Integrand size = 30, antiderivative size = 30 \[ \int \frac {\left (-3+x^2\right ) \left (1-x^2+x^3\right )}{x^6 \sqrt [4]{-x+x^3}} \, dx=-\frac {4 \left (-x+x^3\right )^{3/4} \left (3-3 x^2+7 x^3\right )}{21 x^6} \] Output:
-4/21*(x^3-x)^(3/4)*(7*x^3-3*x^2+3)/x^6
Result contains higher order function than in optimal. Order 5 vs. order 2 in optimal.
Time = 10.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 3.97 \[ \int \frac {\left (-3+x^2\right ) \left (1-x^2+x^3\right )}{x^6 \sqrt [4]{-x+x^3}} \, dx=-\frac {4 \sqrt [4]{1-x^2} \left (-195 \operatorname {Hypergeometric2F1}\left (-\frac {21}{8},\frac {1}{4},-\frac {13}{8},x^2\right )+7 x^2 \left (60 \operatorname {Hypergeometric2F1}\left (-\frac {13}{8},\frac {1}{4},-\frac {5}{8},x^2\right )+13 x \left (-5 \operatorname {Hypergeometric2F1}\left (-\frac {9}{8},\frac {1}{4},-\frac {1}{8},x^2\right )-3 x \operatorname {Hypergeometric2F1}\left (-\frac {5}{8},\frac {1}{4},\frac {3}{8},x^2\right )+15 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},\frac {1}{4},\frac {7}{8},x^2\right )\right )\right )\right )}{1365 x^5 \sqrt [4]{x \left (-1+x^2\right )}} \] Input:
Integrate[((-3 + x^2)*(1 - x^2 + x^3))/(x^6*(-x + x^3)^(1/4)),x]
Output:
(-4*(1 - x^2)^(1/4)*(-195*Hypergeometric2F1[-21/8, 1/4, -13/8, x^2] + 7*x^ 2*(60*Hypergeometric2F1[-13/8, 1/4, -5/8, x^2] + 13*x*(-5*Hypergeometric2F 1[-9/8, 1/4, -1/8, x^2] - 3*x*Hypergeometric2F1[-5/8, 1/4, 3/8, x^2] + 15* x^2*Hypergeometric2F1[-1/8, 1/4, 7/8, x^2]))))/(1365*x^5*(x*(-1 + x^2))^(1 /4))
Time = 0.36 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, 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.
| \(\displaystyle \int \frac {\left (x^2-3\right ) \left (x^3-x^2+1\right )}{x^6 \sqrt [4]{x^3-x}} \, dx\) |
| \(\Big \downarrow \) 2449 |
| \(\displaystyle \int \left (\frac {1}{x \sqrt [4]{x^3-x}}-\frac {3}{x^3 \sqrt [4]{x^3-x}}-\frac {3}{x^6 \sqrt [4]{x^3-x}}+\frac {4}{x^4 \sqrt [4]{x^3-x}}-\frac {1}{x^2 \sqrt [4]{x^3-x}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 \left (x^3-x\right )^{3/4}}{3 x^3}-\frac {4 \left (x^3-x\right )^{3/4}}{7 x^6}+\frac {4 \left (x^3-x\right )^{3/4}}{7 x^4}\) |
Input:
Int[((-3 + x^2)*(1 - x^2 + x^3))/(x^6*(-x + x^3)^(1/4)),x]
Output:
(-4*(-x + x^3)^(3/4))/(7*x^6) + (4*(-x + x^3)^(3/4))/(7*x^4) - (4*(-x + x^ 3)^(3/4))/(3*x^3)
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 = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
| method | result | size |
| trager | \(-\frac {4 \left (x^{3}-x \right )^{\frac {3}{4}} \left (7 x^{3}-3 x^{2}+3\right )}{21 x^{6}}\) | \(27\) |
| gosper | \(-\frac {4 \left (1+x \right ) \left (-1+x \right ) \left (7 x^{3}-3 x^{2}+3\right )}{21 x^{5} \left (x^{3}-x \right )^{\frac {1}{4}}}\) | \(33\) |
| orering | \(-\frac {4 \left (1+x \right ) \left (-1+x \right ) \left (7 x^{3}-3 x^{2}+3\right )}{21 x^{5} \left (x^{3}-x \right )^{\frac {1}{4}}}\) | \(33\) |
| risch | \(-\frac {4 \left (7 x^{5}-3 x^{4}-7 x^{3}+6 x^{2}-3\right )}{21 x^{5} {\left (x \left (x^{2}-1\right )\right )}^{\frac {1}{4}}}\) | \(37\) |
| meijerg | \(\frac {4 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {21}{8}, \frac {1}{4}\right ], \left [-\frac {13}{8}\right ], x^{2}\right )}{7 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {21}{4}}}+\frac {4 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {9}{8}, \frac {1}{4}\right ], \left [-\frac {1}{8}\right ], x^{2}\right )}{3 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {9}{4}}}-\frac {16 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {13}{8}, \frac {1}{4}\right ], \left [-\frac {5}{8}\right ], x^{2}\right )}{13 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {13}{4}}}-\frac {4 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{8}, \frac {1}{4}\right ], \left [\frac {7}{8}\right ], x^{2}\right )}{\operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {1}{4}}}+\frac {4 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {5}{8}, \frac {1}{4}\right ], \left [\frac {3}{8}\right ], x^{2}\right )}{5 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {5}{4}}}\) | \(162\) |
Input:
int((x^2-3)*(x^3-x^2+1)/x^6/(x^3-x)^(1/4),x,method=_RETURNVERBOSE)
Output:
-4/21*(x^3-x)^(3/4)*(7*x^3-3*x^2+3)/x^6
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-3+x^2\right ) \left (1-x^2+x^3\right )}{x^6 \sqrt [4]{-x+x^3}} \, dx=-\frac {4 \, {\left (7 \, x^{3} - 3 \, x^{2} + 3\right )} {\left (x^{3} - x\right )}^{\frac {3}{4}}}{21 \, x^{6}} \] Input:
integrate((x^2-3)*(x^3-x^2+1)/x^6/(x^3-x)^(1/4),x, algorithm="fricas")
Output:
-4/21*(7*x^3 - 3*x^2 + 3)*(x^3 - x)^(3/4)/x^6
\[ \int \frac {\left (-3+x^2\right ) \left (1-x^2+x^3\right )}{x^6 \sqrt [4]{-x+x^3}} \, dx=\int \frac {\left (x^{2} - 3\right ) \left (x^{3} - x^{2} + 1\right )}{x^{6} \sqrt [4]{x \left (x - 1\right ) \left (x + 1\right )}}\, dx \] Input:
integrate((x**2-3)*(x**3-x**2+1)/x**6/(x**3-x)**(1/4),x)
Output:
Integral((x**2 - 3)*(x**3 - x**2 + 1)/(x**6*(x*(x - 1)*(x + 1))**(1/4)), x )
\[ \int \frac {\left (-3+x^2\right ) \left (1-x^2+x^3\right )}{x^6 \sqrt [4]{-x+x^3}} \, dx=\int { \frac {{\left (x^{3} - x^{2} + 1\right )} {\left (x^{2} - 3\right )}}{{\left (x^{3} - x\right )}^{\frac {1}{4}} x^{6}} \,d x } \] Input:
integrate((x^2-3)*(x^3-x^2+1)/x^6/(x^3-x)^(1/4),x, algorithm="maxima")
Output:
integrate((x^3 - x^2 + 1)*(x^2 - 3)/((x^3 - x)^(1/4)*x^6), x)
\[ \int \frac {\left (-3+x^2\right ) \left (1-x^2+x^3\right )}{x^6 \sqrt [4]{-x+x^3}} \, dx=\int { \frac {{\left (x^{3} - x^{2} + 1\right )} {\left (x^{2} - 3\right )}}{{\left (x^{3} - x\right )}^{\frac {1}{4}} x^{6}} \,d x } \] Input:
integrate((x^2-3)*(x^3-x^2+1)/x^6/(x^3-x)^(1/4),x, algorithm="giac")
Output:
integrate((x^3 - x^2 + 1)*(x^2 - 3)/((x^3 - x)^(1/4)*x^6), x)
Time = 6.84 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {\left (-3+x^2\right ) \left (1-x^2+x^3\right )}{x^6 \sqrt [4]{-x+x^3}} \, dx=-\frac {12\,{\left (x^3-x\right )}^{3/4}-12\,x^2\,{\left (x^3-x\right )}^{3/4}+28\,x^3\,{\left (x^3-x\right )}^{3/4}}{21\,x^6} \] Input:
int(((x^2 - 3)*(x^3 - x^2 + 1))/(x^6*(x^3 - x)^(1/4)),x)
Output:
-(12*(x^3 - x)^(3/4) - 12*x^2*(x^3 - x)^(3/4) + 28*x^3*(x^3 - x)^(3/4))/(2 1*x^6)
\[ \int \frac {\left (-3+x^2\right ) \left (1-x^2+x^3\right )}{x^6 \sqrt [4]{-x+x^3}} \, dx=-3 \left (\int \frac {1}{x^{\frac {25}{4}} \left (x^{2}-1\right )^{\frac {1}{4}}}d x \right )+4 \left (\int \frac {1}{x^{\frac {17}{4}} \left (x^{2}-1\right )^{\frac {1}{4}}}d x \right )-3 \left (\int \frac {1}{x^{\frac {13}{4}} \left (x^{2}-1\right )^{\frac {1}{4}}}d x \right )-\left (\int \frac {1}{x^{\frac {9}{4}} \left (x^{2}-1\right )^{\frac {1}{4}}}d x \right )+\int \frac {1}{x^{\frac {5}{4}} \left (x^{2}-1\right )^{\frac {1}{4}}}d x \] Input:
int((x^2-3)*(x^3-x^2+1)/x^6/(x^3-x)^(1/4),x)
Output:
- 3*int(1/(x**(1/4)*(x**2 - 1)**(1/4)*x**6),x) + 4*int(1/(x**(1/4)*(x**2 - 1)**(1/4)*x**4),x) - 3*int(1/(x**(1/4)*(x**2 - 1)**(1/4)*x**3),x) - int( 1/(x**(1/4)*(x**2 - 1)**(1/4)*x**2),x) + int(1/(x**(1/4)*(x**2 - 1)**(1/4) *x),x)