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

Optimal. Leaf size=80 \[ -\frac{\sqrt{1-a x^2}}{2 a x^2 \sqrt{\frac{1}{a x^2+1}}}-\frac{1}{2 a x^2}-\frac{1}{2} \sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \sin ^{-1}\left (a x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.0530153, antiderivative size = 93, normalized size of antiderivative = 1.16, 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 = {6334, 259, 275, 277, 216} \[ -\frac{\sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \sqrt{1-a^2 x^4}}{2 a x^2}-\frac{1}{2 a x^2}-\frac{1}{2} \sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \sin ^{-1}\left (a x^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Rule 6334

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

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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 277

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [A]  time = 0.144055, size = 113, normalized size = 1.41 \[ \frac{\sqrt{\frac{1-a x^2}{a x^2+1}} \left (a x^2+1\right ) \tanh ^{-1}\left (\frac{a x^2}{\sqrt{a^2 x^4-1}}\right )}{2 \sqrt{a^2 x^4-1}}+\sqrt{\frac{1-a x^2}{a x^2+1}} \left (-\frac{1}{2 a x^2}-\frac{1}{2}\right )-\frac{1}{2 a x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.284, size = 103, normalized size = 1.3 \begin{align*} -{\frac{1}{2}\sqrt{-{\frac{a{x}^{2}-1}{a{x}^{2}}}}\sqrt{{\frac{a{x}^{2}+1}{a{x}^{2}}}} \left ( \arctan \left ({{x}^{2}{\frac{1}{\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}}}} \right ){x}^{2}+\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}} \right ){\frac{1}{\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}}}}-{\frac{1}{2\,a{x}^{2}}} \end{align*}

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,x)

[Out]

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

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

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,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.0125, size = 230, normalized size = 2.88 \begin{align*} -\frac{a x^{2} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} - 2 \, a x^{2} \arctan \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^{2}}\right ) + 1}{2 \, a x^{2}} \end{align*}

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,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{x^{3}}\, dx + \int \frac{a \sqrt{-1 + \frac{1}{a x^{2}}} \sqrt{1 + \frac{1}{a x^{2}}}}{x}\, dx}{a} \end{align*}

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,x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

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,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError