Integrand size = 32, antiderivative size = 28 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {4 (-1+2 x) \left (-x^2+x^3\right )^{3/4}}{(-1+x) x^2} \] Output:
4*(-1+2*x)*(x^3-x^2)^(3/4)/(-1+x)/x^2
Result contains higher order function than in optimal. Order 5 vs. order 2 in optimal.
Time = 10.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.50 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=-\frac {4 \sqrt {x} \left (2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},1-x\right )-\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},1-x\right )-2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},1-x\right )\right )}{\sqrt [4]{(-1+x) x^2}} \] Input:
Integrate[(-2 - x + 2*x^2)/((-1 + x)*x*(-x^2 + x^3)^(1/4)),x]
Output:
(-4*Sqrt[x]*(2*Hypergeometric2F1[-1/2, -1/4, 3/4, 1 - x] - Hypergeometric2 F1[-1/4, 1/2, 3/4, 1 - x] - 2*Hypergeometric2F1[-1/4, 3/2, 3/4, 1 - x]))/( (-1 + x)*x^2)^(1/4)
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2467, 25, 1193, 83}
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.
\(\displaystyle \int \frac {2 x^2-x-2}{(x-1) x \sqrt [4]{x^3-x^2}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x-1} \sqrt {x} \int -\frac {-2 x^2+x+2}{(x-1)^{5/4} x^{3/2}}dx}{\sqrt [4]{x^3-x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x-1} \sqrt {x} \int \frac {-2 x^2+x+2}{(x-1)^{5/4} x^{3/2}}dx}{\sqrt [4]{x^3-x^2}}\) |
\(\Big \downarrow \) 1193 |
\(\displaystyle -\frac {\sqrt [4]{x-1} \sqrt {x} \left (2 \int \frac {2-x}{(x-1)^{5/4} \sqrt {x}}dx+\frac {4}{\sqrt [4]{x-1} \sqrt {x}}\right )}{\sqrt [4]{x^3-x^2}}\) |
\(\Big \downarrow \) 83 |
\(\displaystyle -\frac {\left (\frac {4}{\sqrt [4]{x-1} \sqrt {x}}-\frac {8 \sqrt {x}}{\sqrt [4]{x-1}}\right ) \sqrt [4]{x-1} \sqrt {x}}{\sqrt [4]{x^3-x^2}}\) |
Input:
Int[(-2 - x + 2*x^2)/((-1 + x)*x*(-x^2 + x^3)^(1/4)),x]
Output:
-(((4/((-1 + x)^(1/4)*Sqrt[x]) - (8*Sqrt[x])/(-1 + x)^(1/4))*(-1 + x)^(1/4 )*Sqrt[x])/(-x^2 + x^3)^(1/4))
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && EqQ[a*d*f *(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) ), x] + Simp[1/((m + 1)*(e*f - d*g)) Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a , b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] && !IntegerQ[n] && !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
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 = 0.13 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61
method | result | size |
risch | \(\frac {-4+8 x}{\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{4}}}\) | \(17\) |
gosper | \(\frac {-4+8 x}{\left (x^{3}-x^{2}\right )^{\frac {1}{4}}}\) | \(19\) |
orering | \(\frac {-4+8 x}{\left (x^{3}-x^{2}\right )^{\frac {1}{4}}}\) | \(19\) |
trager | \(\frac {4 \left (-1+2 x \right ) \left (x^{3}-x^{2}\right )^{\frac {3}{4}}}{\left (-1+x \right ) x^{2}}\) | \(27\) |
meijerg | \(-\frac {4 \left (-\operatorname {signum}\left (-1+x \right )\right )^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {1}{2}\right ], x\right )}{\operatorname {signum}\left (-1+x \right )^{\frac {1}{4}} \sqrt {x}}-\frac {4 \left (-\operatorname {signum}\left (-1+x \right )\right )^{\frac {1}{4}} x^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [\frac {5}{4}, \frac {3}{2}\right ], \left [\frac {5}{2}\right ], x\right )}{3 \operatorname {signum}\left (-1+x \right )^{\frac {1}{4}}}+\frac {2 \left (-\operatorname {signum}\left (-1+x \right )\right )^{\frac {1}{4}} \sqrt {x}\, \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {3}{2}\right ], x\right )}{\operatorname {signum}\left (-1+x \right )^{\frac {1}{4}}}\) | \(80\) |
Input:
int((2*x^2-x-2)/(-1+x)/x/(x^3-x^2)^(1/4),x,method=_RETURNVERBOSE)
Output:
4*(-1+2*x)/((-1+x)*x^2)^(1/4)
Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {4 \, {\left (2 \, x - 1\right )}}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}}} \] Input:
integrate((2*x^2-x-2)/(-1+x)/x/(x^3-x^2)^(1/4),x, algorithm="fricas")
Output:
4*(2*x - 1)/(x^3 - x^2)^(1/4)
\[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\int \frac {2 x^{2} - x - 2}{x \sqrt [4]{x^{2} \left (x - 1\right )} \left (x - 1\right )}\, dx \] Input:
integrate((2*x**2-x-2)/(-1+x)/x/(x**3-x**2)**(1/4),x)
Output:
Integral((2*x**2 - x - 2)/(x*(x**2*(x - 1))**(1/4)*(x - 1)), x)
\[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} {\left (x - 1\right )} x} \,d x } \] Input:
integrate((2*x^2-x-2)/(-1+x)/x/(x^3-x^2)^(1/4),x, algorithm="maxima")
Output:
integrate((2*x^2 - x - 2)/((x^3 - x^2)^(1/4)*(x - 1)*x), x)
\[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} {\left (x - 1\right )} x} \,d x } \] Input:
integrate((2*x^2-x-2)/(-1+x)/x/(x^3-x^2)^(1/4),x, algorithm="giac")
Output:
integrate((2*x^2 - x - 2)/((x^3 - x^2)^(1/4)*(x - 1)*x), x)
Time = 8.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {8\,x-4}{{\left (x^3-x^2\right )}^{1/4}} \] Input:
int(-(x - 2*x^2 + 2)/(x*(x^3 - x^2)^(1/4)*(x - 1)),x)
Output:
(8*x - 4)/(x^3 - x^2)^(1/4)
Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {4 \sqrt {x}\, \left (-1+2 x \right )}{\left (x -1\right )^{\frac {1}{4}} x} \] Input:
int((2*x^2-x-2)/(-1+x)/x/(x^3-x^2)^(1/4),x)
Output:
(4*sqrt(x)*(2*x - 1))/((x - 1)**(1/4)*x)