\(\int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^2} \, dx\) [71]

Optimal result

Integrand size = 12, antiderivative size = 57 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^2} \, dx=-\frac {4 a}{3 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {2 a}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6472, 45} \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^2} \, dx=\frac {2 a}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}-\frac {4 a}{3 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3} \]

[In]

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

[Out]

(-4*a)/(3*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^3) + (2*a)/(1 - Sqrt[(1 - a*x)/(1 + a*x)])^2

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^2} \, dx=\frac {-2+3 a^2 x^2+2 (-1+a x) \sqrt {\frac {1-a x}{1+a x}} (1+a x)^2}{3 a^2 x^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28

method	result	size
default	\(\frac {-\frac {1}{3 x^{3}}+\frac {a^{2}}{x}}{a^{2}}+\frac {2 \sqrt {\frac {a x +1}{a x}}\, \sqrt {-\frac {a x -1}{a x}}\, \left (a^{2} x^{2}-1\right )}{3 a \,x^{2}}-\frac {1}{3 a^{2} x^{3}}\)	\(73\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^2} \, dx=\frac {3 \, a^{2} x^{2} + 2 \, {\left (a^{3} x^{3} - a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 2}{3 \, a^{2} x^{3}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^2} \, dx=\frac {\int \frac {2}{x^{4}}\, dx + \int \left (- \frac {a^{2}}{x^{2}}\right )\, dx + \int \frac {2 a \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}}{x^{3}}\, dx}{a^{2}} \]

[In]

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

[Out]

(Integral(2/x**4, x) + Integral(-a**2/x**2, x) + Integral(2*a*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x))/x**3, x))/a
**2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^2} \, dx=\frac {1}{x} + \frac {2 \, {\left (a^{2} x^{3} - x\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{3 \, a^{2} x^{4}} - \frac {2}{3 \, a^{2} x^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.76 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^2} \, dx=\frac {a^2\,x^2-\frac {2}{3}}{a^2\,x^3}-\frac {\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {2\,\sqrt {\frac {1}{a\,x}+1}}{3\,a}-\frac {2\,a\,x^2\,\sqrt {\frac {1}{a\,x}+1}}{3}\right )}{x^2} \]

[In]

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

[Out]

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