\(\int \frac {e^{-\text {sech}^{-1}(a x)}}{x^3} \, dx\) [83]

Optimal result

Integrand size = 12, antiderivative size = 116 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^3} \, dx=-\frac {a^2}{2 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {2 a^2}{3 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {a^2}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {a^2}{2 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 116, 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, 1626} \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^3} \, dx=-\frac {a^2}{2 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}-\frac {a^2}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}+\frac {a^2}{\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}-\frac {2 a^2}{3 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3} \]

[In]

Int[1/(E^ArcSech[a*x]*x^3),x]

[Out]

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

Rule 1626

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.37 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^3} \, dx=-\frac {1+(-1+a x) \sqrt {\frac {1-a x}{1+a x}} (1+a x)^2}{3 a x^3} \]

[In]

Integrate[1/(E^ArcSech[a*x]*x^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.50

method	result	size
default	\(a \left (-\frac {1}{3 a^{2} x^{3}}-\frac {\sqrt {\frac {a x +1}{a x}}\, \sqrt {-\frac {a x -1}{a x}}\, \left (a^{2} x^{2}-1\right )}{3 a \,x^{2}}\right )\)	\(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.45 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^3} \, dx=-\frac {{\left (a^{3} x^{3} - a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1}{3 \, a x^{3}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.73 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.50 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^3} \, dx=\frac {\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {x}{3}-\frac {a\,x^2}{3}+\frac {1}{3\,a}-\frac {a^2\,x^3}{3}\right )}{x^3\,\sqrt {\frac {1}{a\,x}+1}}-\frac {1}{3\,a\,x^3} \]

[In]

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

[Out]

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