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

Optimal. Leaf size=40 \[ \frac {c^2}{a^2 x}+\frac {4 c^2 \log (x)}{a}-\frac {8 c^2 \log (a x+1)}{a}+c^2 x \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 40, normalized size of antiderivative = 1.00, 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} \[ \frac {c^2}{a^2 x}+\frac {4 c^2 \log (x)}{a}-\frac {8 c^2 \log (a x+1)}{a}+c^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 42, normalized size = 1.05 \[ \frac {c^2}{a^2 x}+\frac {4 c^2 \log (a x)}{a}-\frac {8 c^2 \log (a x+1)}{a}+c^2 x \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 43, normalized size = 1.08 \[ \frac {a^{2} c^{2} x^{2} - 8 \, a c^{2} x \log \left (a x + 1\right ) + 4 \, a c^{2} x \log \relax (x) + c^{2}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*c^2*x^2 - 8*a*c^2*x*log(a*x + 1) + 4*a*c^2*x*log(x) + c^2)/(a^2*x)

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giac [A]  time = 0.14, size = 42, normalized size = 1.05 \[ c^{2} x - \frac {8 \, c^{2} \log \left ({\left | a x + 1 \right |}\right )}{a} + \frac {4 \, c^{2} \log \left ({\left | x \right |}\right )}{a} + \frac {c^{2}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 41, normalized size = 1.02 \[ \frac {c^{2}}{a^{2} x}+c^{2} x +\frac {4 c^{2} \ln \relax (x )}{a}-\frac {8 c^{2} \ln \left (a x +1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 40, normalized size = 1.00 \[ c^{2} x - \frac {8 \, c^{2} \log \left (a x + 1\right )}{a} + \frac {4 \, c^{2} \log \relax (x)}{a} + \frac {c^{2}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*x - 8*c^2*log(a*x + 1)/a + 4*c^2*log(x)/a + c^2/(a^2*x)

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mupad [B]  time = 1.22, size = 40, normalized size = 1.00 \[ c^2\,x+\frac {c^2}{a^2\,x}+\frac {4\,c^2\,\ln \relax (x)}{a}-\frac {8\,c^2\,\ln \left (a\,x+1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*x + c^2/(a^2*x) + (4*c^2*log(x))/a - (8*c^2*log(a*x + 1))/a

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sympy [A]  time = 0.27, size = 31, normalized size = 0.78 \[ c^{2} x + \frac {4 c^{2} \left (\log {\relax (x )} - 2 \log {\left (x + \frac {1}{a} \right )}\right )}{a} + \frac {c^{2}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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