Integrand size = 28, antiderivative size = 30 \[ \int \frac {\left (3+x^4\right ) \left (-1+x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\frac {4 \left (-3+7 x^3+3 x^4\right ) \left (-x+x^5\right )^{3/4}}{21 x^6} \] Output:
4/21*(3*x^4+7*x^3-3)*(x^5-x)^(3/4)/x^6
Result contains higher order function than in optimal. Order 5 vs. order 2 in optimal.
Time = 10.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 3.90 \[ \int \frac {\left (3+x^4\right ) \left (-1+x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\frac {4 \sqrt [4]{1-x^4} \left (165 \operatorname {Hypergeometric2F1}\left (-\frac {21}{16},\frac {1}{4},-\frac {5}{16},x^4\right )+x^3 \left (-385 \operatorname {Hypergeometric2F1}\left (-\frac {9}{16},\frac {1}{4},\frac {7}{16},x^4\right )-462 x \operatorname {Hypergeometric2F1}\left (-\frac {5}{16},\frac {1}{4},\frac {11}{16},x^4\right )+165 x^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {7}{16},\frac {23}{16},x^4\right )+105 x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {11}{16},\frac {27}{16},x^4\right )\right )\right )}{1155 x^5 \sqrt [4]{x \left (-1+x^4\right )}} \] Input:
Integrate[((3 + x^4)*(-1 + x^3 + x^4))/(x^6*(-x + x^5)^(1/4)),x]
Output:
(4*(1 - x^4)^(1/4)*(165*Hypergeometric2F1[-21/16, 1/4, -5/16, x^4] + x^3*( -385*Hypergeometric2F1[-9/16, 1/4, 7/16, x^4] - 462*x*Hypergeometric2F1[-5 /16, 1/4, 11/16, x^4] + 165*x^4*Hypergeometric2F1[1/4, 7/16, 23/16, x^4] + 105*x^5*Hypergeometric2F1[1/4, 11/16, 27/16, x^4])))/(1155*x^5*(x*(-1 + x ^4))^(1/4))
Time = 0.35 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2449, 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.
| \(\displaystyle \int \frac {\left (x^4+3\right ) \left (x^4+x^3-1\right )}{x^6 \sqrt [4]{x^5-x}} \, dx\) |
| \(\Big \downarrow \) 2449 |
| \(\displaystyle \int \left (\frac {x}{\sqrt [4]{x^5-x}}-\frac {3}{\sqrt [4]{x^5-x} x^6}+\frac {3}{\sqrt [4]{x^5-x} x^3}+\frac {x^2}{\sqrt [4]{x^5-x}}+\frac {2}{\sqrt [4]{x^5-x} x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 \left (x^5-x\right )^{3/4}}{7 x^6}+\frac {4 \left (x^5-x\right )^{3/4}}{3 x^3}+\frac {4 \left (x^5-x\right )^{3/4}}{7 x^2}\) |
Input:
Int[((3 + x^4)*(-1 + x^3 + x^4))/(x^6*(-x + x^5)^(1/4)),x]
Output:
(-4*(-x + x^5)^(3/4))/(7*x^6) + (4*(-x + x^5)^(3/4))/(3*x^3) + (4*(-x + x^ 5)^(3/4))/(7*x^2)
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_S ymbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ [{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) && !Integer Q[p] && NeQ[n, j]
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
| method | result | size |
| trager | \(\frac {4 \left (3 x^{4}+7 x^{3}-3\right ) \left (x^{5}-x \right )^{\frac {3}{4}}}{21 x^{6}}\) | \(27\) |
| pseudoelliptic | \(\frac {4 \left (3 x^{4}+7 x^{3}-3\right ) \left (x^{5}-x \right )^{\frac {3}{4}}}{21 x^{6}}\) | \(27\) |
| risch | \(\frac {\frac {4}{3} x^{7}-\frac {4}{3} x^{3}+\frac {4}{7} x^{8}-\frac {8}{7} x^{4}+\frac {4}{7}}{x^{5} {\left (x \left (x^{4}-1\right )\right )}^{\frac {1}{4}}}\) | \(37\) |
| gosper | \(\frac {4 \left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right ) \left (3 x^{4}+7 x^{3}-3\right )}{21 x^{5} \left (x^{5}-x \right )^{\frac {1}{4}}}\) | \(38\) |
| orering | \(\frac {4 \left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right ) \left (3 x^{4}+7 x^{3}-3\right )}{21 x^{5} \left (x^{5}-x \right )^{\frac {1}{4}}}\) | \(38\) |
| meijerg | \(\frac {4 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {21}{16}, \frac {1}{4}\right ], \left [-\frac {5}{16}\right ], x^{4}\right )}{7 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{\frac {21}{4}}}-\frac {8 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {5}{16}, \frac {1}{4}\right ], \left [\frac {11}{16}\right ], x^{4}\right )}{5 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{\frac {5}{4}}}-\frac {4 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {9}{16}, \frac {1}{4}\right ], \left [\frac {7}{16}\right ], x^{4}\right )}{3 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{\frac {9}{4}}}+\frac {4 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} x^{\frac {11}{4}} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {11}{16}\right ], \left [\frac {27}{16}\right ], x^{4}\right )}{11 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}}}+\frac {4 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} x^{\frac {7}{4}} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {7}{16}\right ], \left [\frac {23}{16}\right ], x^{4}\right )}{7 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}}}\) | \(162\) |
Input:
int((x^4+3)*(x^4+x^3-1)/x^6/(x^5-x)^(1/4),x,method=_RETURNVERBOSE)
Output:
4/21*(3*x^4+7*x^3-3)*(x^5-x)^(3/4)/x^6
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (3+x^4\right ) \left (-1+x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\frac {4 \, {\left (x^{5} - x\right )}^{\frac {3}{4}} {\left (3 \, x^{4} + 7 \, x^{3} - 3\right )}}{21 \, x^{6}} \] Input:
integrate((x^4+3)*(x^4+x^3-1)/x^6/(x^5-x)^(1/4),x, algorithm="fricas")
Output:
4/21*(x^5 - x)^(3/4)*(3*x^4 + 7*x^3 - 3)/x^6
\[ \int \frac {\left (3+x^4\right ) \left (-1+x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\int \frac {\left (x^{4} + 3\right ) \left (x^{4} + x^{3} - 1\right )}{x^{6} \sqrt [4]{x \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}}\, dx \] Input:
integrate((x**4+3)*(x**4+x**3-1)/x**6/(x**5-x)**(1/4),x)
Output:
Integral((x**4 + 3)*(x**4 + x**3 - 1)/(x**6*(x*(x - 1)*(x + 1)*(x**2 + 1)) **(1/4)), x)
\[ \int \frac {\left (3+x^4\right ) \left (-1+x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\int { \frac {{\left (x^{4} + x^{3} - 1\right )} {\left (x^{4} + 3\right )}}{{\left (x^{5} - x\right )}^{\frac {1}{4}} x^{6}} \,d x } \] Input:
integrate((x^4+3)*(x^4+x^3-1)/x^6/(x^5-x)^(1/4),x, algorithm="maxima")
Output:
integrate((x^4 + x^3 - 1)*(x^4 + 3)/((x^5 - x)^(1/4)*x^6), x)
\[ \int \frac {\left (3+x^4\right ) \left (-1+x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\int { \frac {{\left (x^{4} + x^{3} - 1\right )} {\left (x^{4} + 3\right )}}{{\left (x^{5} - x\right )}^{\frac {1}{4}} x^{6}} \,d x } \] Input:
integrate((x^4+3)*(x^4+x^3-1)/x^6/(x^5-x)^(1/4),x, algorithm="giac")
Output:
integrate((x^4 + x^3 - 1)*(x^4 + 3)/((x^5 - x)^(1/4)*x^6), x)
Time = 7.13 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {\left (3+x^4\right ) \left (-1+x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\frac {28\,x^3\,{\left (x^5-x\right )}^{3/4}-12\,{\left (x^5-x\right )}^{3/4}+12\,x^4\,{\left (x^5-x\right )}^{3/4}}{21\,x^6} \] Input:
int(((x^4 + 3)*(x^3 + x^4 - 1))/(x^6*(x^5 - x)^(1/4)),x)
Output:
(28*x^3*(x^5 - x)^(3/4) - 12*(x^5 - x)^(3/4) + 12*x^4*(x^5 - x)^(3/4))/(21 *x^6)
\[ \int \frac {\left (3+x^4\right ) \left (-1+x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\int \frac {x^{\frac {7}{4}}}{\left (x^{4}-1\right )^{\frac {1}{4}}}d x +\int \frac {x^{\frac {3}{4}}}{\left (x^{4}-1\right )^{\frac {1}{4}}}d x -3 \left (\int \frac {1}{x^{\frac {25}{4}} \left (x^{4}-1\right )^{\frac {1}{4}}}d x \right )+3 \left (\int \frac {1}{x^{\frac {13}{4}} \left (x^{4}-1\right )^{\frac {1}{4}}}d x \right )+2 \left (\int \frac {1}{x^{\frac {9}{4}} \left (x^{4}-1\right )^{\frac {1}{4}}}d x \right ) \] Input:
int((x^4+3)*(x^4+x^3-1)/x^6/(x^5-x)^(1/4),x)
Output:
int(x**2/(x**(1/4)*(x**4 - 1)**(1/4)),x) + int(x/(x**(1/4)*(x**4 - 1)**(1/ 4)),x) - 3*int(1/(x**(1/4)*(x**4 - 1)**(1/4)*x**6),x) + 3*int(1/(x**(1/4)* (x**4 - 1)**(1/4)*x**3),x) + 2*int(1/(x**(1/4)*(x**4 - 1)**(1/4)*x**2),x)