Integrand size = 12, antiderivative size = 80 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^9} \, dx=-\frac {1}{15} a^4 \left (-1+\frac {1}{a x^2}\right )^{3/2} \left (1+\frac {1}{a x^2}\right )^{3/2}-\frac {1}{10 a x^{10}}-\frac {a^2 \left (-1+\frac {1}{a x^2}\right )^{3/2} \left (1+\frac {1}{a x^2}\right )^{3/2}}{10 x^4} \] Output:
-1/15*a^4*(-1+1/a/x^2)^(3/2)*(1+1/a/x^2)^(3/2)-1/10/a/x^10-1/10*a^2*(-1+1/ a/x^2)^(3/2)*(1+1/a/x^2)^(3/2)/x^4
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^9} \, dx=\frac {-3+\sqrt {\frac {1-a x^2}{1+a x^2}} \left (1+a x^2\right )^2 \left (-3+3 a x^2-2 a^2 x^4+2 a^3 x^6\right )}{30 a x^{10}} \] Input:
Integrate[E^ArcSech[a*x^2]/x^9,x]
Output:
(-3 + Sqrt[(1 - a*x^2)/(1 + a*x^2)]*(1 + a*x^2)^2*(-3 + 3*a*x^2 - 2*a^2*x^ 4 + 2*a^3*x^6))/(30*a*x^10)
Result contains higher order function than in optimal. Order 3 vs. order 2 in optimal.
Time = 0.52 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.65, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6889, 15, 335, 803, 803, 796}
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.
\(\displaystyle \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^9} \, dx\) |
\(\Big \downarrow \) 6889 |
\(\displaystyle -\frac {\int \frac {1}{x^{11}}dx}{4 a}-\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^{11} \sqrt {1-a x^2} \sqrt {a x^2+1}}dx}{4 a}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{8 x^8}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^{11} \sqrt {1-a x^2} \sqrt {a x^2+1}}dx}{4 a}+\frac {1}{40 a x^{10}}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{8 x^8}\) |
\(\Big \downarrow \) 335 |
\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^{11} \sqrt {1-a^2 x^4}}dx}{4 a}+\frac {1}{40 a x^{10}}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{8 x^8}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {4}{5} a^2 \int \frac {1}{x^7 \sqrt {1-a^2 x^4}}dx-\frac {\sqrt {1-a^2 x^4}}{10 x^{10}}\right )}{4 a}+\frac {1}{40 a x^{10}}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{8 x^8}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {4}{5} a^2 \left (\frac {2}{3} a^2 \int \frac {1}{x^3 \sqrt {1-a^2 x^4}}dx-\frac {\sqrt {1-a^2 x^4}}{6 x^6}\right )-\frac {\sqrt {1-a^2 x^4}}{10 x^{10}}\right )}{4 a}+\frac {1}{40 a x^{10}}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{8 x^8}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {4}{5} a^2 \left (-\frac {\sqrt {1-a^2 x^4}}{6 x^6}-\frac {a^2 \sqrt {1-a^2 x^4}}{3 x^2}\right )-\frac {\sqrt {1-a^2 x^4}}{10 x^{10}}\right )}{4 a}+\frac {1}{40 a x^{10}}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{8 x^8}\) |
Input:
Int[E^ArcSech[a*x^2]/x^9,x]
Output:
1/(40*a*x^10) - E^ArcSech[a*x^2]/(8*x^8) - (Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*(-1/10*Sqrt[1 - a^2*x^4]/x^10 + (4*a^2*(-1/6*Sqrt[1 - a^2*x^4]/x^ 6 - (a^2*Sqrt[1 - a^2*x^4])/(3*x^2)))/5))/(4*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[((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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 ))) Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
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 = 0.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90
method | result | size |
default | \(\frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (a \,x^{2}+1\right ) \left (a \,x^{2}-1\right ) \left (2 a^{2} x^{4}+3\right )}{30 x^{8}}-\frac {1}{10 a \,x^{10}}\) | \(72\) |
orering | \(\frac {\left (\frac {11}{30} a^{4} x^{9}-\frac {11}{60} a^{2} x^{5}-\frac {17}{60} x \right ) \left (\frac {1}{a \,x^{2}}+\sqrt {-1+\frac {1}{a \,x^{2}}}\, \sqrt {1+\frac {1}{a \,x^{2}}}\right )}{x^{9}}+\frac {\left (2 a^{2} x^{4}+1\right ) x^{2} \left (a \,x^{2}-1\right ) \left (a \,x^{2}+1\right ) \left (\frac {-\frac {2}{a \,x^{3}}-\frac {\sqrt {1+\frac {1}{a \,x^{2}}}}{\sqrt {-1+\frac {1}{a \,x^{2}}}\, a \,x^{3}}-\frac {\sqrt {-1+\frac {1}{a \,x^{2}}}}{\sqrt {1+\frac {1}{a \,x^{2}}}\, a \,x^{3}}}{x^{9}}-\frac {9 \left (\frac {1}{a \,x^{2}}+\sqrt {-1+\frac {1}{a \,x^{2}}}\, \sqrt {1+\frac {1}{a \,x^{2}}}\right )}{x^{10}}\right )}{60}\) | \(196\) |
Input:
int((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^9,x,method=_RETURNVER BOSE)
Output:
1/30*(-(a*x^2-1)/a/x^2)^(1/2)/x^8*((a*x^2+1)/a/x^2)^(1/2)*(a*x^2+1)*(a*x^2 -1)*(2*a^2*x^4+3)-1/10/a/x^10
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^9} \, dx=\frac {{\left (2 \, a^{5} x^{10} + a^{3} x^{6} - 3 \, a x^{2}\right )} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} - 3}{30 \, a x^{10}} \] Input:
integrate((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^9,x, algorithm= "fricas")
Output:
1/30*((2*a^5*x^10 + a^3*x^6 - 3*a*x^2)*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a* x^2 - 1)/(a*x^2)) - 3)/(a*x^10)
\[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^9} \, dx=\frac {\int \frac {1}{x^{11}}\, dx + \int \frac {a \sqrt {-1 + \frac {1}{a x^{2}}} \sqrt {1 + \frac {1}{a x^{2}}}}{x^{9}}\, dx}{a} \] Input:
integrate((1/a/x**2+(-1+1/a/x**2)**(1/2)*(1+1/a/x**2)**(1/2))/x**9,x)
Output:
(Integral(x**(-11), x) + Integral(a*sqrt(-1 + 1/(a*x**2))*sqrt(1 + 1/(a*x* *2))/x**9, x))/a
Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^9} \, dx=\frac {{\left (2 \, a^{4} x^{10} + a^{2} x^{6} - 3 \, x^{2}\right )} \sqrt {a x^{2} + 1} \sqrt {-a x^{2} + 1}}{30 \, a x^{12}} - \frac {1}{10 \, a x^{10}} \] Input:
integrate((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^9,x, algorithm= "maxima")
Output:
1/30*(2*a^4*x^10 + a^2*x^6 - 3*x^2)*sqrt(a*x^2 + 1)*sqrt(-a*x^2 + 1)/(a*x^ 12) - 1/10/(a*x^10)
Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (66) = 132\).
Time = 0.23 (sec) , antiderivative size = 326, normalized size of antiderivative = 4.08 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^9} \, dx=-\frac {1}{30} \, a^{9} {\left (\frac {64 \, {\left (5 \, {\left (\frac {\sqrt {2} \sqrt {a} - \sqrt {-a^{2} x^{2} + a}}{\sqrt {a^{2} x^{2} + a}} - \frac {\sqrt {a^{2} x^{2} + a}}{\sqrt {2} \sqrt {a} - \sqrt {-a^{2} x^{2} + a}}\right )}^{7} + 8 \, {\left (\frac {\sqrt {2} \sqrt {a} - \sqrt {-a^{2} x^{2} + a}}{\sqrt {a^{2} x^{2} + a}} - \frac {\sqrt {a^{2} x^{2} + a}}{\sqrt {2} \sqrt {a} - \sqrt {-a^{2} x^{2} + a}}\right )}^{5} + 80 \, {\left (\frac {\sqrt {2} \sqrt {a} - \sqrt {-a^{2} x^{2} + a}}{\sqrt {a^{2} x^{2} + a}} - \frac {\sqrt {a^{2} x^{2} + a}}{\sqrt {2} \sqrt {a} - \sqrt {-a^{2} x^{2} + a}}\right )}^{3}\right )}}{{\left ({\left (\frac {\sqrt {2} \sqrt {a} - \sqrt {-a^{2} x^{2} + a}}{\sqrt {a^{2} x^{2} + a}} - \frac {\sqrt {a^{2} x^{2} + a}}{\sqrt {2} \sqrt {a} - \sqrt {-a^{2} x^{2} + a}}\right )}^{2} - 4\right )}^{5} a^{5}} + \frac {3}{a^{10} x^{10}}\right )} \] Input:
integrate((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^9,x, algorithm= "giac")
Output:
-1/30*a^9*(64*(5*((sqrt(2)*sqrt(a) - sqrt(-a^2*x^2 + a))/sqrt(a^2*x^2 + a) - sqrt(a^2*x^2 + a)/(sqrt(2)*sqrt(a) - sqrt(-a^2*x^2 + a)))^7 + 8*((sqrt( 2)*sqrt(a) - sqrt(-a^2*x^2 + a))/sqrt(a^2*x^2 + a) - sqrt(a^2*x^2 + a)/(sq rt(2)*sqrt(a) - sqrt(-a^2*x^2 + a)))^5 + 80*((sqrt(2)*sqrt(a) - sqrt(-a^2* x^2 + a))/sqrt(a^2*x^2 + a) - sqrt(a^2*x^2 + a)/(sqrt(2)*sqrt(a) - sqrt(-a ^2*x^2 + a)))^3)/((((sqrt(2)*sqrt(a) - sqrt(-a^2*x^2 + a))/sqrt(a^2*x^2 + a) - sqrt(a^2*x^2 + a)/(sqrt(2)*sqrt(a) - sqrt(-a^2*x^2 + a)))^2 - 4)^5*a^ 5) + 3/(a^10*x^10))
Time = 24.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^9} \, dx=\frac {\sqrt {\frac {1}{a\,x^2}-1}\,\left (\frac {a^2\,x^4\,\sqrt {\frac {1}{a\,x^2}+1}}{30}-\frac {\sqrt {\frac {1}{a\,x^2}+1}}{10}+\frac {a^4\,x^8\,\sqrt {\frac {1}{a\,x^2}+1}}{15}\right )}{x^8}-\frac {1}{10\,a\,x^{10}} \] Input:
int(((1/(a*x^2) - 1)^(1/2)*(1/(a*x^2) + 1)^(1/2) + 1/(a*x^2))/x^9,x)
Output:
((1/(a*x^2) - 1)^(1/2)*((a^2*x^4*(1/(a*x^2) + 1)^(1/2))/30 - (1/(a*x^2) + 1)^(1/2)/10 + (a^4*x^8*(1/(a*x^2) + 1)^(1/2))/15))/x^8 - 1/(10*a*x^10)
Time = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^9} \, dx=\frac {2 \sqrt {a \,x^{2}+1}\, \sqrt {-a \,x^{2}+1}\, a^{4} x^{8}+\sqrt {a \,x^{2}+1}\, \sqrt {-a \,x^{2}+1}\, a^{2} x^{4}-3 \sqrt {a \,x^{2}+1}\, \sqrt {-a \,x^{2}+1}-3}{30 a \,x^{10}} \] Input:
int((1/a/x^2+(-1+1/a/x^2)^(1/2)*(1+1/a/x^2)^(1/2))/x^9,x)
Output:
(2*sqrt(a*x**2 + 1)*sqrt( - a*x**2 + 1)*a**4*x**8 + sqrt(a*x**2 + 1)*sqrt( - a*x**2 + 1)*a**2*x**4 - 3*sqrt(a*x**2 + 1)*sqrt( - a*x**2 + 1) - 3)/(30 *a*x**10)