\(\int \frac {e^{4 \coth ^{-1}(a x)}}{x^2} \, dx\) [30]

Optimal result

Integrand size = 12, antiderivative size = 32 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {1}{x}+\frac {4 a}{1-a x}+4 a \log (x)-4 a \log (1-a x) \]

[Out]

-1/x+4*a/(-a*x+1)+4*a*ln(x)-4*a*ln(-a*x+1)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, 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} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^2} \, dx=\frac {4 a}{1-a x}+4 a \log (x)-4 a \log (1-a x)-\frac {1}{x} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {1}{x}+\frac {4 a}{1-a x}+4 a \log (x)-4 a \log (1-a x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {1}{x}+4 a \ln \left (x \right )-\frac {4 a}{a x -1}-4 a \ln \left (a x -1\right )\)	\(31\)
risch	\(\frac {-5 a x +1}{\left (a x -1\right ) x}+4 a \ln \left (-x \right )-4 a \ln \left (a x -1\right )\)	\(35\)
norman	\(\frac {-5 a^{2} x^{2}+1}{\left (a x -1\right ) x}+4 a \ln \left (x \right )-4 a \ln \left (a x -1\right )\)	\(37\)
parallelrisch	\(\frac {4 a^{2} \ln \left (x \right ) x^{2}-4 a^{2} \ln \left (a x -1\right ) x^{2}-5 a^{2} x^{2}+1-4 a \ln \left (x \right ) x +4 a \ln \left (a x -1\right ) x}{\left (a x -1\right ) x}\)	\(62\)
meijerg	\(\frac {a^{2} x}{-a x +1}+2 a \left (\frac {2 a x}{-2 a x +2}-\ln \left (-a x +1\right )+1+\ln \left (x \right )+\ln \left (-a \right )\right )-a \left (-\frac {3 a x}{-3 a x +3}+2 \ln \left (-a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (-a \right )+\frac {1}{a x}\right )\)	\(90\)

[In]

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

[Out]

-1/x+4*a*ln(x)-4*a/(a*x-1)-4*a*ln(a*x-1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {5 \, a x + 4 \, {\left (a^{2} x^{2} - a x\right )} \log \left (a x - 1\right ) - 4 \, {\left (a^{2} x^{2} - a x\right )} \log \left (x\right ) - 1}{a x^{2} - x} \]

[In]

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

[Out]

-(5*a*x + 4*(a^2*x^2 - a*x)*log(a*x - 1) - 4*(a^2*x^2 - a*x)*log(x) - 1)/(a*x^2 - x)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^2} \, dx=4 a \left (\log {\left (x \right )} - \log {\left (x - \frac {1}{a} \right )}\right ) + \frac {- 5 a x + 1}{a x^{2} - x} \]

[In]

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

[Out]

4*a*(log(x) - log(x - 1/a)) + (-5*a*x + 1)/(a*x**2 - x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^2} \, dx=-4 \, a \log \left (a x - 1\right ) + 4 \, a \log \left (x\right ) - \frac {5 \, a x - 1}{a x^{2} - x} \]

[In]

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

[Out]

-4*a*log(a*x - 1) + 4*a*log(x) - (5*a*x - 1)/(a*x^2 - x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^2} \, dx=4 \, a \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right ) - \frac {4 \, a}{a x - 1} + \frac {a}{\frac {1}{a x - 1} + 1} \]

[In]

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

[Out]

4*a*log(abs(-1/(a*x - 1) - 1)) - 4*a/(a*x - 1) + a/(1/(a*x - 1) + 1)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{x^2} \, dx=8\,a\,\mathrm {atanh}\left (2\,a\,x-1\right )+\frac {5\,a\,x-1}{x-a\,x^2} \]

[In]

int((a*x + 1)^2/(x^2*(a*x - 1)^2),x)

[Out]

8*a*atanh(2*a*x - 1) + (5*a*x - 1)/(x - a*x^2)