### 3.819 $$\int \frac{e^{-2 \coth ^{-1}(a x)}}{(c-\frac{c}{a^2 x^2})^2} \, dx$$

Optimal. Leaf size=73 $-\frac{7}{4 a c^2 (a x+1)}+\frac{1}{4 a c^2 (a x+1)^2}+\frac{\log (1-a x)}{8 a c^2}-\frac{17 \log (a x+1)}{8 a c^2}+\frac{x}{c^2}$

[Out]

x/c^2 + 1/(4*a*c^2*(1 + a*x)^2) - 7/(4*a*c^2*(1 + a*x)) + Log[1 - a*x]/(8*a*c^2) - (17*Log[1 + a*x])/(8*a*c^2)

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Rubi [A]  time = 0.17431, antiderivative size = 73, 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, 88} $-\frac{7}{4 a c^2 (a x+1)}+\frac{1}{4 a c^2 (a x+1)^2}+\frac{\log (1-a x)}{8 a c^2}-\frac{17 \log (a x+1)}{8 a c^2}+\frac{x}{c^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - c/(a^2*x^2))^2),x]

[Out]

x/c^2 + 1/(4*a*c^2*(1 + a*x)^2) - 7/(4*a*c^2*(1 + a*x)) + Log[1 - a*x]/(8*a*c^2) - (17*Log[1 + a*x])/(8*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^2} \, dx &=-\int \frac{e^{-2 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^2} \, dx\\ &=-\frac{a^4 \int \frac{e^{-2 \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) (1+a x)^3} \, dx}{c^2}\\ &=-\frac{a^4 \int \left (-\frac{1}{a^4}-\frac{1}{8 a^4 (-1+a x)}+\frac{1}{2 a^4 (1+a x)^3}-\frac{7}{4 a^4 (1+a x)^2}+\frac{17}{8 a^4 (1+a x)}\right ) \, dx}{c^2}\\ &=\frac{x}{c^2}+\frac{1}{4 a c^2 (1+a x)^2}-\frac{7}{4 a c^2 (1+a x)}+\frac{\log (1-a x)}{8 a c^2}-\frac{17 \log (1+a x)}{8 a c^2}\\ \end{align*}

Mathematica [A]  time = 0.0468059, size = 70, normalized size = 0.96 $\frac{2 \left (4 a^3 x^3+8 a^2 x^2-3 a x-6\right )+(a x+1)^2 \log (1-a x)-17 (a x+1)^2 \log (a x+1)}{8 a (a c x+c)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - c/(a^2*x^2))^2),x]

[Out]

(2*(-6 - 3*a*x + 8*a^2*x^2 + 4*a^3*x^3) + (1 + a*x)^2*Log[1 - a*x] - 17*(1 + a*x)^2*Log[1 + a*x])/(8*a*(c + a*
c*x)^2)

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Maple [A]  time = 0.05, size = 65, normalized size = 0.9 \begin{align*}{\frac{x}{{c}^{2}}}+{\frac{1}{4\,a{c}^{2} \left ( ax+1 \right ) ^{2}}}-{\frac{7}{4\,a{c}^{2} \left ( ax+1 \right ) }}-{\frac{17\,\ln \left ( ax+1 \right ) }{8\,a{c}^{2}}}+{\frac{\ln \left ( ax-1 \right ) }{8\,a{c}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c^2+1/4/a/c^2/(a*x+1)^2-7/4/a/c^2/(a*x+1)-17/8*ln(a*x+1)/a/c^2+1/8/a/c^2*ln(a*x-1)

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Maxima [A]  time = 1.04798, size = 93, normalized size = 1.27 \begin{align*} -\frac{7 \, a x + 6}{4 \,{\left (a^{3} c^{2} x^{2} + 2 \, a^{2} c^{2} x + a c^{2}\right )}} + \frac{x}{c^{2}} - \frac{17 \, \log \left (a x + 1\right )}{8 \, a c^{2}} + \frac{\log \left (a x - 1\right )}{8 \, a c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(7*a*x + 6)/(a^3*c^2*x^2 + 2*a^2*c^2*x + a*c^2) + x/c^2 - 17/8*log(a*x + 1)/(a*c^2) + 1/8*log(a*x - 1)/(a
*c^2)

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Fricas [A]  time = 1.38272, size = 211, normalized size = 2.89 \begin{align*} \frac{8 \, a^{3} x^{3} + 16 \, a^{2} x^{2} - 6 \, a x - 17 \,{\left (a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x + 1\right ) +{\left (a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 12}{8 \,{\left (a^{3} c^{2} x^{2} + 2 \, a^{2} c^{2} x + a c^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*a^3*x^3 + 16*a^2*x^2 - 6*a*x - 17*(a^2*x^2 + 2*a*x + 1)*log(a*x + 1) + (a^2*x^2 + 2*a*x + 1)*log(a*x -
1) - 12)/(a^3*c^2*x^2 + 2*a^2*c^2*x + a*c^2)

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Sympy [A]  time = 0.599817, size = 73, normalized size = 1. \begin{align*} a^{4} \left (- \frac{7 a x + 6}{4 a^{7} c^{2} x^{2} + 8 a^{6} c^{2} x + 4 a^{5} c^{2}} + \frac{x}{a^{4} c^{2}} + \frac{\frac{\log{\left (x - \frac{1}{a} \right )}}{8} - \frac{17 \log{\left (x + \frac{1}{a} \right )}}{8}}{a^{5} c^{2}}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*(-(7*a*x + 6)/(4*a**7*c**2*x**2 + 8*a**6*c**2*x + 4*a**5*c**2) + x/(a**4*c**2) + (log(x - 1/a)/8 - 17*log
(x + 1/a)/8)/(a**5*c**2))

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Giac [A]  time = 1.14497, size = 77, normalized size = 1.05 \begin{align*} \frac{x}{c^{2}} - \frac{17 \, \log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{2}} + \frac{\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{2}} - \frac{7 \, a x + 6}{4 \,{\left (a x + 1\right )}^{2} a c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c^2 - 17/8*log(abs(a*x + 1))/(a*c^2) + 1/8*log(abs(a*x - 1))/(a*c^2) - 1/4*(7*a*x + 6)/((a*x + 1)^2*a*c^2)