\(\int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx\) [70]

Optimal result

Integrand size = 12, antiderivative size = 86 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=-\frac {2}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {2}{1-\sqrt {\frac {1-a x}{1+a x}}}-\log (1+a x)-2 \log \left (1-\sqrt {\frac {1-a x}{1+a x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6472, 1643, 266} \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\frac {2}{1-\sqrt {\frac {1-a x}{a x+1}}}-\frac {2}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}-\log (a x+1)-2 \log \left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \]

[In]

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

[Out]

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 6472

Int[E^(ArcSech[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[(1 - u)/(1 + u)] + (1/u)*Sqrt[(1 - u)/(
1 + u)])^n, x] /; FreeQ[m, x] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (\frac {1}{a x}+\sqrt {\frac {1-a x}{1+a x}}+\frac {\sqrt {\frac {1-a x}{1+a x}}}{a x}\right )^2}{x} \, dx \\ & = 4 \text {Subst}\left (\int \frac {x (1+x)}{(-1+x)^3 \left (1+x^2\right )} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right ) \\ & = 4 \text {Subst}\left (\int \left (\frac {1}{(-1+x)^3}+\frac {1}{2 (-1+x)^2}-\frac {1}{2 (-1+x)}+\frac {x}{2 \left (1+x^2\right )}\right ) \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right ) \\ & = -\frac {2}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {2}{1-\sqrt {\frac {1-a x}{1+a x}}}-2 \log \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )+2 \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right ) \\ & = -\frac {2}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {2}{1-\sqrt {\frac {1-a x}{1+a x}}}-\log (1+a x)-2 \log \left (1-\sqrt {\frac {1-a x}{1+a x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=-\frac {1}{a^2 x^2}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{a^2 x^2}-2 \log (x)+\log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.28

method	result	size
default	\(\frac {-a^{2} \ln \left (x \right )-\frac {1}{2 x^{2}}}{a^{2}}+\frac {\sqrt {\frac {a x +1}{a x}}\, \sqrt {-\frac {a x -1}{a x}}\, \left (a^{2} x^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\sqrt {-a^{2} x^{2}+1}\right )}{a x \sqrt {-a^{2} x^{2}+1}}-\frac {1}{2 a^{2} x^{2}}\)	\(110\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.60 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\frac {a^{2} x^{2} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - a^{2} x^{2} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) - 2 \, a^{2} x^{2} \log \left (x\right ) - 2 \, a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 2}{2 \, a^{2} x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.89 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\frac {- 2 a^{2} \cdot \left (2 \sqrt {-1 + \frac {1}{a x}} \left (\frac {\left (1 + \frac {1}{a x}\right )^{\frac {3}{2}}}{4} - \frac {\sqrt {1 + \frac {1}{a x}}}{4}\right ) - \log {\left (2 \sqrt {-1 + \frac {1}{a x}} + 2 \sqrt {1 + \frac {1}{a x}} \right )}\right ) - a^{2} \log {\left (x \right )} - \frac {1}{x^{2}}}{a^{2}} \]

[In]

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

[Out]

(-2*a**2*(2*sqrt(-1 + 1/(a*x))*((1 + 1/(a*x))**(3/2)/4 - sqrt(1 + 1/(a*x))/4) - log(2*sqrt(-1 + 1/(a*x)) + 2*s
qrt(1 + 1/(a*x)))) - a**2*log(x) - 1/x**2)/a**2

Maxima [F]

\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\int { \frac {{\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2}}{x} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\int { \frac {{\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2}}{x} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.88 (sec) , antiderivative size = 323, normalized size of antiderivative = 3.76 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\ln \left (\frac {1}{x}\right )-4\,\mathrm {atanh}\left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )+2\,\mathrm {acosh}\left (\frac {1}{a\,x}\right )+\frac {\frac {28\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^3}+\frac {28\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^5}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^5}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^7}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^7}+\frac {4\,\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}{\sqrt {\frac {1}{a\,x}+1}-1}}{1+\frac {6\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}-\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}-\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}}-\frac {1}{a^2\,x^2} \]

[In]

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

[Out]

log(1/x) - 4*atanh(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1)) + 2*acosh(1/(a*x)) + ((28*((1/(a*x) -
1)^(1/2) - 1i)^3)/((1/(a*x) + 1)^(1/2) - 1)^3 + (28*((1/(a*x) - 1)^(1/2) - 1i)^5)/((1/(a*x) + 1)^(1/2) - 1)^5
+ (4*((1/(a*x) - 1)^(1/2) - 1i)^7)/((1/(a*x) + 1)^(1/2) - 1)^7 + (4*((1/(a*x) - 1)^(1/2) - 1i))/((1/(a*x) + 1
)^(1/2) - 1))/((6*((1/(a*x) - 1)^(1/2) - 1i)^4)/((1/(a*x) + 1)^(1/2) - 1)^4 - (4*((1/(a*x) - 1)^(1/2) - 1i)^2)
/((1/(a*x) + 1)^(1/2) - 1)^2 - (4*((1/(a*x) - 1)^(1/2) - 1i)^6)/((1/(a*x) + 1)^(1/2) - 1)^6 + ((1/(a*x) - 1)^(
1/2) - 1i)^8/((1/(a*x) + 1)^(1/2) - 1)^8 + 1) - 1/(a^2*x^2)