### 3.803 $$\int \frac{e^{4 \coth ^{-1}(a x)}}{(c-\frac{c}{a^2 x^2})^2} \, dx$$

Optimal. Leaf size=71 $\frac{6}{a c^2 (1-a x)}-\frac{2}{a c^2 (1-a x)^2}+\frac{1}{3 a c^2 (1-a x)^3}+\frac{4 \log (1-a x)}{a c^2}+\frac{x}{c^2}$

[Out]

x/c^2 + 1/(3*a*c^2*(1 - a*x)^3) - 2/(a*c^2*(1 - a*x)^2) + 6/(a*c^2*(1 - a*x)) + (4*Log[1 - a*x])/(a*c^2)

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Rubi [A]  time = 0.172337, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {6167, 6157, 6150, 43} $\frac{6}{a c^2 (1-a x)}-\frac{2}{a c^2 (1-a x)^2}+\frac{1}{3 a c^2 (1-a x)^3}+\frac{4 \log (1-a x)}{a c^2}+\frac{x}{c^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/c^2 + 1/(3*a*c^2*(1 - a*x)^3) - 2/(a*c^2*(1 - a*x)^2) + 6/(a*c^2*(1 - a*x)) + (4*Log[1 - a*x])/(a*c^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 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0379981, size = 63, normalized size = 0.89 $\frac{3 a^4 x^4-9 a^3 x^3-9 a^2 x^2+27 a x+12 (a x-1)^3 \log (1-a x)-13}{3 a c^2 (a x-1)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-13 + 27*a*x - 9*a^2*x^2 - 9*a^3*x^3 + 3*a^4*x^4 + 12*(-1 + a*x)^3*Log[1 - a*x])/(3*a*c^2*(-1 + a*x)^3)

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Maple [A]  time = 0.056, size = 66, normalized size = 0.9 \begin{align*}{\frac{x}{{c}^{2}}}-2\,{\frac{1}{a{c}^{2} \left ( ax-1 \right ) ^{2}}}-{\frac{1}{3\,a{c}^{2} \left ( ax-1 \right ) ^{3}}}+4\,{\frac{\ln \left ( ax-1 \right ) }{a{c}^{2}}}-6\,{\frac{1}{a{c}^{2} \left ( ax-1 \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c^2-2/a/c^2/(a*x-1)^2-1/3/a/c^2/(a*x-1)^3+4/a/c^2*ln(a*x-1)-6/a/c^2/(a*x-1)

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Maxima [A]  time = 1.03598, size = 101, normalized size = 1.42 \begin{align*} -\frac{18 \, a^{2} x^{2} - 30 \, a x + 13}{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 )}} + \frac{x}{c^{2}} + \frac{4 \, \log \left (a x - 1\right )}{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/(c-c/a^2/x^2)^2,x, algorithm="maxima")

[Out]

-1/3*(18*a^2*x^2 - 30*a*x + 13)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2) + x/c^2 + 4*log(a*x - 1)/(
a*c^2)

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Fricas [A]  time = 1.50434, size = 215, normalized size = 3.03 \begin{align*} \frac{3 \, a^{4} x^{4} - 9 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 27 \, a x + 12 \,{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) - 13}{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/(c-c/a^2/x^2)^2,x, algorithm="fricas")

[Out]

1/3*(3*a^4*x^4 - 9*a^3*x^3 - 9*a^2*x^2 + 27*a*x + 12*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x - 1) - 13)/(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 [A]  time = 0.533337, size = 83, normalized size = 1.17 \begin{align*} a^{4} \left (- \frac{18 a^{2} x^{2} - 30 a x + 13}{3 a^{8} c^{2} x^{3} - 9 a^{7} c^{2} x^{2} + 9 a^{6} c^{2} x - 3 a^{5} c^{2}} + \frac{x}{a^{4} c^{2}} + \frac{4 \log{\left (a x - 1 \right )}}{a^{5} 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/(c-c/a**2/x**2)**2,x)

[Out]

a**4*(-(18*a**2*x**2 - 30*a*x + 13)/(3*a**8*c**2*x**3 - 9*a**7*c**2*x**2 + 9*a**6*c**2*x - 3*a**5*c**2) + x/(a
**4*c**2) + 4*log(a*x - 1)/(a**5*c**2))

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Giac [A]  time = 1.12164, size = 126, normalized size = 1.77 \begin{align*} \frac{a x - 1}{a c^{2}} - \frac{4 \, \log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | a \right |}}\right )}{a c^{2}} - \frac{\frac{18 \, a^{5} c^{4}}{a x - 1} + \frac{6 \, a^{5} c^{4}}{{\left (a x - 1\right )}^{2}} + \frac{a^{5} c^{4}}{{\left (a x - 1\right )}^{3}}}{3 \, a^{6} c^{6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x - 1)/(a*c^2) - 4*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/(a*c^2) - 1/3*(18*a^5*c^4/(a*x - 1) + 6*a^5*c^4/(
a*x - 1)^2 + a^5*c^4/(a*x - 1)^3)/(a^6*c^6)