\(\int \frac {e^{4 \coth ^{-1}(a x)}}{x^3} \, dx\) [31]

Optimal result

Integrand size = 12, antiderivative size = 46 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^3} \, dx=-\frac {1}{2 x^2}-\frac {4 a}{x}+\frac {4 a^2}{1-a x}+8 a^2 \log (x)-8 a^2 \log (1-a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, 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, 90} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^3} \, dx=\frac {4 a^2}{1-a x}+8 a^2 \log (x)-8 a^2 \log (1-a x)-\frac {4 a}{x}-\frac {1}{2 x^2} \]

[In]

Int[E^(4*ArcCoth[a*x])/x^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^3} \, dx=-\frac {1}{2 x^2}-\frac {4 a}{x}+\frac {4 a^2}{1-a x}+8 a^2 \log (x)-8 a^2 \log (1-a x) \]

[In]

Integrate[E^(4*ArcCoth[a*x])/x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {1}{2 x^{2}}-\frac {4 a}{x}+8 a^{2} \ln \left (x \right )-\frac {4 a^{2}}{a x -1}-8 a^{2} \ln \left (a x -1\right )\)	\(43\)
norman	\(\frac {\frac {1}{2}-8 a^{3} x^{3}+\frac {7}{2} a x}{\left (a x -1\right ) x^{2}}+8 a^{2} \ln \left (x \right )-8 a^{2} \ln \left (a x -1\right )\)	\(45\)
risch	\(\frac {-8 a^{2} x^{2}+\frac {7}{2} a x +\frac {1}{2}}{x^{2} \left (a x -1\right )}-8 a^{2} \ln \left (a x -1\right )+8 a^{2} \ln \left (-x \right )\)	\(47\)
parallelrisch	\(\frac {16 a^{3} \ln \left (x \right ) x^{3}-16 a^{3} \ln \left (a x -1\right ) x^{3}-16 a^{3} x^{3}-16 a^{2} \ln \left (x \right ) x^{2}+16 a^{2} \ln \left (a x -1\right ) x^{2}+1+7 a x}{2 x^{2} \left (a x -1\right )}\)	\(75\)
meijerg	\(a^{2} \left (\frac {2 a x}{-2 a x +2}-\ln \left (-a x +1\right )+1+\ln \left (x \right )+\ln \left (-a \right )\right )-2 a^{2} \left (-\frac {3 a x}{-3 a x +3}+2 \ln \left (-a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (-a \right )+\frac {1}{a x}\right )+a^{2} \left (\frac {4 a x}{-4 a x +4}-3 \ln \left (-a x +1\right )+1+3 \ln \left (x \right )+3 \ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {2}{a x}\right )\)	\(133\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.59 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^3} \, dx=-\frac {16 \, a^{2} x^{2} - 7 \, a x + 16 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (x\right ) - 1}{2 \, {\left (a x^{3} - x^{2}\right )}} \]

[In]

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

[Out]

-1/2*(16*a^2*x^2 - 7*a*x + 16*(a^3*x^3 - a^2*x^2)*log(a*x - 1) - 16*(a^3*x^3 - a^2*x^2)*log(x) - 1)/(a*x^3 - x
^2)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^3} \, dx=8 a^{2} \left (\log {\left (x \right )} - \log {\left (x - \frac {1}{a} \right )}\right ) + \frac {- 16 a^{2} x^{2} + 7 a x + 1}{2 a x^{3} - 2 x^{2}} \]

[In]

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

[Out]

8*a**2*(log(x) - log(x - 1/a)) + (-16*a**2*x**2 + 7*a*x + 1)/(2*a*x**3 - 2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^3} \, dx=-8 \, a^{2} \log \left (a x - 1\right ) + 8 \, a^{2} \log \left (x\right ) - \frac {16 \, a^{2} x^{2} - 7 \, a x - 1}{2 \, {\left (a x^{3} - x^{2}\right )}} \]

[In]

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

[Out]

-8*a^2*log(a*x - 1) + 8*a^2*log(x) - 1/2*(16*a^2*x^2 - 7*a*x - 1)/(a*x^3 - x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.35 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^3} \, dx=8 \, a^{2} \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right ) - \frac {4 \, a^{2}}{a x - 1} + \frac {9 \, a^{2} + \frac {10 \, a^{2}}{a x - 1}}{2 \, {\left (\frac {1}{a x - 1} + 1\right )}^{2}} \]

[In]

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

[Out]

8*a^2*log(abs(-1/(a*x - 1) - 1)) - 4*a^2/(a*x - 1) + 1/2*(9*a^2 + 10*a^2/(a*x - 1))/(1/(a*x - 1) + 1)^2

Mupad [B] (verification not implemented)

Time = 4.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^3} \, dx=16\,a^2\,\mathrm {atanh}\left (2\,a\,x-1\right )+\frac {-8\,a^2\,x^2+\frac {7\,a\,x}{2}+\frac {1}{2}}{a\,x^3-x^2} \]

[In]

int((a*x + 1)^2/(x^3*(a*x - 1)^2),x)

[Out]

16*a^2*atanh(2*a*x - 1) + ((7*a*x)/2 - 8*a^2*x^2 + 1/2)/(a*x^3 - x^2)