3.428 $$\int \frac{e^{-2 \coth ^{-1}(a x)}}{(c-\frac{c}{a x})^4} \, dx$$

Optimal. Leaf size=75 $\frac{7}{4 a c^4 (1-a x)}-\frac{1}{4 a c^4 (1-a x)^2}+\frac{17 \log (1-a x)}{8 a c^4}-\frac{\log (a x+1)}{8 a c^4}+\frac{x}{c^4}$

[Out]

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

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Rubi [A]  time = 0.162013, antiderivative size = 75, 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} $\frac{7}{4 a c^4 (1-a x)}-\frac{1}{4 a c^4 (1-a x)^2}+\frac{17 \log (1-a x)}{8 a c^4}-\frac{\log (a x+1)}{8 a c^4}+\frac{x}{c^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^4),x]

[Out]

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

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

Mathematica [A]  time = 0.146722, size = 73, normalized size = 0.97 $-\frac{7}{4 a c^4 (a x-1)}-\frac{1}{4 a c^4 (a x-1)^2}+\frac{17 \log (1-a x)}{8 a c^4}-\frac{\log (a x+1)}{8 a c^4}+\frac{x}{c^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^4),x]

[Out]

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

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Maple [A]  time = 0.046, size = 65, normalized size = 0.9 \begin{align*}{\frac{x}{{c}^{4}}}-{\frac{\ln \left ( ax+1 \right ) }{8\,a{c}^{4}}}-{\frac{1}{4\,a{c}^{4} \left ( ax-1 \right ) ^{2}}}-{\frac{7}{4\,a{c}^{4} \left ( ax-1 \right ) }}+{\frac{17\,\ln \left ( ax-1 \right ) }{8\,a{c}^{4}}} \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)^4,x)

[Out]

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

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Maxima [A]  time = 1.03637, size = 93, normalized size = 1.24 \begin{align*} -\frac{7 \, a x - 6}{4 \,{\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} + \frac{x}{c^{4}} - \frac{\log \left (a x + 1\right )}{8 \, a c^{4}} + \frac{17 \, \log \left (a x - 1\right )}{8 \, a c^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.67704, size = 211, normalized size = 2.81 \begin{align*} \frac{8 \, a^{3} x^{3} - 16 \, a^{2} x^{2} - 6 \, a x -{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 17 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 12}{8 \,{\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\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)^4,x, algorithm="fricas")

[Out]

1/8*(8*a^3*x^3 - 16*a^2*x^2 - 6*a*x - (a^2*x^2 - 2*a*x + 1)*log(a*x + 1) + 17*(a^2*x^2 - 2*a*x + 1)*log(a*x -
1) + 12)/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4)

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Sympy [A]  time = 0.614067, size = 73, normalized size = 0.97 \begin{align*} a^{4} \left (- \frac{7 a x - 6}{4 a^{7} c^{4} x^{2} - 8 a^{6} c^{4} x + 4 a^{5} c^{4}} + \frac{x}{a^{4} c^{4}} + \frac{\frac{17 \log{\left (x - \frac{1}{a} \right )}}{8} - \frac{\log{\left (x + \frac{1}{a} \right )}}{8}}{a^{5} c^{4}}\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)**4,x)

[Out]

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

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Giac [A]  time = 1.15018, size = 77, normalized size = 1.03 \begin{align*} \frac{x}{c^{4}} - \frac{\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{4}} + \frac{17 \, \log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{4}} - \frac{7 \, a x - 6}{4 \,{\left (a x - 1\right )}^{2} a c^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)^4,x, algorithm="giac")

[Out]

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