\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^3} \, dx\) [820]

Optimal result

Integrand size = 22, antiderivative size = 108 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {x}{c^3}+\frac {1}{16 a c^3 (1-a x)}-\frac {1}{12 a c^3 (1+a x)^3}+\frac {5}{8 a c^3 (1+a x)^2}-\frac {39}{16 a c^3 (1+a x)}+\frac {\log (1-a x)}{4 a c^3}-\frac {9 \log (1+a x)}{4 a c^3} \]

[Out]

x/c^3+1/16/a/c^3/(-a*x+1)-1/12/a/c^3/(a*x+1)^3+5/8/a/c^3/(a*x+1)^2-39/16/a/c^3/(a*x+1)+1/4*ln(-a*x+1)/a/c^3-9/
4*ln(a*x+1)/a/c^3

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 108, 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} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {1}{16 a c^3 (1-a x)}-\frac {39}{16 a c^3 (a x+1)}+\frac {5}{8 a c^3 (a x+1)^2}-\frac {1}{12 a c^3 (a x+1)^3}+\frac {\log (1-a x)}{4 a c^3}-\frac {9 \log (a x+1)}{4 a c^3}+\frac {x}{c^3} \]

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - c/(a^2*x^2))^3),x]

[Out]

x/c^3 + 1/(16*a*c^3*(1 - a*x)) - 1/(12*a*c^3*(1 + a*x)^3) + 5/(8*a*c^3*(1 + a*x)^2) - 39/(16*a*c^3*(1 + a*x))
+ Log[1 - a*x]/(4*a*c^3) - (9*Log[1 + a*x])/(4*a*c^3)

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 )^3} \, dx \\ & = \frac {a^6 \int \frac {e^{-2 \text {arctanh}(a x)} x^6}{\left (1-a^2 x^2\right )^3} \, dx}{c^3} \\ & = \frac {a^6 \int \frac {x^6}{(1-a x)^2 (1+a x)^4} \, dx}{c^3} \\ & = \frac {a^6 \int \left (\frac {1}{a^6}+\frac {1}{16 a^6 (-1+a x)^2}+\frac {1}{4 a^6 (-1+a x)}+\frac {1}{4 a^6 (1+a x)^4}-\frac {5}{4 a^6 (1+a x)^3}+\frac {39}{16 a^6 (1+a x)^2}-\frac {9}{4 a^6 (1+a x)}\right ) \, dx}{c^3} \\ & = \frac {x}{c^3}+\frac {1}{16 a c^3 (1-a x)}-\frac {1}{12 a c^3 (1+a x)^3}+\frac {5}{8 a c^3 (1+a x)^2}-\frac {39}{16 a c^3 (1+a x)}+\frac {\log (1-a x)}{4 a c^3}-\frac {9 \log (1+a x)}{4 a c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {2 \left (11+7 a x-24 a^2 x^2-15 a^3 x^3+12 a^4 x^4+6 a^5 x^5\right )+3 (-1+a x) (1+a x)^3 \log (1-a x)-27 (-1+a x) (1+a x)^3 \log (1+a x)}{12 a (-1+a x) (c+a c x)^3} \]

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - c/(a^2*x^2))^3),x]

[Out]

(2*(11 + 7*a*x - 24*a^2*x^2 - 15*a^3*x^3 + 12*a^4*x^4 + 6*a^5*x^5) + 3*(-1 + a*x)*(1 + a*x)^3*Log[1 - a*x] - 2
7*(-1 + a*x)*(1 + a*x)^3*Log[1 + a*x])/(12*a*(-1 + a*x)*(c + a*c*x)^3)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {a^{6} \left (-\frac {9 \ln \left (a x +1\right )}{4 a^{7}}-\frac {1}{12 a^{7} \left (a x +1\right )^{3}}+\frac {5}{8 a^{7} \left (a x +1\right )^{2}}-\frac {39}{16 a^{7} \left (a x +1\right )}+\frac {x}{a^{6}}-\frac {1}{16 a^{7} \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{4 a^{7}}\right )}{c^{3}}\)	\(84\)
risch	\(\frac {x}{c^{3}}+\frac {-\frac {5 a^{2} c^{3} x^{3}}{2}-2 a \,c^{3} x^{2}+\frac {13 c^{3} x}{6}+\frac {11 c^{3}}{6 a}}{c^{6} \left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right )}+\frac {\ln \left (-a x +1\right )}{4 a \,c^{3}}-\frac {9 \ln \left (a x +1\right )}{4 a \,c^{3}}\)	\(93\)
norman	\(\frac {\frac {a^{5} x^{6}}{c}+\frac {5 x}{2 c}+\frac {3 a \,x^{2}}{2 c}-\frac {31 a^{2} x^{3}}{6 c}-\frac {8 a^{3} x^{4}}{3 c}+\frac {17 a^{4} x^{5}}{6 c}}{c^{2} \left (a x +1\right )^{3} \left (a x -1\right )^{2}}+\frac {\ln \left (a x -1\right )}{4 a \,c^{3}}-\frac {9 \ln \left (a x +1\right )}{4 a \,c^{3}}\)	\(107\)
parallelrisch	\(\frac {12 a^{5} x^{5}+3 \ln \left (a x -1\right ) x^{4} a^{4}-27 \ln \left (a x +1\right ) x^{4} a^{4}+46 a^{4} x^{4}+6 a^{3} \ln \left (a x -1\right ) x^{3}-54 a^{3} \ln \left (a x +1\right ) x^{3}+14 a^{3} x^{3}-48 a^{2} x^{2}-6 a \ln \left (a x -1\right ) x +54 a \ln \left (a x +1\right ) x -30 a x -3 \ln \left (a x -1\right )+27 \ln \left (a x +1\right )}{12 c^{3} \left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right ) a}\)	\(156\)

[In]

int((a*x-1)/(a*x+1)/(c-c/a^2/x^2)^3,x,method=_RETURNVERBOSE)

[Out]

a^6/c^3*(-9/4*ln(a*x+1)/a^7-1/12/a^7/(a*x+1)^3+5/8/a^7/(a*x+1)^2-39/16/a^7/(a*x+1)+x/a^6-1/16/a^7/(a*x-1)+1/4/
a^7*ln(a*x-1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {12 \, a^{5} x^{5} + 24 \, a^{4} x^{4} - 30 \, a^{3} x^{3} - 48 \, a^{2} x^{2} + 14 \, a x - 27 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x - 1\right ) + 22}{12 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} \]

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a^2/x^2)^3,x, algorithm="fricas")

[Out]

1/12*(12*a^5*x^5 + 24*a^4*x^4 - 30*a^3*x^3 - 48*a^2*x^2 + 14*a*x - 27*(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)*log(a*
x + 1) + 3*(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)*log(a*x - 1) + 22)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a
*c^3)

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=a^{6} \left (\frac {- 15 a^{3} x^{3} - 12 a^{2} x^{2} + 13 a x + 11}{6 a^{11} c^{3} x^{4} + 12 a^{10} c^{3} x^{3} - 12 a^{8} c^{3} x - 6 a^{7} c^{3}} + \frac {x}{a^{6} c^{3}} + \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{4} - \frac {9 \log {\left (x + \frac {1}{a} \right )}}{4}}{a^{7} c^{3}}\right ) \]

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a**2/x**2)**3,x)

[Out]

a**6*((-15*a**3*x**3 - 12*a**2*x**2 + 13*a*x + 11)/(6*a**11*c**3*x**4 + 12*a**10*c**3*x**3 - 12*a**8*c**3*x -
6*a**7*c**3) + x/(a**6*c**3) + (log(x - 1/a)/4 - 9*log(x + 1/a)/4)/(a**7*c**3))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=-\frac {15 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 13 \, a x - 11}{6 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} + \frac {x}{c^{3}} - \frac {9 \, \log \left (a x + 1\right )}{4 \, a c^{3}} + \frac {\log \left (a x - 1\right )}{4 \, a c^{3}} \]

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a^2/x^2)^3,x, algorithm="maxima")

[Out]

-1/6*(15*a^3*x^3 + 12*a^2*x^2 - 13*a*x - 11)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3) + x/c^3 - 9/4
*log(a*x + 1)/(a*c^3) + 1/4*log(a*x - 1)/(a*c^3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {x}{c^{3}} - \frac {9 \, \log \left ({\left | a x + 1 \right |}\right )}{4 \, a c^{3}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{4 \, a c^{3}} - \frac {15 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 13 \, a x - 11}{6 \, {\left (a x + 1\right )}^{3} {\left (a x - 1\right )} a c^{3}} \]

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a^2/x^2)^3,x, algorithm="giac")

[Out]

x/c^3 - 9/4*log(abs(a*x + 1))/(a*c^3) + 1/4*log(abs(a*x - 1))/(a*c^3) - 1/6*(15*a^3*x^3 + 12*a^2*x^2 - 13*a*x
- 11)/((a*x + 1)^3*(a*x - 1)*a*c^3)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {x}{c^3}-\frac {\frac {13\,x}{6}-2\,a\,x^2+\frac {11}{6\,a}-\frac {5\,a^2\,x^3}{2}}{-a^4\,c^3\,x^4-2\,a^3\,c^3\,x^3+2\,a\,c^3\,x+c^3}+\frac {\ln \left (a\,x-1\right )}{4\,a\,c^3}-\frac {9\,\ln \left (a\,x+1\right )}{4\,a\,c^3} \]

[In]

int((a*x - 1)/((c - c/(a^2*x^2))^3*(a*x + 1)),x)

[Out]

x/c^3 - ((13*x)/6 - 2*a*x^2 + 11/(6*a) - (5*a^2*x^3)/2)/(c^3 - 2*a^3*c^3*x^3 - a^4*c^3*x^4 + 2*a*c^3*x) + log(
a*x - 1)/(4*a*c^3) - (9*log(a*x + 1))/(4*a*c^3)