Integrand size = 13, antiderivative size = 52 \[ \int \frac {1}{x \left (-2+x^2\right )^{5/2}} \, dx=-\frac {1}{6 \left (-2+x^2\right )^{3/2}}+\frac {1}{4 \sqrt {-2+x^2}}+\frac {\arctan \left (\frac {\sqrt {-2+x^2}}{\sqrt {2}}\right )}{4 \sqrt {2}} \] Output:
-1/6/(x^2-2)^(3/2)+1/8*arctan(1/2*(x^2-2)^(1/2)*2^(1/2))*2^(1/2)+1/4/(x^2- 2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (-2+x^2\right )^{5/2}} \, dx=\frac {-8+3 x^2}{12 \left (-2+x^2\right )^{3/2}}+\frac {\arctan \left (\frac {\sqrt {-2+x^2}}{\sqrt {2}}\right )}{4 \sqrt {2}} \] Input:
Integrate[1/(x*(-2 + x^2)^(5/2)),x]
Output:
(-8 + 3*x^2)/(12*(-2 + x^2)^(3/2)) + ArcTan[Sqrt[-2 + x^2]/Sqrt[2]]/(4*Sqr t[2])
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {243, 61, 61, 73, 216}
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.
\(\displaystyle \int \frac {1}{x \left (x^2-2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^2 \left (x^2-2\right )^{5/2}}dx^2\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2 \left (x^2-2\right )^{3/2}}dx^2-\frac {1}{3 \left (x^2-2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2 \sqrt {x^2-2}}dx^2+\frac {1}{\sqrt {x^2-2}}\right )-\frac {1}{3 \left (x^2-2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\int \frac {1}{x^4+2}d\sqrt {x^2-2}+\frac {1}{\sqrt {x^2-2}}\right )-\frac {1}{3 \left (x^2-2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {x^2-2}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {1}{\sqrt {x^2-2}}\right )-\frac {1}{3 \left (x^2-2\right )^{3/2}}\right )\) |
Input:
Int[1/(x*(-2 + x^2)^(5/2)),x]
Output:
(-1/3*1/(-2 + x^2)^(3/2) + (1/Sqrt[-2 + x^2] + ArcTan[Sqrt[-2 + x^2]/Sqrt[ 2]]/Sqrt[2])/2)/2
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] - Simp[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] && 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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Time = 0.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67
method | result | size |
risch | \(\frac {3 x^{2}-8}{12 \left (x^{2}-2\right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}}{\sqrt {x^{2}-2}}\right )}{8}\) | \(35\) |
default | \(-\frac {1}{6 \left (x^{2}-2\right )^{\frac {3}{2}}}+\frac {1}{4 \sqrt {x^{2}-2}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}}{\sqrt {x^{2}-2}}\right )}{8}\) | \(37\) |
pseudoelliptic | \(\frac {\arctan \left (\frac {\sqrt {x^{2}-2}\, \sqrt {2}}{2}\right ) \sqrt {2}\, \left (x^{2}-2\right )^{\frac {3}{2}}+2 x^{2}-\frac {16}{3}}{8 \left (x^{2}-2\right )^{\frac {3}{2}}}\) | \(41\) |
trager | \(\frac {3 x^{2}-8}{12 \left (x^{2}-2\right )^{\frac {3}{2}}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )-\sqrt {x^{2}-2}}{x}\right )}{8}\) | \(48\) |
meijerg | \(\frac {\sqrt {2}\, {\left (-\operatorname {signum}\left (-1+\frac {x^{2}}{2}\right )\right )}^{\frac {5}{2}} \left (\frac {3 \left (\frac {8}{3}-3 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{4}-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-6 x^{2}+16\right )}{8 \left (-\frac {x^{2}}{2}+1\right )^{\frac {3}{2}}}-\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-\frac {x^{2}}{2}+1}}{2}\right )}{2}\right )}{12 \sqrt {\pi }\, \operatorname {signum}\left (-1+\frac {x^{2}}{2}\right )^{\frac {5}{2}}}\) | \(96\) |
Input:
int(1/x/(x^2-2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/12*(3*x^2-8)/(x^2-2)^(3/2)-1/8*2^(1/2)*arctan(1/(x^2-2)^(1/2)*2^(1/2))
Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x \left (-2+x^2\right )^{5/2}} \, dx=\frac {3 \, \sqrt {2} {\left (x^{4} - 4 \, x^{2} + 4\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} x + \frac {1}{2} \, \sqrt {2} \sqrt {x^{2} - 2}\right ) + {\left (3 \, x^{2} - 8\right )} \sqrt {x^{2} - 2}}{12 \, {\left (x^{4} - 4 \, x^{2} + 4\right )}} \] Input:
integrate(1/x/(x^2-2)^(5/2),x, algorithm="fricas")
Output:
1/12*(3*sqrt(2)*(x^4 - 4*x^2 + 4)*arctan(-1/2*sqrt(2)*x + 1/2*sqrt(2)*sqrt (x^2 - 2)) + (3*x^2 - 8)*sqrt(x^2 - 2))/(x^4 - 4*x^2 + 4)
Result contains complex when optimal does not.
Time = 1.63 (sec) , antiderivative size = 984, normalized size of antiderivative = 18.92 \[ \int \frac {1}{x \left (-2+x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(1/x/(x**2-2)**(5/2),x)
Output:
Piecewise((6*I*x**4*log(x)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2) ) - 3*I*x**4*log(x**2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) - 6*x**4*asin(sqrt(2)/x)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) + 6*sqrt(2)*x**2*sqrt(x**2 - 2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt (2)) - 24*I*x**2*log(x)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) + 12*I*x**2*log(x**2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) + 24 *x**2*asin(sqrt(2)/x)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) - 1 6*sqrt(2)*sqrt(x**2 - 2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) + 24*I*log(x)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) - 12*I*log( x**2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) - 24*asin(sqrt(2)/x )/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)), Abs(x**2) > 2), (-3*I* x**4*log(x**2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) + 6*I*x**4 *log(sqrt(1 - x**2/2) + 1)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2) ) - 3*pi*x**4/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) + 3*I*x**4* log(2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) + 6*sqrt(2)*I*x**2 *sqrt(2 - x**2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) + 12*I*x* *2*log(x**2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) - 24*I*x**2* log(sqrt(1 - x**2/2) + 1)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) + 12*pi*x**2/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) - 12*I*x**2 *log(2)/(24*sqrt(2)*x**4 - 96*sqrt(2)*x**2 + 96*sqrt(2)) - 16*sqrt(2)*I...
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x \left (-2+x^2\right )^{5/2}} \, dx=-\frac {1}{8} \, \sqrt {2} \arcsin \left (\frac {\sqrt {2}}{{\left | x \right |}}\right ) + \frac {1}{4 \, \sqrt {x^{2} - 2}} - \frac {1}{6 \, {\left (x^{2} - 2\right )}^{\frac {3}{2}}} \] Input:
integrate(1/x/(x^2-2)^(5/2),x, algorithm="maxima")
Output:
-1/8*sqrt(2)*arcsin(sqrt(2)/abs(x)) + 1/4/sqrt(x^2 - 2) - 1/6/(x^2 - 2)^(3 /2)
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x \left (-2+x^2\right )^{5/2}} \, dx=\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x^{2} - 2}\right ) + \frac {3 \, x^{2} - 8}{12 \, {\left (x^{2} - 2\right )}^{\frac {3}{2}}} \] Input:
integrate(1/x/(x^2-2)^(5/2),x, algorithm="giac")
Output:
1/8*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(x^2 - 2)) + 1/12*(3*x^2 - 8)/(x^2 - 2) ^(3/2)
Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x \left (-2+x^2\right )^{5/2}} \, dx=\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {x^2-2}}{2}\right )}{8}+\frac {\frac {x^2}{4}-\frac {2}{3}}{{\left (x^2-2\right )}^{3/2}} \] Input:
int(1/(x*(x^2 - 2)^(5/2)),x)
Output:
(2^(1/2)*atan((2^(1/2)*(x^2 - 2)^(1/2))/2))/8 + (x^2/4 - 2/3)/(x^2 - 2)^(3 /2)
Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.83 \[ \int \frac {1}{x \left (-2+x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {x^{2}-2}+x}{\sqrt {2}}\right ) x^{4}-12 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {x^{2}-2}+x}{\sqrt {2}}\right ) x^{2}+12 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {x^{2}-2}+x}{\sqrt {2}}\right )+3 \sqrt {x^{2}-2}\, x^{2}-8 \sqrt {x^{2}-2}}{12 x^{4}-48 x^{2}+48} \] Input:
int(1/x/(x^2-2)^(5/2),x)
Output:
(3*sqrt(2)*atan((sqrt(x**2 - 2) + x)/sqrt(2))*x**4 - 12*sqrt(2)*atan((sqrt (x**2 - 2) + x)/sqrt(2))*x**2 + 12*sqrt(2)*atan((sqrt(x**2 - 2) + x)/sqrt( 2)) + 3*sqrt(x**2 - 2)*x**2 - 8*sqrt(x**2 - 2))/(12*(x**4 - 4*x**2 + 4))