3.427 \(\int \frac{e^{-2 \coth ^{-1}(a x)}}{(c-\frac{c}{a x})^3} \, dx\)

Optimal. Leaf size=57 \[ \frac{1}{2 a c^3 (1-a x)}+\frac{5 \log (1-a x)}{4 a c^3}-\frac{\log (a x+1)}{4 a c^3}+\frac{x}{c^3} \]

[Out]

x/c^3 + 1/(2*a*c^3*(1 - a*x)) + (5*Log[1 - a*x])/(4*a*c^3) - Log[1 + a*x]/(4*a*c^3)

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Rubi [A]  time = 0.150284, antiderivative size = 57, 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, 6131, 6129, 88} \[ \frac{1}{2 a c^3 (1-a x)}+\frac{5 \log (1-a x)}{4 a c^3}-\frac{\log (a x+1)}{4 a c^3}+\frac{x}{c^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^3),x]

[Out]

x/c^3 + 1/(2*a*c^3*(1 - a*x)) + (5*Log[1 - a*x])/(4*a*c^3) - Log[1 + a*x]/(4*a*c^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 6131

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

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 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 x}\right )^3} \, dx &=-\int \frac{e^{-2 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^3} \, dx\\ &=\frac{a^3 \int \frac{e^{-2 \tanh ^{-1}(a x)} x^3}{(1-a x)^3} \, dx}{c^3}\\ &=\frac{a^3 \int \frac{x^3}{(1-a x)^2 (1+a x)} \, dx}{c^3}\\ &=\frac{a^3 \int \left (\frac{1}{a^3}+\frac{1}{2 a^3 (-1+a x)^2}+\frac{5}{4 a^3 (-1+a x)}-\frac{1}{4 a^3 (1+a x)}\right ) \, dx}{c^3}\\ &=\frac{x}{c^3}+\frac{1}{2 a c^3 (1-a x)}+\frac{5 \log (1-a x)}{4 a c^3}-\frac{\log (1+a x)}{4 a c^3}\\ \end{align*}

Mathematica [A]  time = 0.110389, size = 56, normalized size = 0.98 \[ -\frac{1}{2 a c^3 (a x-1)}+\frac{5 \log (1-a x)}{4 a c^3}-\frac{\log (a x+1)}{4 a c^3}+\frac{x}{c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^3),x]

[Out]

x/c^3 - 1/(2*a*c^3*(-1 + a*x)) + (5*Log[1 - a*x])/(4*a*c^3) - Log[1 + a*x]/(4*a*c^3)

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Maple [A]  time = 0.044, size = 50, normalized size = 0.9 \begin{align*}{\frac{x}{{c}^{3}}}-{\frac{\ln \left ( ax+1 \right ) }{4\,a{c}^{3}}}-{\frac{1}{2\,a{c}^{3} \left ( ax-1 \right ) }}+{\frac{5\,\ln \left ( ax-1 \right ) }{4\,a{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(a*x-1)/(c-c/a/x)^3,x)

[Out]

x/c^3-1/4*ln(a*x+1)/a/c^3-1/2/a/c^3/(a*x-1)+5/4/a/c^3*ln(a*x-1)

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Maxima [A]  time = 1.03942, size = 72, normalized size = 1.26 \begin{align*} -\frac{1}{2 \,{\left (a^{2} c^{3} x - a c^{3}\right )}} + \frac{x}{c^{3}} - \frac{\log \left (a x + 1\right )}{4 \, a c^{3}} + \frac{5 \, \log \left (a x - 1\right )}{4 \, a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.80426, size = 136, normalized size = 2.39 \begin{align*} \frac{4 \, a^{2} x^{2} - 4 \, a x -{\left (a x - 1\right )} \log \left (a x + 1\right ) + 5 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{4 \,{\left (a^{2} c^{3} x - a c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.505827, size = 56, normalized size = 0.98 \begin{align*} a^{3} \left (- \frac{1}{2 a^{5} c^{3} x - 2 a^{4} c^{3}} + \frac{x}{a^{3} c^{3}} + \frac{\frac{5 \log{\left (x - \frac{1}{a} \right )}}{4} - \frac{\log{\left (x + \frac{1}{a} \right )}}{4}}{a^{4} c^{3}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)**3,x)

[Out]

a**3*(-1/(2*a**5*c**3*x - 2*a**4*c**3) + x/(a**3*c**3) + (5*log(x - 1/a)/4 - log(x + 1/a)/4)/(a**4*c**3))

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Giac [A]  time = 1.15894, size = 69, normalized size = 1.21 \begin{align*} \frac{x}{c^{3}} - \frac{\log \left ({\left | a x + 1 \right |}\right )}{4 \, a c^{3}} + \frac{5 \, \log \left ({\left | a x - 1 \right |}\right )}{4 \, a c^{3}} - \frac{1}{2 \,{\left (a x - 1\right )} a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)^3,x, algorithm="giac")

[Out]

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