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

Optimal. Leaf size=51 \[ -\frac{2 c^2}{a^3 x^2}-\frac{c^2}{3 a^4 x^3}-\frac{6 c^2}{a^2 x}+\frac{4 c^2 \log (x)}{a}+c^2 x \]

[Out]

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

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Rubi [A]  time = 0.147742, antiderivative size = 51, 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} \[ -\frac{2 c^2}{a^3 x^2}-\frac{c^2}{3 a^4 x^3}-\frac{6 c^2}{a^2 x}+\frac{4 c^2 \log (x)}{a}+c^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0205453, size = 51, normalized size = 1. \[ -\frac{2 c^2}{a^3 x^2}-\frac{c^2}{3 a^4 x^3}-\frac{6 c^2}{a^2 x}+\frac{4 c^2 \log (x)}{a}+c^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 50, normalized size = 1. \begin{align*} -{\frac{{c}^{2}}{3\,{a}^{4}{x}^{3}}}-2\,{\frac{{c}^{2}}{{x}^{2}{a}^{3}}}-6\,{\frac{{c}^{2}}{{a}^{2}x}}+x{c}^{2}+4\,{\frac{{c}^{2}\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)^2,x)

[Out]

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

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Maxima [A]  time = 1.0086, size = 62, normalized size = 1.22 \begin{align*} c^{2} x + \frac{4 \, c^{2} \log \left (x\right )}{a} - \frac{18 \, a^{2} c^{2} x^{2} + 6 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \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]

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

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Fricas [A]  time = 1.49503, size = 122, normalized size = 2.39 \begin{align*} \frac{3 \, a^{4} c^{2} x^{4} + 12 \, a^{3} c^{2} x^{3} \log \left (x\right ) - 18 \, a^{2} c^{2} x^{2} - 6 \, a c^{2} x - c^{2}}{3 \, a^{4} x^{3}} \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*c^2*x^4 + 12*a^3*c^2*x^3*log(x) - 18*a^2*c^2*x^2 - 6*a*c^2*x - c^2)/(a^4*x^3)

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Sympy [A]  time = 0.397397, size = 51, normalized size = 1. \begin{align*} \frac{a^{4} c^{2} x + 4 a^{3} c^{2} \log{\left (x \right )} - \frac{18 a^{2} c^{2} x^{2} + 6 a c^{2} x + c^{2}}{3 x^{3}}}{a^{4}} \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*c**2*x + 4*a**3*c**2*log(x) - (18*a**2*c**2*x**2 + 6*a*c**2*x + c**2)/(3*x**3))/a**4

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Giac [B]  time = 1.169, size = 151, normalized size = 2.96 \begin{align*} -\frac{4 \, c^{2} \log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | a \right |}}\right )}{a} + \frac{4 \, c^{2} \log \left ({\left | -\frac{1}{a x - 1} - 1 \right |}\right )}{a} + \frac{{\left (3 \, c^{2} + \frac{34 \, c^{2}}{a x - 1} + \frac{66 \, c^{2}}{{\left (a x - 1\right )}^{2}} + \frac{36 \, c^{2}}{{\left (a x - 1\right )}^{3}}\right )}{\left (a x - 1\right )}}{3 \, a{\left (\frac{1}{a x - 1} + 1\right )}^{3}} \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]

-4*c^2*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a + 4*c^2*log(abs(-1/(a*x - 1) - 1))/a + 1/3*(3*c^2 + 34*c^2/(a*
x - 1) + 66*c^2/(a*x - 1)^2 + 36*c^2/(a*x - 1)^3)*(a*x - 1)/(a*(1/(a*x - 1) + 1)^3)