3.55 \(\int \frac{e^{\text{sech}^{-1}(a x^2)}}{x^3} \, dx\)

Optimal. Leaf size=118 \[ \frac{\sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \sqrt{1-a^2 x^4}}{4 a x^4}+\frac{1}{4} a \sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \tanh ^{-1}\left (\sqrt{1-a^2 x^4}\right )+\frac{1}{4 a x^4}-\frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0611727, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {6335, 30, 259, 266, 51, 63, 208} \[ \frac{\sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \sqrt{1-a^2 x^4}}{4 a x^4}+\frac{1}{4} a \sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \tanh ^{-1}\left (\sqrt{1-a^2 x^4}\right )+\frac{1}{4 a x^4}-\frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSech[a*x^2]/x^3,x]

[Out]

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

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]

Rule 30

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

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 266

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{x^3} \, dx &=-\frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{2 x^2}-\frac{\int \frac{1}{x^5} \, dx}{a}-\frac{\left (\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \int \frac{1}{x^5 \sqrt{1-a x^2} \sqrt{1+a x^2}} \, dx}{a}\\ &=\frac{1}{4 a x^4}-\frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{2 x^2}-\frac{\left (\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \int \frac{1}{x^5 \sqrt{1-a^2 x^4}} \, dx}{a}\\ &=\frac{1}{4 a x^4}-\frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{2 x^2}-\frac{\left (\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-a^2 x}} \, dx,x,x^4\right )}{4 a}\\ &=\frac{1}{4 a x^4}-\frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{2 x^2}+\frac{\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2} \sqrt{1-a^2 x^4}}{4 a x^4}-\frac{1}{8} \left (a \sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^4\right )\\ &=\frac{1}{4 a x^4}-\frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{2 x^2}+\frac{\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2} \sqrt{1-a^2 x^4}}{4 a x^4}+\frac{\left (\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^4}\right )}{4 a}\\ &=\frac{1}{4 a x^4}-\frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{2 x^2}+\frac{\sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2} \sqrt{1-a^2 x^4}}{4 a x^4}+\frac{1}{4} a \sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^4}\right )\\ \end{align*}

Mathematica [A]  time = 0.255935, size = 105, normalized size = 0.89 \[ -\frac{-\frac{a^2 \sqrt{\frac{1-a x^2}{a x^2+1}} \left (a x^2+1\right ) \tan ^{-1}\left (\sqrt{a^2 x^4-1}\right )}{\sqrt{a^2 x^4-1}}+\frac{\sqrt{\frac{1-a x^2}{a x^2+1}} \left (a x^2+1\right )}{x^4}+\frac{1}{x^4}}{4 a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSech[a*x^2]/x^3,x]

[Out]

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

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Maple [C]  time = 0.299, size = 129, normalized size = 1.1 \begin{align*}{\frac{{\it csgn} \left ({a}^{-1} \right ) }{4\,{x}^{2}}\sqrt{-{\frac{a{x}^{2}-1}{a{x}^{2}}}}\sqrt{{\frac{a{x}^{2}+1}{a{x}^{2}}}} \left ( \ln \left ( 2\,{\frac{1}{{a}^{2}{x}^{2}} \left ({\it csgn} \left ({a}^{-1} \right ) a\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}+1 \right ) } \right ){x}^{4}a-\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}{\it csgn} \left ({a}^{-1} \right ) \right ){\frac{1}{\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}}}}-{\frac{1}{4\,{x}^{4}a}} \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^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\frac{1}{4} \, a^{2} \log \left (\frac{2 \, \sqrt{-a^{2} x^{4} + 1}}{x^{2}} + \frac{2}{x^{2}}\right ) - \frac{1}{4} \, \sqrt{-a^{2} x^{4} + 1} a^{2} - \frac{{\left (-a^{2} x^{4} + 1\right )}^{\frac{3}{2}}}{4 \, x^{4}}}{a} - \frac{1}{4 \, a x^{4}} \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^3,x, algorithm="maxima")

[Out]

integrate(sqrt(a*x^2 + 1)*sqrt(-a*x^2 + 1)/x^5, x)/a - 1/4/(a*x^4)

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Fricas [A]  time = 2.03432, size = 319, normalized size = 2.7 \begin{align*} \frac{a^{2} x^{4} \log \left (a x^{2} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} + 1\right ) - a^{2} x^{4} \log \left (a x^{2} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} - 1\right ) - 2 \, a x^{2} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} - 2}{8 \, a x^{4}} \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^3,x, algorithm="fricas")

[Out]

1/8*(a^2*x^4*log(a*x^2*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) + 1) - a^2*x^4*log(a*x^2*sqrt((a*x
^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) - 1) - 2*a*x^2*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)
) - 2)/(a*x^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{x^{5}}\, dx + \int \frac{a \sqrt{-1 + \frac{1}{a x^{2}}} \sqrt{1 + \frac{1}{a x^{2}}}}{x^{3}}\, 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**3,x)

[Out]

(Integral(x**(-5), x) + Integral(a*sqrt(-1 + 1/(a*x**2))*sqrt(1 + 1/(a*x**2))/x**3, x))/a

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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^3,x, algorithm="giac")

[Out]

Exception raised: TypeError