### 3.194 $$\int \frac{e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx$$

Optimal. Leaf size=25 $\frac{(a x+1)^3}{6 a c^2 (1-a x)^3}$

[Out]

(1 + a*x)^3/(6*a*c^2*(1 - a*x)^3)

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Rubi [A]  time = 0.0504695, antiderivative size = 25, 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, 37} $\frac{(a x+1)^3}{6 a c^2 (1-a x)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(1 + a*x)^3/(6*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 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 37

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

Rubi steps

\begin{align*} \int \frac{e^{4 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx &=\int \frac{e^{4 \tanh ^{-1}(a x)}}{(c-a c x)^2} \, dx\\ &=\frac{\int \frac{(1+a x)^2}{(1-a x)^4} \, dx}{c^2}\\ &=\frac{(1+a x)^3}{6 a c^2 (1-a x)^3}\\ \end{align*}

Mathematica [A]  time = 0.0089624, size = 25, normalized size = 1. $\frac{(a x+1)^3}{6 a c^2 (1-a x)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(1 + a*x)^3/(6*a*c^2*(1 - a*x)^3)

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Maple [A]  time = 0.048, size = 42, normalized size = 1.7 \begin{align*}{\frac{1}{{c}^{2}} \left ( -{\frac{4}{3\,a \left ( ax-1 \right ) ^{3}}}-2\,{\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*c*x+c)^2,x)

[Out]

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

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Maxima [B]  time = 1.02094, size = 69, normalized size = 2.76 \begin{align*} -\frac{3 \, a^{2} x^{2} + 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*c*x+c)^2,x, algorithm="maxima")

[Out]

-1/3*(3*a^2*x^2 + 1)/(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 = 1.43907, size = 100, normalized size = 4. \begin{align*} -\frac{3 \, a^{2} x^{2} + 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*c*x+c)^2,x, algorithm="fricas")

[Out]

-1/3*(3*a^2*x^2 + 1)/(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 = 1.1261, size = 51, normalized size = 2.04 \begin{align*} - \frac{3 a^{2} x^{2} + 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*c*x+c)**2,x)

[Out]

-(3*a**2*x**2 + 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 [B]  time = 1.15261, size = 68, normalized size = 2.72 \begin{align*} -\frac{2}{{\left (a c x - c\right )}^{2} a} - \frac{1}{{\left (a c x - c\right )} a c} - \frac{4 \, c}{3 \,{\left (a c x - c\right )}^{3} a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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