Integrand size = 15, antiderivative size = 71 \[ \int \frac {x^{14}}{\left (1-x^4\right )^{3/2}} \, dx=\frac {x^{11}}{2 \sqrt {1-x^4}}+\frac {77}{90} x^3 \sqrt {1-x^4}+\frac {11}{18} x^7 \sqrt {1-x^4}-\frac {77}{30} E(\arcsin (x)|-1)+\frac {77}{30} \operatorname {EllipticF}(\arcsin (x),-1) \]
-77/30*EllipticE(x,I)+77/30*EllipticF(x,I)+1/2*x^11/(-x^4+1)^(1/2)+77/90*x ^3*(-x^4+1)^(1/2)+11/18*x^7*(-x^4+1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79 \[ \int \frac {x^{14}}{\left (1-x^4\right )^{3/2}} \, dx=-\frac {x^3 \left (77+11 x^4+5 x^8-77 \sqrt {1-x^4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},x^4\right )\right )}{45 \sqrt {1-x^4}} \]
-1/45*(x^3*(77 + 11*x^4 + 5*x^8 - 77*Sqrt[1 - x^4]*Hypergeometric2F1[3/4, 3/2, 7/4, x^4]))/Sqrt[1 - x^4]
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {817, 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.
\(\displaystyle \int \frac {x^{14}}{\left (1-x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {x^{11}}{2 \sqrt {1-x^4}}-\frac {11}{2} \int \frac {x^{10}}{\sqrt {1-x^4}}dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {x^{11}}{2 \sqrt {1-x^4}}-\frac {11}{2} \left (\frac {7}{9} \int \frac {x^6}{\sqrt {1-x^4}}dx-\frac {1}{9} x^7 \sqrt {1-x^4}\right )\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {x^{11}}{2 \sqrt {1-x^4}}-\frac {11}{2} \left (\frac {7}{9} \left (\frac {3}{5} \int \frac {x^2}{\sqrt {1-x^4}}dx-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\right )\) |
\(\Big \downarrow \) 836 |
\(\displaystyle \frac {x^{11}}{2 \sqrt {1-x^4}}-\frac {11}{2} \left (\frac {7}{9} \left (\frac {3}{5} \left (\int \frac {x^2+1}{\sqrt {1-x^4}}dx-\int \frac {1}{\sqrt {1-x^4}}dx\right )-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\right )\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {x^{11}}{2 \sqrt {1-x^4}}-\frac {11}{2} \left (\frac {7}{9} \left (\frac {3}{5} \left (\int \frac {x^2+1}{\sqrt {1-x^4}}dx-\operatorname {EllipticF}(\arcsin (x),-1)\right )-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\right )\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \frac {x^{11}}{2 \sqrt {1-x^4}}-\frac {11}{2} \left (\frac {7}{9} \left (\frac {3}{5} \left (\int \frac {\sqrt {x^2+1}}{\sqrt {1-x^2}}dx-\operatorname {EllipticF}(\arcsin (x),-1)\right )-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\right )\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {x^{11}}{2 \sqrt {1-x^4}}-\frac {11}{2} \left (\frac {7}{9} \left (\frac {3}{5} (E(\arcsin (x)|-1)-\operatorname {EllipticF}(\arcsin (x),-1))-\frac {1}{5} x^3 \sqrt {1-x^4}\right )-\frac {1}{9} x^7 \sqrt {1-x^4}\right )\) |
x^11/(2*Sqrt[1 - x^4]) - (11*(-1/9*(x^7*Sqrt[1 - x^4]) + (7*(-1/5*(x^3*Sqr t[1 - x^4]) + (3*(EllipticE[ArcSin[x], -1] - EllipticF[ArcSin[x], -1]))/5) )/9))/2
3.10.10.3.1 Defintions of rubi rules used
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[((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*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
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 = 5.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.21
method | result | size |
meijerg | \(\frac {x^{15} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{2},\frac {15}{4};\frac {19}{4};x^{4}\right )}{15}\) | \(15\) |
risch | \(-\frac {x^{3} \left (10 x^{8}+22 x^{4}-77\right )}{90 \sqrt {-x^{4}+1}}+\frac {77 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{30 \sqrt {-x^{4}+1}}\) | \(66\) |
default | \(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}+\frac {x^{7} \sqrt {-x^{4}+1}}{9}+\frac {16 x^{3} \sqrt {-x^{4}+1}}{45}+\frac {77 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{30 \sqrt {-x^{4}+1}}\) | \(82\) |
elliptic | \(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}+\frac {x^{7} \sqrt {-x^{4}+1}}{9}+\frac {16 x^{3} \sqrt {-x^{4}+1}}{45}+\frac {77 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{30 \sqrt {-x^{4}+1}}\) | \(82\) |
Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04 \[ \int \frac {x^{14}}{\left (1-x^4\right )^{3/2}} \, dx=-\frac {231 \, {\left (-i \, x^{5} + i \, x\right )} E(\arcsin \left (\frac {1}{x}\right )\,|\,-1) + 231 \, {\left (i \, x^{5} - i \, x\right )} F(\arcsin \left (\frac {1}{x}\right )\,|\,-1) - {\left (10 \, x^{12} + 22 \, x^{8} + 154 \, x^{4} - 231\right )} \sqrt {-x^{4} + 1}}{90 \, {\left (x^{5} - x\right )}} \]
-1/90*(231*(-I*x^5 + I*x)*elliptic_e(arcsin(1/x), -1) + 231*(I*x^5 - I*x)* elliptic_f(arcsin(1/x), -1) - (10*x^12 + 22*x^8 + 154*x^4 - 231)*sqrt(-x^4 + 1))/(x^5 - x)
Time = 0.72 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.44 \[ \int \frac {x^{14}}{\left (1-x^4\right )^{3/2}} \, dx=\frac {x^{15} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {19}{4}\right )} \]
\[ \int \frac {x^{14}}{\left (1-x^4\right )^{3/2}} \, dx=\int { \frac {x^{14}}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^{14}}{\left (1-x^4\right )^{3/2}} \, dx=\int { \frac {x^{14}}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^{14}}{\left (1-x^4\right )^{3/2}} \, dx=\int \frac {x^{14}}{{\left (1-x^4\right )}^{3/2}} \,d x \]