### 3.799 $$\int e^{4 \coth ^{-1}(a x)} (c-\frac{c}{a^2 x^2})^3 \, dx$$

Optimal. Leaf size=63 $\frac{5 c^3}{3 a^4 x^3}+\frac{c^3}{a^5 x^4}+\frac{c^3}{5 a^6 x^5}-\frac{5 c^3}{a^2 x}+\frac{4 c^3 \log (x)}{a}+c^3 x$

[Out]

c^3/(5*a^6*x^5) + c^3/(a^5*x^4) + (5*c^3)/(3*a^4*x^3) - (5*c^3)/(a^2*x) + c^3*x + (4*c^3*Log[x])/a

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Rubi [A]  time = 0.151695, antiderivative size = 63, 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, 75} $\frac{5 c^3}{3 a^4 x^3}+\frac{c^3}{a^5 x^4}+\frac{c^3}{5 a^6 x^5}-\frac{5 c^3}{a^2 x}+\frac{4 c^3 \log (x)}{a}+c^3 x$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(4*ArcCoth[a*x])*(c - c/(a^2*x^2))^3,x]

[Out]

c^3/(5*a^6*x^5) + c^3/(a^5*x^4) + (5*c^3)/(3*a^4*x^3) - (5*c^3)/(a^2*x) + c^3*x + (4*c^3*Log[x])/a

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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

\begin{align*} \int e^{4 \coth ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right )^3 \, dx &=\int e^{4 \tanh ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right )^3 \, dx\\ &=-\frac{c^3 \int \frac{e^{4 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^3}{x^6} \, dx}{a^6}\\ &=-\frac{c^3 \int \frac{(1-a x) (1+a x)^5}{x^6} \, dx}{a^6}\\ &=-\frac{c^3 \int \left (-a^6+\frac{1}{x^6}+\frac{4 a}{x^5}+\frac{5 a^2}{x^4}-\frac{5 a^4}{x^2}-\frac{4 a^5}{x}\right ) \, dx}{a^6}\\ &=\frac{c^3}{5 a^6 x^5}+\frac{c^3}{a^5 x^4}+\frac{5 c^3}{3 a^4 x^3}-\frac{5 c^3}{a^2 x}+c^3 x+\frac{4 c^3 \log (x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.0231702, size = 63, normalized size = 1. $\frac{5 c^3}{3 a^4 x^3}+\frac{c^3}{a^5 x^4}+\frac{c^3}{5 a^6 x^5}-\frac{5 c^3}{a^2 x}+\frac{4 c^3 \log (x)}{a}+c^3 x$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - c/(a^2*x^2))^3,x]

[Out]

c^3/(5*a^6*x^5) + c^3/(a^5*x^4) + (5*c^3)/(3*a^4*x^3) - (5*c^3)/(a^2*x) + c^3*x + (4*c^3*Log[x])/a

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Maple [A]  time = 0.046, size = 60, normalized size = 1. \begin{align*}{\frac{{c}^{3}}{5\,{a}^{6}{x}^{5}}}+{\frac{{c}^{3}}{{a}^{5}{x}^{4}}}+{\frac{5\,{c}^{3}}{3\,{a}^{4}{x}^{3}}}-5\,{\frac{{c}^{3}}{{a}^{2}x}}+{c}^{3}x+4\,{\frac{{c}^{3}\ln \left ( x \right ) }{a}} \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)^3,x)

[Out]

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

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Maxima [A]  time = 1.07938, size = 80, normalized size = 1.27 \begin{align*} c^{3} x + \frac{4 \, c^{3} \log \left (x\right )}{a} - \frac{75 \, a^{4} c^{3} x^{4} - 25 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 3 \, c^{3}}{15 \, a^{6} x^{5}} \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)^3,x, algorithm="maxima")

[Out]

c^3*x + 4*c^3*log(x)/a - 1/15*(75*a^4*c^3*x^4 - 25*a^2*c^3*x^2 - 15*a*c^3*x - 3*c^3)/(a^6*x^5)

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Fricas [A]  time = 1.57881, size = 151, normalized size = 2.4 \begin{align*} \frac{15 \, a^{6} c^{3} x^{6} + 60 \, a^{5} c^{3} x^{5} \log \left (x\right ) - 75 \, a^{4} c^{3} x^{4} + 25 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x + 3 \, c^{3}}{15 \, a^{6} x^{5}} \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)^3,x, algorithm="fricas")

[Out]

1/15*(15*a^6*c^3*x^6 + 60*a^5*c^3*x^5*log(x) - 75*a^4*c^3*x^4 + 25*a^2*c^3*x^2 + 15*a*c^3*x + 3*c^3)/(a^6*x^5)

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Sympy [A]  time = 0.498356, size = 65, normalized size = 1.03 \begin{align*} \frac{a^{6} c^{3} x + 4 a^{5} c^{3} \log{\left (x \right )} - \frac{75 a^{4} c^{3} x^{4} - 25 a^{2} c^{3} x^{2} - 15 a c^{3} x - 3 c^{3}}{15 x^{5}}}{a^{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)**3,x)

[Out]

(a**6*c**3*x + 4*a**5*c**3*log(x) - (75*a**4*c**3*x**4 - 25*a**2*c**3*x**2 - 15*a*c**3*x - 3*c**3)/(15*x**5))/
a**6

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Giac [B]  time = 1.12317, size = 184, normalized size = 2.92 \begin{align*} -\frac{4 \, c^{3} \log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | a \right |}}\right )}{a} + \frac{4 \, c^{3} \log \left ({\left | -\frac{1}{a x - 1} - 1 \right |}\right )}{a} + \frac{{\left (15 \, c^{3} + \frac{107 \, c^{3}}{a x - 1} + \frac{235 \, c^{3}}{{\left (a x - 1\right )}^{2}} + \frac{170 \, c^{3}}{{\left (a x - 1\right )}^{3}} - \frac{30 \, c^{3}}{{\left (a x - 1\right )}^{4}} - \frac{60 \, c^{3}}{{\left (a x - 1\right )}^{5}}\right )}{\left (a x - 1\right )}}{15 \, a{\left (\frac{1}{a x - 1} + 1\right )}^{5}} \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)^3,x, algorithm="giac")

[Out]

-4*c^3*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a + 4*c^3*log(abs(-1/(a*x - 1) - 1))/a + 1/15*(15*c^3 + 107*c^3/
(a*x - 1) + 235*c^3/(a*x - 1)^2 + 170*c^3/(a*x - 1)^3 - 30*c^3/(a*x - 1)^4 - 60*c^3/(a*x - 1)^5)*(a*x - 1)/(a*
(1/(a*x - 1) + 1)^5)