\(\int e^{-2 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^2 \, dx\) [816]

Optimal result

Integrand size = 22, antiderivative size = 40 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {c^2}{3 a^4 x^3}-\frac {c^2}{a^3 x^2}+c^2 x-\frac {2 c^2 \log (x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6302, 6292, 6285, 76} \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {c^2}{3 a^4 x^3}-\frac {c^2}{a^3 x^2}-\frac {2 c^2 \log (x)}{a}+c^2 x \]

[In]

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

[Out]

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

Rule 76

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])

Rule 6285

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 6292

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

Rule 6302

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {c^2}{3 a^4 x^3}-\frac {c^2}{a^3 x^2}+c^2 x-\frac {2 c^2 \log (x)}{a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {c^{2} \left (a^{4} x -2 a^{3} \ln \left (x \right )+\frac {1}{3 x^{3}}-\frac {a}{x^{2}}\right )}{a^{4}}\)	\(32\)
risch	\(c^{2} x +\frac {-a \,c^{2} x +\frac {1}{3} c^{2}}{a^{4} x^{3}}-\frac {2 c^{2} \ln \left (x \right )}{a}\)	\(37\)
norman	\(\frac {a^{3} c^{2} x^{4}+\frac {c^{2}}{3 a}-c^{2} x}{a^{3} x^{3}}-\frac {2 c^{2} \ln \left (x \right )}{a}\)	\(44\)
parallelrisch	\(-\frac {-3 a^{4} c^{2} x^{4}+6 c^{2} \ln \left (x \right ) a^{3} x^{3}+3 a \,c^{2} x -c^{2}}{3 a^{4} x^{3}}\)	\(46\)
meijerg	\(\frac {c^{2} \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {c^{2} \ln \left (a x +1\right )}{a}-\frac {2 c^{2} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )\right )}{a}+\frac {2 c^{2} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{a x}\right )}{a}+\frac {c^{2} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )-\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}-\frac {c^{2} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}\)	\(169\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.08 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {3 \, a^{4} c^{2} x^{4} - 6 \, a^{3} c^{2} x^{3} \log \left (x\right ) - 3 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.98 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {a^{4} c^{2} x - 2 a^{3} c^{2} \log {\left (x \right )} + \frac {- 3 a c^{2} x + c^{2}}{3 x^{3}}}{a^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=c^{2} x - \frac {2 \, c^{2} \log \left (x\right )}{a} - \frac {3 \, a c^{2} x - c^{2}}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=c^{2} x - \frac {2 \, c^{2} \log \left ({\left | x \right |}\right )}{a} - \frac {3 \, a c^{2} x - c^{2}}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=-\frac {c^2\,\left (3\,a\,x-3\,a^4\,x^4+6\,a^3\,x^3\,\ln \left (x\right )-1\right )}{3\,a^4\,x^3} \]

[In]

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

[Out]

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