Integrand size = 15, antiderivative size = 53 \[ \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx=-\frac {7}{45} x^3 \sqrt {1-x^4}-\frac {1}{9} x^7 \sqrt {1-x^4}+\frac {7}{15} E(\arcsin (x)|-1)-\frac {7}{15} \operatorname {EllipticF}(\arcsin (x),-1) \] Output:
-7/45*x^3*(-x^4+1)^(1/2)-1/9*x^7*(-x^4+1)^(1/2)+7/15*EllipticE(x,I)-7/15*E llipticF(x,I)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx=\frac {1}{45} x^3 \left (-\sqrt {1-x^4} \left (7+5 x^4\right )+7 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},x^4\right )\right ) \] Input:
Integrate[x^10/Sqrt[1 - x^4],x]
Output:
(x^3*(-(Sqrt[1 - x^4]*(7 + 5*x^4)) + 7*Hypergeometric2F1[1/2, 3/4, 7/4, x^ 4]))/45
Time = 0.35 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {843, 843, 836, 762, 1388, 327}
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.
\(\displaystyle \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {7}{9} \int \frac {x^6}{\sqrt {1-x^4}}dx-\frac {1}{9} x^7 \sqrt {1-x^4}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {7}{9} \left (\frac {3}{5} \int \frac {x^2}{\sqrt {1-x^4}}dx-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\) |
\(\Big \downarrow \) 836 |
\(\displaystyle \frac {7}{9} \left (\frac {3}{5} \left (\int \frac {x^2+1}{\sqrt {1-x^4}}dx-\int \frac {1}{\sqrt {1-x^4}}dx\right )-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {7}{9} \left (\frac {3}{5} \left (\int \frac {x^2+1}{\sqrt {1-x^4}}dx-\operatorname {EllipticF}(\arcsin (x),-1)\right )-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \frac {7}{9} \left (\frac {3}{5} \left (\int \frac {\sqrt {x^2+1}}{\sqrt {1-x^2}}dx-\operatorname {EllipticF}(\arcsin (x),-1)\right )-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {7}{9} \left (\frac {3}{5} (E(\arcsin (x)|-1)-\operatorname {EllipticF}(\arcsin (x),-1))-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\) |
Input:
Int[x^10/Sqrt[1 - x^4],x]
Output:
-1/9*(x^7*Sqrt[1 - x^4]) + (7*(-1/5*(x^3*Sqrt[1 - x^4]) + (3*(EllipticE[Ar cSin[x], -1] - EllipticF[ArcSin[x], -1]))/5))/9
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) )], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 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[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Simp[-q^(-1) Int[1/Sqrt[a + b*x^4], x], x] + Simp[1/q Int[(1 + q*x^2)/S qrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 1.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.28
method | result | size |
meijerg | \(\frac {x^{11} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {11}{4}\right ], \left [\frac {15}{4}\right ], x^{4}\right )}{11}\) | \(15\) |
risch | \(\frac {x^{3} \left (5 x^{4}+7\right ) \left (x^{4}-1\right )}{45 \sqrt {-x^{4}+1}}-\frac {7 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (x , i\right )-\operatorname {EllipticE}\left (x , i\right )\right )}{15 \sqrt {-x^{4}+1}}\) | \(66\) |
default | \(-\frac {x^{7} \sqrt {-x^{4}+1}}{9}-\frac {7 x^{3} \sqrt {-x^{4}+1}}{45}-\frac {7 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (x , i\right )-\operatorname {EllipticE}\left (x , i\right )\right )}{15 \sqrt {-x^{4}+1}}\) | \(68\) |
elliptic | \(-\frac {x^{7} \sqrt {-x^{4}+1}}{9}-\frac {7 x^{3} \sqrt {-x^{4}+1}}{45}-\frac {7 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (x , i\right )-\operatorname {EllipticE}\left (x , i\right )\right )}{15 \sqrt {-x^{4}+1}}\) | \(68\) |
Input:
int(x^10/(-x^4+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/11*x^11*hypergeom([1/2,11/4],[15/4],x^4)
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx=\frac {-21 i \, x E(\arcsin \left (\frac {1}{x}\right )\,|\,-1) + 21 i \, x F(\arcsin \left (\frac {1}{x}\right )\,|\,-1) - {\left (5 \, x^{8} + 7 \, x^{4} + 21\right )} \sqrt {-x^{4} + 1}}{45 \, x} \] Input:
integrate(x^10/(-x^4+1)^(1/2),x, algorithm="fricas")
Output:
1/45*(-21*I*x*elliptic_e(arcsin(1/x), -1) + 21*I*x*elliptic_f(arcsin(1/x), -1) - (5*x^8 + 7*x^4 + 21)*sqrt(-x^4 + 1))/x
Time = 0.49 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.58 \[ \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx=\frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} \] Input:
integrate(x**10/(-x**4+1)**(1/2),x)
Output:
x**11*gamma(11/4)*hyper((1/2, 11/4), (15/4,), x**4*exp_polar(2*I*pi))/(4*g amma(15/4))
\[ \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx=\int { \frac {x^{10}}{\sqrt {-x^{4} + 1}} \,d x } \] Input:
integrate(x^10/(-x^4+1)^(1/2),x, algorithm="maxima")
Output:
integrate(x^10/sqrt(-x^4 + 1), x)
\[ \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx=\int { \frac {x^{10}}{\sqrt {-x^{4} + 1}} \,d x } \] Input:
integrate(x^10/(-x^4+1)^(1/2),x, algorithm="giac")
Output:
integrate(x^10/sqrt(-x^4 + 1), x)
Timed out. \[ \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx=\int \frac {x^{10}}{\sqrt {1-x^4}} \,d x \] Input:
int(x^10/(1 - x^4)^(1/2),x)
Output:
int(x^10/(1 - x^4)^(1/2), x)
\[ \int \frac {x^{10}}{\sqrt {1-x^4}} \, dx=-\frac {\sqrt {-x^{4}+1}\, x^{7}}{9}-\frac {7 \sqrt {-x^{4}+1}\, x^{3}}{45}-\frac {7 \left (\int \frac {\sqrt {-x^{4}+1}\, x^{2}}{x^{4}-1}d x \right )}{15} \] Input:
int(x^10/(-x^4+1)^(1/2),x)
Output:
( - 5*sqrt( - x**4 + 1)*x**7 - 7*sqrt( - x**4 + 1)*x**3 - 21*int((sqrt( - x**4 + 1)*x**2)/(x**4 - 1),x))/45