\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^2} \, dx\) [819]

Optimal result

Integrand size = 22, antiderivative size = 73 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {x}{c^2}+\frac {1}{4 a c^2 (1+a x)^2}-\frac {7}{4 a c^2 (1+a x)}+\frac {\log (1-a x)}{8 a c^2}-\frac {17 \log (1+a x)}{8 a c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 73, 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, 90} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=-\frac {7}{4 a c^2 (a x+1)}+\frac {1}{4 a c^2 (a x+1)^2}+\frac {\log (1-a x)}{8 a c^2}-\frac {17 \log (a x+1)}{8 a c^2}+\frac {x}{c^2} \]

[In]

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

[Out]

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

Rule 90

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {2 \left (-6-3 a x+8 a^2 x^2+4 a^3 x^3\right )+(1+a x)^2 \log (1-a x)-17 (1+a x)^2 \log (1+a x)}{8 a (c+a c x)^2} \]

[In]

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

[Out]

(2*(-6 - 3*a*x + 8*a^2*x^2 + 4*a^3*x^3) + (1 + a*x)^2*Log[1 - a*x] - 17*(1 + a*x)^2*Log[1 + a*x])/(8*a*(c + a*
c*x)^2)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {a^{4} \left (-\frac {17 \ln \left (a x +1\right )}{8 a^{5}}+\frac {1}{4 a^{5} \left (a x +1\right )^{2}}-\frac {7}{4 a^{5} \left (a x +1\right )}+\frac {x}{a^{4}}+\frac {\ln \left (a x -1\right )}{8 a^{5}}\right )}{c^{2}}\)	\(60\)
risch	\(\frac {x}{c^{2}}+\frac {-\frac {7 c^{2} x}{4}-\frac {3 c^{2}}{2 a}}{c^{4} \left (a x +1\right )^{2}}-\frac {17 \ln \left (a x +1\right )}{8 a \,c^{2}}+\frac {\ln \left (-a x +1\right )}{8 a \,c^{2}}\)	\(62\)
norman	\(\frac {\frac {a^{3} x^{4}}{c}-\frac {9 x}{4 c}-\frac {5 a \,x^{2}}{4 c}+\frac {5 a^{2} x^{3}}{2 c}}{\left (a x +1\right )^{2} c \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{8 a \,c^{2}}-\frac {17 \ln \left (a x +1\right )}{8 a \,c^{2}}\)	\(85\)
parallelrisch	\(\frac {8 a^{3} x^{3}+a^{2} \ln \left (a x -1\right ) x^{2}-17 a^{2} \ln \left (a x +1\right ) x^{2}+28 a^{2} x^{2}+2 a \ln \left (a x -1\right ) x -34 a \ln \left (a x +1\right ) x +18 a x +\ln \left (a x -1\right )-17 \ln \left (a x +1\right )}{8 c^{2} \left (a x +1\right )^{2} a}\)	\(98\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {8 \, a^{3} x^{3} + 16 \, a^{2} x^{2} - 6 \, a x - 17 \, {\left (a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x + 1\right ) + {\left (a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 12}{8 \, {\left (a^{3} c^{2} x^{2} + 2 \, a^{2} c^{2} x + a c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=a^{4} \left (\frac {- 7 a x - 6}{4 a^{7} c^{2} x^{2} + 8 a^{6} c^{2} x + 4 a^{5} c^{2}} + \frac {x}{a^{4} c^{2}} + \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{8} - \frac {17 \log {\left (x + \frac {1}{a} \right )}}{8}}{a^{5} c^{2}}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=-\frac {7 \, a x + 6}{4 \, {\left (a^{3} c^{2} x^{2} + 2 \, a^{2} c^{2} x + a c^{2}\right )}} + \frac {x}{c^{2}} - \frac {17 \, \log \left (a x + 1\right )}{8 \, a c^{2}} + \frac {\log \left (a x - 1\right )}{8 \, a c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {x}{c^{2}} - \frac {17 \, \log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{2}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{2}} - \frac {7 \, a x + 6}{4 \, {\left (a x + 1\right )}^{2} a c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {x}{c^2}-\frac {\frac {7\,x}{4}+\frac {3}{2\,a}}{a^2\,c^2\,x^2+2\,a\,c^2\,x+c^2}+\frac {\ln \left (a\,x-1\right )}{8\,a\,c^2}-\frac {17\,\ln \left (a\,x+1\right )}{8\,a\,c^2} \]

[In]

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

[Out]

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