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \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)^2,x, algorithm="fricas")

[Out]

-1/3/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2)

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Sympy [B]  time = 0.400754, size = 42, normalized size = 2.33 \begin{align*} - \frac{1}{3 a^{4} c^{2} x^{3} - 9 a^{3} c^{2} x^{2} + 9 a^{2} c^{2} x - 3 a c^{2}} \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)**2,x)

[Out]

-1/(3*a**4*c**2*x**3 - 9*a**3*c**2*x**2 + 9*a**2*c**2*x - 3*a*c**2)

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Giac [A]  time = 1.13769, size = 20, normalized size = 1.11 \begin{align*} -\frac{1}{3 \,{\left (a x - 1\right )}^{3} a c^{2}} \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)^2,x, algorithm="giac")

[Out]

-1/3/((a*x - 1)^3*a*c^2)