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

Optimal. Leaf size=115 \[ \frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sqrt {1-a^2 x^4}}{3 a x^3}+\frac {2}{3 a x^3}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x}-\frac {2}{3} \sqrt {a} \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right ) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6335, 30, 259, 325, 221} \[ \frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sqrt {1-a^2 x^4}}{3 a x^3}+\frac {2}{3 a x^3}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x}-\frac {2}{3} \sqrt {a} \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2/(3*a*x^3) - E^ArcSech[a*x^2]/x + (2*Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*Sqrt[1 - a^2*x^4])/(3*a*x^3) - (2
*Sqrt[a]*Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*EllipticF[ArcSin[Sqrt[a]*x], -1])/3

Rule 30

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 259

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[(c*x)
^m*(a1*a2 + b1*b2*x^(2*n))^p, x] /; FreeQ[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (Intege
rQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

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Mathematica [C]  time = 0.20, size = 123, normalized size = 1.07 \[ \frac {2 i \sqrt {-a} \sqrt {\frac {1-a x^2}{a x^2+1}} \sqrt {1-a^2 x^4} F\left (\left .i \sinh ^{-1}\left (\sqrt {-a} x\right )\right |-1\right )}{3 a x^2-3}-\frac {1}{3 a x^3}-\frac {\sqrt {\frac {1-a x^2}{a x^2+1}} \left (a x^2+1\right )}{3 a x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} + 1}{a x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {1}{a x^{2}} + 1} \sqrt {\frac {1}{a x^{2}} - 1} + \frac {1}{a x^{2}}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 104, normalized size = 0.90 \[ \frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (2 \sqrt {-a \,x^{2}+1}\, \sqrt {a \,x^{2}+1}\, \EllipticF \left (x \sqrt {a}, i\right ) x^{3} a^{\frac {3}{2}}-a^{2} x^{4}+1\right )}{3 x \left (a^{2} x^{4}-1\right )}-\frac {1}{3 x^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {a x^{2} + 1} \sqrt {-a x^{2} + 1}}{x^{4}}\,{d x}}{a} - \frac {1}{3 \, a x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {1}{a\,x^2}-1}\,\sqrt {\frac {1}{a\,x^2}+1}+\frac {1}{a\,x^2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{x^{4}}\, dx + \int \frac {a \sqrt {-1 + \frac {1}{a x^{2}}} \sqrt {1 + \frac {1}{a x^{2}}}}{x^{2}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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