Integrand size = 13, antiderivative size = 89 \[ \int \frac {1}{x^6 \sqrt {-1+x^4}} \, dx=\frac {\sqrt {-1+x^4}}{5 x^5}+\frac {3 \sqrt {-1+x^4}}{5 x}-\frac {3 \sqrt {1-x^4} E(\arcsin (x)|-1)}{5 \sqrt {-1+x^4}}+\frac {3 \sqrt {1-x^4} \operatorname {EllipticF}(\arcsin (x),-1)}{5 \sqrt {-1+x^4}} \] Output:
1/5*(x^4-1)^(1/2)/x^5+3/5*(x^4-1)^(1/2)/x-3/5*(-x^4+1)^(1/2)*EllipticE(x,I )/(x^4-1)^(1/2)+3/5*(-x^4+1)^(1/2)*EllipticF(x,I)/(x^4-1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^6 \sqrt {-1+x^4}} \, dx=-\frac {\sqrt {1-x^4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},-\frac {1}{4},x^4\right )}{5 x^5 \sqrt {-1+x^4}} \] Input:
Integrate[1/(x^6*Sqrt[-1 + x^4]),x]
Output:
-1/5*(Sqrt[1 - x^4]*Hypergeometric2F1[-5/4, 1/2, -1/4, x^4])/(x^5*Sqrt[-1 + x^4])
Time = 0.44 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.81, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {847, 847, 835, 763, 1499}
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.
\(\displaystyle \int \frac {1}{x^6 \sqrt {x^4-1}} \, dx\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {3}{5} \int \frac {1}{x^2 \sqrt {x^4-1}}dx+\frac {\sqrt {x^4-1}}{5 x^5}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {3}{5} \left (\frac {\sqrt {x^4-1}}{x}-\int \frac {x^2}{\sqrt {x^4-1}}dx\right )+\frac {\sqrt {x^4-1}}{5 x^5}\) |
\(\Big \downarrow \) 835 |
\(\displaystyle \frac {3}{5} \left (-\int \frac {1}{\sqrt {x^4-1}}dx+\int \frac {1-x^2}{\sqrt {x^4-1}}dx+\frac {\sqrt {x^4-1}}{x}\right )+\frac {\sqrt {x^4-1}}{5 x^5}\) |
\(\Big \downarrow \) 763 |
\(\displaystyle \frac {3}{5} \left (\int \frac {1-x^2}{\sqrt {x^4-1}}dx-\frac {\sqrt {x^2-1} \sqrt {x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right ),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4-1}}+\frac {\sqrt {x^4-1}}{x}\right )+\frac {\sqrt {x^4-1}}{5 x^5}\) |
\(\Big \downarrow \) 1499 |
\(\displaystyle \frac {3}{5} \left (-\frac {\sqrt {x^2-1} \sqrt {x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right ),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4-1}}+\frac {\sqrt {2} \sqrt {x^2-1} \sqrt {x^2+1} E\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right )|\frac {1}{2}\right )}{\sqrt {x^4-1}}+\frac {\sqrt {x^4-1}}{x}-\frac {x \left (x^2+1\right )}{\sqrt {x^4-1}}\right )+\frac {\sqrt {x^4-1}}{5 x^5}\) |
Input:
Int[1/(x^6*Sqrt[-1 + x^4]),x]
Output:
Sqrt[-1 + x^4]/(5*x^5) + (3*(-((x*(1 + x^2))/Sqrt[-1 + x^4]) + Sqrt[-1 + x ^4]/x + (Sqrt[2]*Sqrt[-1 + x^2]*Sqrt[1 + x^2]*EllipticE[ArcSin[(Sqrt[2]*x) /Sqrt[-1 + x^2]], 1/2])/Sqrt[-1 + x^4] - (Sqrt[-1 + x^2]*Sqrt[1 + x^2]*Ell ipticF[ArcSin[(Sqrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/(Sqrt[2]*Sqrt[-1 + x^4])) )/5
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*b, 2]}, Sim p[Sqrt[-a + q*x^2]*(Sqrt[(a + q*x^2)/q]/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4])) *EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2], x] /; IntegerQ[q]] /; F reeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Simp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[ a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 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[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[e*x*((q + c*x^2)/(c*Sqrt[a + c*x^4])), x] - Simp[Sqrt [2]*e*q*Sqrt[-a + q*x^2]*(Sqrt[(a + q*x^2)/q]/(Sqrt[-a]*c*Sqrt[a + c*x^4])) *EllipticE[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2], x] /; EqQ[c*d + e*q, 0] && IntegerQ[q]] /; FreeQ[{a, c, d, e}, x] && LtQ[a, 0] && GtQ[c, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.78 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.37
method | result | size |
meijerg | \(-\frac {\sqrt {-\operatorname {signum}\left (x^{4}-1\right )}\, \operatorname {hypergeom}\left (\left [-\frac {5}{4}, \frac {1}{2}\right ], \left [-\frac {1}{4}\right ], x^{4}\right )}{5 \sqrt {\operatorname {signum}\left (x^{4}-1\right )}\, x^{5}}\) | \(33\) |
default | \(\frac {\sqrt {x^{4}-1}}{5 x^{5}}+\frac {3 \sqrt {x^{4}-1}}{5 x}+\frac {3 i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (\operatorname {EllipticF}\left (i x , i\right )-\operatorname {EllipticE}\left (i x , i\right )\right )}{5 \sqrt {x^{4}-1}}\) | \(69\) |
risch | \(\frac {3 x^{8}-2 x^{4}-1}{5 x^{5} \sqrt {x^{4}-1}}+\frac {3 i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (\operatorname {EllipticF}\left (i x , i\right )-\operatorname {EllipticE}\left (i x , i\right )\right )}{5 \sqrt {x^{4}-1}}\) | \(69\) |
elliptic | \(\frac {\sqrt {x^{4}-1}}{5 x^{5}}+\frac {3 \sqrt {x^{4}-1}}{5 x}+\frac {3 i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (\operatorname {EllipticF}\left (i x , i\right )-\operatorname {EllipticE}\left (i x , i\right )\right )}{5 \sqrt {x^{4}-1}}\) | \(69\) |
Input:
int(1/x^6/(x^4-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/5/signum(x^4-1)^(1/2)*(-signum(x^4-1))^(1/2)/x^5*hypergeom([-5/4,1/2],[ -1/4],x^4)
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^6 \sqrt {-1+x^4}} \, dx=\frac {3 i \, x^{5} E(\arcsin \left (x\right )\,|\,-1) - 3 i \, x^{5} F(\arcsin \left (x\right )\,|\,-1) + {\left (3 \, x^{4} + 1\right )} \sqrt {x^{4} - 1}}{5 \, x^{5}} \] Input:
integrate(1/x^6/(x^4-1)^(1/2),x, algorithm="fricas")
Output:
1/5*(3*I*x^5*elliptic_e(arcsin(x), -1) - 3*I*x^5*elliptic_f(arcsin(x), -1) + (3*x^4 + 1)*sqrt(x^4 - 1))/x^5
Time = 0.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^6 \sqrt {-1+x^4}} \, dx=- \frac {i \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} \] Input:
integrate(1/x**6/(x**4-1)**(1/2),x)
Output:
-I*gamma(-5/4)*hyper((-5/4, 1/2), (-1/4,), x**4)/(4*x**5*gamma(-1/4))
\[ \int \frac {1}{x^6 \sqrt {-1+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} - 1} x^{6}} \,d x } \] Input:
integrate(1/x^6/(x^4-1)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(x^4 - 1)*x^6), x)
\[ \int \frac {1}{x^6 \sqrt {-1+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} - 1} x^{6}} \,d x } \] Input:
integrate(1/x^6/(x^4-1)^(1/2),x, algorithm="giac")
Output:
integrate(1/(sqrt(x^4 - 1)*x^6), x)
Timed out. \[ \int \frac {1}{x^6 \sqrt {-1+x^4}} \, dx=\int \frac {1}{x^6\,\sqrt {x^4-1}} \,d x \] Input:
int(1/(x^6*(x^4 - 1)^(1/2)),x)
Output:
int(1/(x^6*(x^4 - 1)^(1/2)), x)
\[ \int \frac {1}{x^6 \sqrt {-1+x^4}} \, dx=\int \frac {\sqrt {x^{4}-1}}{x^{10}-x^{6}}d x \] Input:
int(1/x^6/(x^4-1)^(1/2),x)
Output:
int(sqrt(x**4 - 1)/(x**10 - x**6),x)