Integrand size = 45, antiderivative size = 48 \[ \int \frac {\left (-3+x^4\right ) \left (1-2 x^3+x^4\right ) \left (1-x^3+x^4\right )}{x^6 \left (1+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\frac {4 \left (x+x^5\right )^{3/4} \left (1-7 x^3+2 x^4-14 x^6-7 x^7+x^8\right )}{7 x^6 \left (1+x^4\right )} \]
Result contains higher order function than in optimal. Order 5 vs. order 2 in optimal.
Time = 10.13 (sec) , antiderivative size = 203, normalized size of antiderivative = 4.23 \[ \int \frac {\left (-3+x^4\right ) \left (1-2 x^3+x^4\right ) \left (1-x^3+x^4\right )}{x^6 \left (1+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\frac {4 \sqrt [4]{1+x^4} \left (129789 \operatorname {Hypergeometric2F1}\left (-\frac {21}{16},\frac {5}{4},-\frac {5}{16},-x^4\right )+x^3 \left (-908523 \operatorname {Hypergeometric2F1}\left (-\frac {9}{16},\frac {5}{4},\frac {7}{16},-x^4\right )+908523 x \operatorname {Hypergeometric2F1}\left (-\frac {5}{16},\frac {5}{4},\frac {11}{16},-x^4\right )-1817046 x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{16},\frac {5}{4},\frac {19}{16},-x^4\right )+778734 x^4 \operatorname {Hypergeometric2F1}\left (\frac {7}{16},\frac {5}{4},\frac {23}{16},-x^4\right )-82593 x^5 \operatorname {Hypergeometric2F1}\left (\frac {11}{16},\frac {5}{4},\frac {27}{16},-x^4\right )+95634 x^7 \operatorname {Hypergeometric2F1}\left (\frac {19}{16},\frac {5}{4},\frac {35}{16},-x^4\right )-118503 x^8 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {23}{16},\frac {39}{16},-x^4\right )+33649 x^9 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {27}{16},\frac {43}{16},-x^4\right )\right )\right )}{908523 x^5 \sqrt [4]{x+x^5}} \]
(4*(1 + x^4)^(1/4)*(129789*Hypergeometric2F1[-21/16, 5/4, -5/16, -x^4] + x ^3*(-908523*Hypergeometric2F1[-9/16, 5/4, 7/16, -x^4] + 908523*x*Hypergeom etric2F1[-5/16, 5/4, 11/16, -x^4] - 1817046*x^3*Hypergeometric2F1[3/16, 5/ 4, 19/16, -x^4] + 778734*x^4*Hypergeometric2F1[7/16, 5/4, 23/16, -x^4] - 8 2593*x^5*Hypergeometric2F1[11/16, 5/4, 27/16, -x^4] + 95634*x^7*Hypergeome tric2F1[19/16, 5/4, 35/16, -x^4] - 118503*x^8*Hypergeometric2F1[5/4, 23/16 , 39/16, -x^4] + 33649*x^9*Hypergeometric2F1[5/4, 27/16, 43/16, -x^4])))/( 908523*x^5*(x + x^5)^(1/4))
Time = 1.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.56, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2467, 25, 2035, 2368, 27, 2372, 2009}
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.
| \(\displaystyle \int \frac {\left (x^4-3\right ) \left (x^4-2 x^3+1\right ) \left (x^4-x^3+1\right )}{x^6 \left (x^4+1\right ) \sqrt [4]{x^5+x}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{x^4+1} \int -\frac {\left (3-x^4\right ) \left (x^4-2 x^3+1\right ) \left (x^4-x^3+1\right )}{x^{25/4} \left (x^4+1\right )^{5/4}}dx}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [4]{x} \sqrt [4]{x^4+1} \int \frac {\left (3-x^4\right ) \left (x^4-2 x^3+1\right ) \left (x^4-x^3+1\right )}{x^{25/4} \left (x^4+1\right )^{5/4}}dx}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{x^4+1} \int \frac {\left (3-x^4\right ) \left (x^4-2 x^3+1\right ) \left (x^4-x^3+1\right )}{x^{11/2} \left (x^4+1\right )^{5/4}}d\sqrt [4]{x}}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 2368 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{x^4+1} \left (\frac {2 x^{3/4}}{\sqrt [4]{x^4+1}}-\frac {1}{4} \int -\frac {4 \left (-x^8+3 x^7+2 x^4-9 x^3+3\right )}{x^{11/2} \sqrt [4]{x^4+1}}d\sqrt [4]{x}\right )}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{x^4+1} \left (\int \frac {-x^8+3 x^7+2 x^4-9 x^3+3}{x^{11/2} \sqrt [4]{x^4+1}}d\sqrt [4]{x}+\frac {2 x^{3/4}}{\sqrt [4]{x^4+1}}\right )}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{x^4+1} \left (\int \left (\frac {3 x^6-9 x^2}{x^{9/2} \sqrt [4]{x^4+1}}+\frac {-x^8+2 x^4+3}{x^{11/2} \sqrt [4]{x^4+1}}\right )d\sqrt [4]{x}+\frac {2 x^{3/4}}{\sqrt [4]{x^4+1}}\right )}{\sqrt [4]{x^5+x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{x^4+1} \left (-\frac {\left (x^4+1\right )^{7/4}}{7 x^{21/4}}+\frac {\left (x^4+1\right )^{3/4}}{x^{9/4}}+\frac {2 x^{3/4}}{\sqrt [4]{x^4+1}}\right )}{\sqrt [4]{x^5+x}}\) |
(-4*x^(1/4)*(1 + x^4)^(1/4)*((2*x^(3/4))/(1 + x^4)^(1/4) + (1 + x^4)^(3/4) /x^(9/4) - (1 + x^4)^(7/4)/(7*x^(21/4))))/(x + x^5)^(1/4)
3.7.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^ m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1)*x^m *Pq, a + b*x^n, x], i}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a^2*n*(p + 1)*b^( Floor[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) Int[x^m*(a + b*x^n)^(p + 1)*ExpandToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)/a)*Coeff[R, x, i]*x^(i - m), {i, 0, n - 1}], x], x], x]]] /; F reeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Mo dule[{q = Expon[Pq, x], j, k}, Int[Sum[((c*x)^(m + j)/c^j)*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2*((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0 ] && !PolyQ[Pq, x^(n/2)]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 1.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79
| method | result | size |
| gosper | \(\frac {\frac {4}{7} x^{8}+\frac {8}{7} x^{4}+\frac {4}{7}-8 x^{6}-4 x^{7}-4 x^{3}}{\left (x^{5}+x \right )^{\frac {1}{4}} x^{5}}\) | \(38\) |
| risch | \(\frac {\frac {4}{7} x^{8}+\frac {8}{7} x^{4}+\frac {4}{7}-8 x^{6}-4 x^{7}-4 x^{3}}{x^{5} {\left (x \left (x^{4}+1\right )\right )}^{\frac {1}{4}}}\) | \(40\) |
| pseudoelliptic | \(\frac {\frac {4}{7} x^{8}+\frac {8}{7} x^{4}+\frac {4}{7}-8 x^{6}-4 x^{7}-4 x^{3}}{x^{5} {\left (x \left (x^{4}+1\right )\right )}^{\frac {1}{4}}}\) | \(40\) |
| trager | \(\frac {4 \left (x^{5}+x \right )^{\frac {3}{4}} \left (x^{8}-7 x^{7}-14 x^{6}+2 x^{4}-7 x^{3}+1\right )}{7 x^{6} \left (x^{4}+1\right )}\) | \(45\) |
| meijerg | \(\frac {4 \operatorname {hypergeom}\left (\left [-\frac {21}{16}, \frac {5}{4}\right ], \left [-\frac {5}{16}\right ], -x^{4}\right )}{7 x^{\frac {21}{4}}}+\frac {4 \operatorname {hypergeom}\left (\left [-\frac {5}{16}, \frac {5}{4}\right ], \left [\frac {11}{16}\right ], -x^{4}\right )}{x^{\frac {5}{4}}}-\frac {4 \operatorname {hypergeom}\left (\left [-\frac {9}{16}, \frac {5}{4}\right ], \left [\frac {7}{16}\right ], -x^{4}\right )}{x^{\frac {9}{4}}}-\frac {4 x^{\frac {11}{4}} \operatorname {hypergeom}\left (\left [\frac {11}{16}, \frac {5}{4}\right ], \left [\frac {27}{16}\right ], -x^{4}\right )}{11}+\frac {24 x^{\frac {7}{4}} \operatorname {hypergeom}\left (\left [\frac {7}{16}, \frac {5}{4}\right ], \left [\frac {23}{16}\right ], -x^{4}\right )}{7}-8 x^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [\frac {3}{16}, \frac {5}{4}\right ], \left [\frac {19}{16}\right ], -x^{4}\right )+\frac {4 x^{\frac {27}{4}} \operatorname {hypergeom}\left (\left [\frac {5}{4}, \frac {27}{16}\right ], \left [\frac {43}{16}\right ], -x^{4}\right )}{27}-\frac {12 x^{\frac {23}{4}} \operatorname {hypergeom}\left (\left [\frac {5}{4}, \frac {23}{16}\right ], \left [\frac {39}{16}\right ], -x^{4}\right )}{23}+\frac {8 x^{\frac {19}{4}} \operatorname {hypergeom}\left (\left [\frac {19}{16}, \frac {5}{4}\right ], \left [\frac {35}{16}\right ], -x^{4}\right )}{19}\) | \(146\) |
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-3+x^4\right ) \left (1-2 x^3+x^4\right ) \left (1-x^3+x^4\right )}{x^6 \left (1+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\frac {4 \, {\left (x^{8} - 7 \, x^{7} - 14 \, x^{6} + 2 \, x^{4} - 7 \, x^{3} + 1\right )} {\left (x^{5} + x\right )}^{\frac {3}{4}}}{7 \, {\left (x^{10} + x^{6}\right )}} \]
\[ \int \frac {\left (-3+x^4\right ) \left (1-2 x^3+x^4\right ) \left (1-x^3+x^4\right )}{x^6 \left (1+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\int \frac {\left (x - 1\right ) \left (x^{4} - 3\right ) \left (x^{4} - x^{3} + 1\right ) \left (x^{3} - x^{2} - x - 1\right )}{x^{6} \sqrt [4]{x \left (x^{4} + 1\right )} \left (x^{4} + 1\right )}\, dx \]
Integral((x - 1)*(x**4 - 3)*(x**4 - x**3 + 1)*(x**3 - x**2 - x - 1)/(x**6* (x*(x**4 + 1))**(1/4)*(x**4 + 1)), x)
\[ \int \frac {\left (-3+x^4\right ) \left (1-2 x^3+x^4\right ) \left (1-x^3+x^4\right )}{x^6 \left (1+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\int { \frac {{\left (x^{4} - x^{3} + 1\right )} {\left (x^{4} - 2 \, x^{3} + 1\right )} {\left (x^{4} - 3\right )}}{{\left (x^{5} + x\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )} x^{6}} \,d x } \]
\[ \int \frac {\left (-3+x^4\right ) \left (1-2 x^3+x^4\right ) \left (1-x^3+x^4\right )}{x^6 \left (1+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\int { \frac {{\left (x^{4} - x^{3} + 1\right )} {\left (x^{4} - 2 \, x^{3} + 1\right )} {\left (x^{4} - 3\right )}}{{\left (x^{5} + x\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )} x^{6}} \,d x } \]
Time = 5.78 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {\left (-3+x^4\right ) \left (1-2 x^3+x^4\right ) \left (1-x^3+x^4\right )}{x^6 \left (1+x^4\right ) \sqrt [4]{x+x^5}} \, dx=\frac {4\,{\left (x^5+x\right )}^{3/4}}{7\,x^2}-\frac {8\,{\left (x^5+x\right )}^{3/4}}{x^4+1}-\frac {4\,{\left (x^5+x\right )}^{3/4}}{x^3}+\frac {4\,{\left (x^5+x\right )}^{3/4}}{7\,x^6} \]