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

Optimal. Leaf size=55 \[ -\frac {8 a^2 \left (\frac {1-a x}{a x+1}\right )^{3/2}}{3 \left (1-\frac {1-a x}{a x+1}\right )^3}-\frac {1}{3 a x^3} \]

[Out]

-1/3/a/x^3-8/3*a^2*((-a*x+1)/(a*x+1))^(3/2)/(1+(a*x-1)/(a*x+1))^3

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Rubi [C]  time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.53, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6335, 30, 103, 12, 95} \[ \frac {\sqrt {1-a x}}{6 a x^3 \sqrt {\frac {1}{a x+1}}}+\frac {1}{6 a x^3}-\frac {e^{\text {sech}^{-1}(a x)}}{2 x^2}+\frac {a \sqrt {1-a x}}{3 x \sqrt {\frac {1}{a x+1}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

1/(6*a*x^3) - E^ArcSech[a*x]/(2*x^2) + Sqrt[1 - a*x]/(6*a*x^3*Sqrt[(1 + a*x)^(-1)]) + (a*Sqrt[1 - a*x])/(3*x*S
qrt[(1 + a*x)^(-1)])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 95

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] /; FreeQ[{a, b, c, d,
e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0
] && NeQ[m, -1]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 6335

Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*E^ArcSech[a*x^p])/(m + 1), x] + (Dist
[p/(a*(m + 1)), Int[x^(m - p), x], x] + Dist[(p*Sqrt[1 + a*x^p]*Sqrt[1/(1 + a*x^p)])/(a*(m + 1)), Int[x^(m - p
)/(Sqrt[1 + a*x^p]*Sqrt[1 - a*x^p]), x], x]) /; FreeQ[{a, m, p}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.78 \[ \frac {(a x-1) \sqrt {\frac {1-a x}{a x+1}} (a x+1)^2-1}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 52, normalized size = 0.95 \[ \frac {{\left (a^{3} x^{3} - a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1}{3 \, a x^{3}} \]

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))/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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}}{x^{3}}\,{d x} \]

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))/x^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 53, normalized size = 0.96 \[ \frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \left (a^{2} x^{2}-1\right )}{3 x^{2}}-\frac {1}{3 x^{3} a} \]

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))/x^3,x)

[Out]

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

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maxima [A]  time = 0.37, size = 43, normalized size = 0.78 \[ \frac {{\left (a^{2} x^{3} - x\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{3 \, a x^{4}} - \frac {1}{3 \, a x^{3}} \]

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))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.47, size = 58, normalized size = 1.05 \[ -\frac {1}{3\,a\,x^3}-\frac {\left (\frac {\sqrt {\frac {1}{a\,x}+1}}{3}-\frac {a^2\,x^2\,\sqrt {\frac {1}{a\,x}+1}}{3}\right )\,\sqrt {\frac {1}{a\,x}-1}}{x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{x^{4}}\, dx + \int \frac {a \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}}{x^{3}}\, dx}{a} \]

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))/x**3,x)

[Out]

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

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