Integrand size = 12, antiderivative size = 105 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^2} \, dx=-\frac {1}{3 a x^3}-\frac {\sqrt {1-\frac {1}{a^2 x^4}} \sqrt {-1+\frac {1}{a x^2}}}{3 \sqrt {1-\frac {1}{a x^2}} x}-\frac {2 \sqrt {a} \sqrt {-1+\frac {1}{a x^2}} \operatorname {EllipticF}\left (\csc ^{-1}\left (\sqrt {a} x\right ),-1\right )}{3 \sqrt {1-\frac {1}{a x^2}}} \] Output:
-1/3/a/x^3-1/3*(1-1/a^2/x^4)^(1/2)*(-1+1/a/x^2)^(1/2)/(1-1/a/x^2)^(1/2)/x- 2/3*a^(1/2)*(-1+1/a/x^2)^(1/2)*InverseJacobiAM(arccsc(a^(1/2)*x),I)/(1-1/a /x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^2} \, dx=-\frac {a \sqrt {1+e^{2 \text {sech}^{-1}\left (a x^2\right )}} \sqrt {\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{2+2 e^{2 \text {sech}^{-1}\left (a x^2\right )}}} x \left (\sqrt {1+e^{2 \text {sech}^{-1}\left (a x^2\right )}}-4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 \text {sech}^{-1}\left (a x^2\right )}\right )\right )}{3 \sqrt {a x^2}} \] Input:
Integrate[E^ArcSech[a*x^2]/x^2,x]
Output:
-1/3*(a*Sqrt[1 + E^(2*ArcSech[a*x^2])]*Sqrt[E^ArcSech[a*x^2]/(2 + 2*E^(2*A rcSech[a*x^2]))]*x*(Sqrt[1 + E^(2*ArcSech[a*x^2])] - 4*Hypergeometric2F1[1 /4, 1/2, 5/4, -E^(2*ArcSech[a*x^2])]))/Sqrt[a*x^2]
Time = 0.47 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6889, 15, 335, 847, 762}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^2} \, dx\) |
\(\Big \downarrow \) 6889 |
\(\displaystyle -\frac {2 \int \frac {1}{x^4}dx}{a}-\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^4 \sqrt {1-a x^2} \sqrt {a x^2+1}}dx}{a}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^4 \sqrt {1-a x^2} \sqrt {a x^2+1}}dx}{a}+\frac {2}{3 a x^3}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x}\) |
\(\Big \downarrow \) 335 |
\(\displaystyle -\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^4 \sqrt {1-a^2 x^4}}dx}{a}+\frac {2}{3 a x^3}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle -\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {1}{3} a^2 \int \frac {1}{\sqrt {1-a^2 x^4}}dx-\frac {\sqrt {1-a^2 x^4}}{3 x^3}\right )}{a}+\frac {2}{3 a x^3}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle -\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {1}{3} a^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {a} x\right ),-1\right )-\frac {\sqrt {1-a^2 x^4}}{3 x^3}\right )}{a}+\frac {2}{3 a x^3}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x}\) |
Input:
Int[E^ArcSech[a*x^2]/x^2,x]
Output:
2/(3*a*x^3) - E^ArcSech[a*x^2]/x - (2*Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^ 2]*(-1/3*Sqrt[1 - a^2*x^4]/x^3 + (a^(3/2)*EllipticF[ArcSin[Sqrt[a]*x], -1] )/3))/a
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p _.), x_Symbol] :> Int[(e*x)^m*(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, e , m, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] ))
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
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] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(E^ ArcSech[a*x^p]/(m + 1)), x] + (Simp[p/(a*(m + 1)) Int[x^(m - p), x], x] + Simp[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]
Time = 0.70 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.99
method | result | size |
default | \(\frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (2 \sqrt {-a \,x^{2}+1}\, \sqrt {a \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {a}\, x , i\right ) x^{3} a^{\frac {3}{2}}-a^{2} x^{4}+1\right )}{3 x \left (a^{2} x^{4}-1\right )}-\frac {1}{3 a \,x^{3}}\) | \(104\) |
Input:
int((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^2,x,method=_RETURNVER BOSE)
Output:
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(a^(1/2)*x,I)*x^3*a^(3/2)-a^2*x^4+1)/(a^2*x^4-1) -1/3/a/x^3
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.61 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^2} \, dx=-\frac {2 \, a^{\frac {3}{2}} x^{3} F(\arcsin \left (\sqrt {a} x\right )\,|\,-1) + a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} + 1}{3 \, a x^{3}} \] Input:
integrate((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^2,x, algorithm= "fricas")
Output:
-1/3*(2*a^(3/2)*x^3*elliptic_f(arcsin(sqrt(a)*x), -1) + a*x^2*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) + 1)/(a*x^3)
\[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^2} \, dx=\frac {\int \frac {1}{x^{4}}\, dx + \int \frac {a \sqrt {-1 + \frac {1}{a x^{2}}} \sqrt {1 + \frac {1}{a x^{2}}}}{x^{2}}\, dx}{a} \] Input:
integrate((1/a/x**2+(-1+1/a/x**2)**(1/2)*(1+1/a/x**2)**(1/2))/x**2,x)
Output:
(Integral(x**(-4), x) + Integral(a*sqrt(-1 + 1/(a*x**2))*sqrt(1 + 1/(a*x** 2))/x**2, x))/a
\[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^2} \, dx=\int { \frac {\sqrt {\frac {1}{a x^{2}} + 1} \sqrt {\frac {1}{a x^{2}} - 1} + \frac {1}{a x^{2}}}{x^{2}} \,d x } \] Input:
integrate((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^2,x, algorithm= "maxima")
Output:
integrate(sqrt(a*x^2 + 1)*sqrt(-a*x^2 + 1)/x^4, x)/a - 1/3/(a*x^3)
\[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^2} \, dx=\int { \frac {\sqrt {\frac {1}{a x^{2}} + 1} \sqrt {\frac {1}{a x^{2}} - 1} + \frac {1}{a x^{2}}}{x^{2}} \,d x } \] Input:
integrate((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^2,x, algorithm= "giac")
Output:
integrate((sqrt(1/(a*x^2) + 1)*sqrt(1/(a*x^2) - 1) + 1/(a*x^2))/x^2, x)
Timed out. \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^2} \, dx=\int \frac {\sqrt {\frac {1}{a\,x^2}-1}\,\sqrt {\frac {1}{a\,x^2}+1}+\frac {1}{a\,x^2}}{x^2} \,d x \] Input:
int(((1/(a*x^2) - 1)^(1/2)*(1/(a*x^2) + 1)^(1/2) + 1/(a*x^2))/x^2,x)
Output:
int(((1/(a*x^2) - 1)^(1/2)*(1/(a*x^2) + 1)^(1/2) + 1/(a*x^2))/x^2, x)
\[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^2} \, dx=\frac {-3 \sqrt {a \,x^{2}+1}\, \sqrt {-a \,x^{2}+1}+6 \left (\int \frac {\sqrt {a \,x^{2}+1}\, \sqrt {-a \,x^{2}+1}}{a^{2} x^{8}-x^{4}}d x \right ) x^{3}-1}{3 a \,x^{3}} \] Input:
int((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^2,x)
Output:
( - 3*sqrt(a*x**2 + 1)*sqrt( - a*x**2 + 1) + 6*int((sqrt(a*x**2 + 1)*sqrt( - a*x**2 + 1))/(a**2*x**8 - x**4),x)*x**3 - 1)/(3*a*x**3)