\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{x^4} \, dx\) [49]

Optimal result

Integrand size = 12, antiderivative size = 40 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^4} \, dx=\frac {1}{3 x^3}-\frac {a}{x^2}+\frac {2 a^2}{x}+2 a^3 \log (x)-2 a^3 \log (1+a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6302, 6261, 78} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^4} \, dx=2 a^3 \log (x)-2 a^3 \log (a x+1)+\frac {2 a^2}{x}-\frac {a}{x^2}+\frac {1}{3 x^3} \]

[In]

Int[1/(E^(2*ArcCoth[a*x])*x^4),x]

[Out]

1/(3*x^3) - a/x^2 + (2*a^2)/x + 2*a^3*Log[x] - 2*a^3*Log[1 + a*x]

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 6261

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^4} \, dx=\frac {1}{3 x^3}-\frac {a}{x^2}+\frac {2 a^2}{x}+2 a^3 \log (x)-2 a^3 \log (1+a x) \]

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*x^4),x]

[Out]

1/(3*x^3) - a/x^2 + (2*a^2)/x + 2*a^3*Log[x] - 2*a^3*Log[1 + a*x]

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95

method	result	size
norman	\(\frac {\frac {1}{3}+2 a^{2} x^{2}-a x}{x^{3}}+2 a^{3} \ln \left (x \right )-2 a^{3} \ln \left (a x +1\right )\)	\(38\)
default	\(\frac {1}{3 x^{3}}-\frac {a}{x^{2}}+\frac {2 a^{2}}{x}+2 a^{3} \ln \left (x \right )-2 a^{3} \ln \left (a x +1\right )\)	\(39\)
risch	\(\frac {\frac {1}{3}+2 a^{2} x^{2}-a x}{x^{3}}+2 a^{3} \ln \left (-x \right )-2 a^{3} \ln \left (a x +1\right )\)	\(40\)
parallelrisch	\(\frac {6 a^{3} \ln \left (x \right ) x^{3}-6 a^{3} \ln \left (a x +1\right ) x^{3}+1+6 a^{2} x^{2}-3 a x}{3 x^{3}}\)	\(44\)
meijerg	\(a^{3} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )-\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )-a^{3} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )\)	\(78\)

[In]

int((a*x-1)/(a*x+1)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^4} \, dx=-\frac {6 \, a^{3} x^{3} \log \left (a x + 1\right ) - 6 \, a^{3} x^{3} \log \left (x\right ) - 6 \, a^{2} x^{2} + 3 \, a x - 1}{3 \, x^{3}} \]

[In]

integrate((a*x-1)/(a*x+1)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^4} \, dx=2 a^{3} \left (\log {\left (x \right )} - \log {\left (x + \frac {1}{a} \right )}\right ) + \frac {6 a^{2} x^{2} - 3 a x + 1}{3 x^{3}} \]

[In]

integrate((a*x-1)/(a*x+1)/x**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^4} \, dx=-2 \, a^{3} \log \left (a x + 1\right ) + 2 \, a^{3} \log \left (x\right ) + \frac {6 \, a^{2} x^{2} - 3 \, a x + 1}{3 \, x^{3}} \]

[In]

integrate((a*x-1)/(a*x+1)/x^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^4} \, dx=-2 \, a^{3} \log \left ({\left | a x + 1 \right |}\right ) + 2 \, a^{3} \log \left ({\left | x \right |}\right ) + \frac {6 \, a^{2} x^{2} - 3 \, a x + 1}{3 \, x^{3}} \]

[In]

integrate((a*x-1)/(a*x+1)/x^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^4} \, dx=\frac {2\,a^2\,x^2-a\,x+\frac {1}{3}}{x^3}-4\,a^3\,\mathrm {atanh}\left (2\,a\,x+1\right ) \]

[In]

int((a*x - 1)/(x^4*(a*x + 1)),x)

[Out]

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