3.1.55 \(\int \frac {e^{\text {sech}^{-1}(a x^2)}}{x^3} \, dx\) [55]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 118, normalized size of antiderivative = 1.00, 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 = {6470, 30, 265, 272, 44, 65, 214} \begin {gather*} \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} \end {gather*}

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 30

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

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

Rule 265

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 272

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 6470

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]/(a*(m + 1)))*Sqrt[1/(1 + a*x^p)], 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^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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.19, size = 105, normalized size = 0.89 \begin {gather*} -\frac {\frac {1}{x^4}+\frac {\sqrt {\frac {1-a x^2}{1+a x^2}} \left (1+a x^2\right )}{x^4}-\frac {a^2 \sqrt {\frac {1-a x^2}{1+a x^2}} \left (1+a x^2\right ) \text {ArcTan}\left (\sqrt {-1+a^2 x^4}\right )}{\sqrt {-1+a^2 x^4}}}{4 a} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.08, size = 129, normalized size = 1.09

method	result	size
default	\(-\frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (-\ln \left (\frac {2 \,\mathrm {csgn}\left (\frac {1}{a}\right ) a \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}+2}{a^{2} x^{2}}\right ) x^{4} a +\sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}\, \mathrm {csgn}\left (\frac {1}{a}\right )\right ) \mathrm {csgn}\left (\frac {1}{a}\right )}{4 x^{2} \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}}-\frac {1}{4 a \,x^{4}}\)	\(129\)

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,method=_RETURNVERBOSE)

[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/a/x^4

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.43, size = 146, normalized size = 1.24 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 2.02, size = 71, normalized size = 0.60 \begin {gather*} \frac {a\,\ln \left (\sqrt {\frac {1}{a\,x^2}-1}\,\sqrt {\frac {1}{a\,x^2}+1}+\frac {1}{a\,x^2}\right )}{4}-\frac {1}{4\,a\,x^4}-\frac {\sqrt {\frac {1}{a\,x^2}-1}\,\sqrt {\frac {1}{a\,x^2}+1}}{4\,x^2} \end {gather*}

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^3,x)

[Out]

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

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