3.73 \(\int \frac{e^{2 \text{sech}^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=183 \[ -\frac{a^3}{4 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )}-\frac{a^3}{4 \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )}+\frac{3 a^3}{2 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^2}-\frac{7 a^3}{3 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^3}+\frac{2 a^3}{\left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^4}-\frac{4 a^3}{5 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^5} \]

[Out]

(-4*a^3)/(5*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^5) + (2*a^3)/(1 - Sqrt[(1 - a*x)/(1 + a*x)])^4 - (7*a^3)/(3*(1 - S
qrt[(1 - a*x)/(1 + a*x)])^3) + (3*a^3)/(2*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^2) - a^3/(4*(1 - Sqrt[(1 - a*x)/(1 +
a*x)])) - a^3/(4*(1 + Sqrt[(1 - a*x)/(1 + a*x)]))

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Rubi [A]  time = 0.504922, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6337, 1612} \[ -\frac{a^3}{4 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )}-\frac{a^3}{4 \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )}+\frac{3 a^3}{2 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^2}-\frac{7 a^3}{3 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^3}+\frac{2 a^3}{\left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^4}-\frac{4 a^3}{5 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^3)/(5*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^5) + (2*a^3)/(1 - Sqrt[(1 - a*x)/(1 + a*x)])^4 - (7*a^3)/(3*(1 - S
qrt[(1 - a*x)/(1 + a*x)])^3) + (3*a^3)/(2*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^2) - a^3/(4*(1 - Sqrt[(1 - a*x)/(1 +
a*x)])) - a^3/(4*(1 + Sqrt[(1 - a*x)/(1 + a*x)]))

Rule 6337

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

Rule 1612

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]

Rubi steps

\begin{align*} \int \frac{e^{2 \text{sech}^{-1}(a x)}}{x^4} \, dx &=\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^4} \, dx\\ &=-\left ((4 a) \operatorname{Subst}\left (\int \frac{x \left (a+a x^2\right )^2}{(-1+x)^6 (1+x)^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )\right )\\ &=-\left ((4 a) \operatorname{Subst}\left (\int \left (\frac{a^2}{(-1+x)^6}+\frac{2 a^2}{(-1+x)^5}+\frac{7 a^2}{4 (-1+x)^4}+\frac{3 a^2}{4 (-1+x)^3}+\frac{a^2}{16 (-1+x)^2}-\frac{a^2}{16 (1+x)^2}\right ) \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )\right )\\ &=-\frac{4 a^3}{5 \left (1-\sqrt{\frac{1-a x}{1+a x}}\right )^5}+\frac{2 a^3}{\left (1-\sqrt{\frac{1-a x}{1+a x}}\right )^4}-\frac{7 a^3}{3 \left (1-\sqrt{\frac{1-a x}{1+a x}}\right )^3}+\frac{3 a^3}{2 \left (1-\sqrt{\frac{1-a x}{1+a x}}\right )^2}-\frac{a^3}{4 \left (1-\sqrt{\frac{1-a x}{1+a x}}\right )}-\frac{a^3}{4 \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )}\\ \end{align*}

Mathematica [A]  time = 0.0786524, size = 69, normalized size = 0.38 \[ \frac{5 a^2 x^2+2 \sqrt{\frac{1-a x}{a x+1}} (a x+1)^2 \left (2 a^3 x^3-2 a^2 x^2+3 a x-3\right )-6}{15 a^2 x^5} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-6 + 5*a^2*x^2 + 2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^2*(-3 + 3*a*x - 2*a^2*x^2 + 2*a^3*x^3))/(15*a^2*x^5)

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Maple [A]  time = 0.196, size = 84, normalized size = 0.5 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{1}{5\,{x}^{5}}}+{\frac{{a}^{2}}{3\,{x}^{3}}} \right ) }+{\frac{ \left ( 2\,{a}^{2}{x}^{2}-2 \right ) \left ( 2\,{a}^{2}{x}^{2}+3 \right ) }{15\,{x}^{4}a}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}-{\frac{1}{5\,{x}^{5}{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(-1/5/x^5+1/3*a^2/x^3)+2/15/a*(-(a*x-1)/a/x)^(1/2)/x^4*((a*x+1)/a/x)^(1/2)*(a^2*x^2-1)*(2*a^2*x^2+3)-1/5
/a^2/x^5

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Maxima [A]  time = 1.04593, size = 76, normalized size = 0.42 \begin{align*} \frac{1}{3 \, x^{3}} + \frac{2 \,{\left (2 \, a^{4} x^{5} + a^{2} x^{3} - 3 \, x\right )} \sqrt{a x + 1} \sqrt{-a x + 1}}{15 \, a^{2} x^{6}} - \frac{2}{5 \, a^{2} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2/x^4,x, algorithm="maxima")

[Out]

1/3/x^3 + 2/15*(2*a^4*x^5 + a^2*x^3 - 3*x)*sqrt(a*x + 1)*sqrt(-a*x + 1)/(a^2*x^6) - 2/5/(a^2*x^5)

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Fricas [A]  time = 2.16558, size = 150, normalized size = 0.82 \begin{align*} \frac{5 \, a^{2} x^{2} + 2 \,{\left (2 \, a^{5} x^{5} + a^{3} x^{3} - 3 \, a x\right )} \sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}} - 6}{15 \, a^{2} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2/x^4,x, algorithm="fricas")

[Out]

1/15*(5*a^2*x^2 + 2*(2*a^5*x^5 + a^3*x^3 - 3*a*x)*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - 6)/(a^2*x^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{2}{x^{6}}\, dx + \int - \frac{a^{2}}{x^{4}}\, dx + \int \frac{2 a \sqrt{-1 + \frac{1}{a x}} \sqrt{1 + \frac{1}{a x}}}{x^{5}}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1/a/x-1)**(1/2)*(1+1/a/x)**(1/2))**2/x**4,x)

[Out]

(Integral(2/x**6, x) + Integral(-a**2/x**4, x) + Integral(2*a*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x))/x**5, x))/a
**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}\right )}^{2}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2/x^4,x, algorithm="giac")

[Out]

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