\(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-\frac {c}{a x})^3} \, dx\) [394]

Optimal result

Integrand size = 22, antiderivative size = 73 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {x}{c^3}+\frac {2}{3 a c^3 (1-a x)^3}-\frac {7}{2 a c^3 (1-a x)^2}+\frac {9}{a c^3 (1-a x)}+\frac {5 \log (1-a x)}{a c^3} \]

[Out]

x/c^3+2/3/a/c^3/(-a*x+1)^3-7/2/a/c^3/(-a*x+1)^2+9/a/c^3/(-a*x+1)+5*ln(-a*x+1)/a/c^3

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, 6266, 6264, 78} \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {9}{a c^3 (1-a x)}-\frac {7}{2 a c^3 (1-a x)^2}+\frac {2}{3 a c^3 (1-a x)^3}+\frac {5 \log (1-a x)}{a c^3}+\frac {x}{c^3} \]

[In]

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

[Out]

x/c^3 + 2/(3*a*c^3*(1 - a*x)^3) - 7/(2*a*c^3*(1 - a*x)^2) + 9/(a*c^3*(1 - a*x)) + (5*Log[1 - a*x])/(a*c^3)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 6264

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 6266

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 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 x}\right )^3} \, dx \\ & = \frac {a^3 \int \frac {e^{2 \text {arctanh}(a x)} x^3}{(1-a x)^3} \, dx}{c^3} \\ & = \frac {a^3 \int \frac {x^3 (1+a x)}{(1-a x)^4} \, dx}{c^3} \\ & = \frac {a^3 \int \left (\frac {1}{a^3}+\frac {2}{a^3 (-1+a x)^4}+\frac {7}{a^3 (-1+a x)^3}+\frac {9}{a^3 (-1+a x)^2}+\frac {5}{a^3 (-1+a x)}\right ) \, dx}{c^3} \\ & = \frac {x}{c^3}+\frac {2}{3 a c^3 (1-a x)^3}-\frac {7}{2 a c^3 (1-a x)^2}+\frac {9}{a c^3 (1-a x)}+\frac {5 \log (1-a x)}{a c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {-37+81 a x-36 a^2 x^2-18 a^3 x^3+6 a^4 x^4+30 (-1+a x)^3 \log (1-a x)}{6 a c^3 (-1+a x)^3} \]

[In]

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

[Out]

(-37 + 81*a*x - 36*a^2*x^2 - 18*a^3*x^3 + 6*a^4*x^4 + 30*(-1 + a*x)^3*Log[1 - a*x])/(6*a*c^3*(-1 + a*x)^3)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {x}{c^{3}}+\frac {-9 a \,c^{3} x^{2}+\frac {29 c^{3} x}{2}-\frac {37 c^{3}}{6 a}}{c^{6} \left (a x -1\right )^{3}}+\frac {5 \ln \left (a x -1\right )}{a \,c^{3}}\)	\(56\)
default	\(\frac {a^{3} \left (\frac {x}{a^{3}}-\frac {7}{2 a^{4} \left (a x -1\right )^{2}}-\frac {9}{a^{4} \left (a x -1\right )}-\frac {2}{3 a^{4} \left (a x -1\right )^{3}}+\frac {5 \ln \left (a x -1\right )}{a^{4}}\right )}{c^{3}}\)	\(61\)
norman	\(\frac {\frac {a^{3} x^{4}}{c}-\frac {5 x}{c}+\frac {25 a \,x^{2}}{2 c}-\frac {55 a^{2} x^{3}}{6 c}}{\left (a x -1\right )^{3} c^{2}}+\frac {5 \ln \left (a x -1\right )}{a \,c^{3}}\)	\(64\)
parallelrisch	\(\frac {6 a^{4} x^{4}+30 a^{3} \ln \left (a x -1\right ) x^{3}-55 a^{3} x^{3}-90 a^{2} \ln \left (a x -1\right ) x^{2}+75 a^{2} x^{2}+90 a \ln \left (a x -1\right ) x -30 a x -30 \ln \left (a x -1\right )}{6 c^{3} \left (a x -1\right )^{3} a}\)	\(91\)

[In]

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

[Out]

x/c^3+(-9*a*c^3*x^2+29/2*c^3*x-37/6*c^3/a)/c^6/(a*x-1)^3+5/a/c^3*ln(a*x-1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.37 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {6 \, a^{4} x^{4} - 18 \, a^{3} x^{3} - 36 \, a^{2} x^{2} + 81 \, a x + 30 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) - 37}{6 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \]

[In]

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

[Out]

1/6*(6*a^4*x^4 - 18*a^3*x^3 - 36*a^2*x^2 + 81*a*x + 30*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x - 1) - 37)/(a
^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {- 54 a^{2} x^{2} + 87 a x - 37}{6 a^{4} c^{3} x^{3} - 18 a^{3} c^{3} x^{2} + 18 a^{2} c^{3} x - 6 a c^{3}} + \frac {x}{c^{3}} + \frac {5 \log {\left (a x - 1 \right )}}{a c^{3}} \]

[In]

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

[Out]

(-54*a**2*x**2 + 87*a*x - 37)/(6*a**4*c**3*x**3 - 18*a**3*c**3*x**2 + 18*a**2*c**3*x - 6*a*c**3) + x/c**3 + 5*
log(a*x - 1)/(a*c**3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=-\frac {54 \, a^{2} x^{2} - 87 \, a x + 37}{6 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} + \frac {x}{c^{3}} + \frac {5 \, \log \left (a x - 1\right )}{a c^{3}} \]

[In]

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

[Out]

-1/6*(54*a^2*x^2 - 87*a*x + 37)/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3) + x/c^3 + 5*log(a*x - 1)/(
a*c^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {x}{c^{3}} + \frac {5 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{3}} - \frac {54 \, a^{2} x^{2} - 87 \, a x + 37}{6 \, {\left (a x - 1\right )}^{3} a c^{3}} \]

[In]

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

[Out]

x/c^3 + 5*log(abs(a*x - 1))/(a*c^3) - 1/6*(54*a^2*x^2 - 87*a*x + 37)/((a*x - 1)^3*a*c^3)

Mupad [B] (verification not implemented)

Time = 3.84 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {9\,a\,x^2-\frac {29\,x}{2}+\frac {37}{6\,a}}{-a^3\,c^3\,x^3+3\,a^2\,c^3\,x^2-3\,a\,c^3\,x+c^3}+\frac {x}{c^3}+\frac {5\,\ln \left (a\,x-1\right )}{a\,c^3} \]

[In]

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

[Out]

(9*a*x^2 - (29*x)/2 + 37/(6*a))/(c^3 + 3*a^2*c^3*x^2 - a^3*c^3*x^3 - 3*a*c^3*x) + x/c^3 + (5*log(a*x - 1))/(a*
c^3)