\(\int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx\) [312]

Optimal result

Integrand size = 32, antiderivative size = 28 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {4 (-1+2 x) \left (-x^2+x^3\right )^{3/4}}{(-1+x) x^2} \]

[Out]

4*(-1+2*x)*(x^3-x^2)^(3/4)/(-1+x)/x^2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2081, 963, 75} \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {4}{\sqrt [4]{x^3-x^2}}-\frac {8 (1-x)}{\sqrt [4]{x^3-x^2}} \]

[In]

Int[(-2 - x + 2*x^2)/((-1 + x)*x*(-x^2 + x^3)^(1/4)),x]

[Out]

4/(-x^2 + x^3)^(1/4) - (8*(1 - x))/(-x^2 + x^3)^(1/4)

Rule 75

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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]

Rule 963

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] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& IGtQ[p, 0] && LtQ[m, -1]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{-1+x} \sqrt {x}\right ) \int \frac {-2-x+2 x^2}{(-1+x)^{5/4} x^{3/2}} \, dx}{\sqrt [4]{-x^2+x^3}} \\ & = \frac {4}{\sqrt [4]{-x^2+x^3}}-\frac {\left (4 \sqrt [4]{-1+x} \sqrt {x}\right ) \int \frac {-1-\frac {x}{2}}{\sqrt [4]{-1+x} x^{3/2}} \, dx}{\sqrt [4]{-x^2+x^3}} \\ & = \frac {4}{\sqrt [4]{-x^2+x^3}}-\frac {8 (1-x)}{\sqrt [4]{-x^2+x^3}} \\ \end{align*}

Mathematica [C] (verified)

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 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=-\frac {4 \sqrt {x} \left (2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},1-x\right )-\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},1-x\right )-2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},1-x\right )\right )}{\sqrt [4]{(-1+x) x^2}} \]

[In]

Integrate[(-2 - x + 2*x^2)/((-1 + x)*x*(-x^2 + x^3)^(1/4)),x]

[Out]

(-4*Sqrt[x]*(2*Hypergeometric2F1[-1/2, -1/4, 3/4, 1 - x] - Hypergeometric2F1[-1/4, 1/2, 3/4, 1 - x] - 2*Hyperg
eometric2F1[-1/4, 3/2, 3/4, 1 - x]))/((-1 + x)*x^2)^(1/4)

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {-4+8 x}{\left (\left (x -1\right ) x^{2}\right )^{\frac {1}{4}}}\)	\(17\)
gosper	\(\frac {-4+8 x}{\left (x^{3}-x^{2}\right )^{\frac {1}{4}}}\)	\(19\)
trager	\(\frac {4 \left (-1+2 x \right ) \left (x^{3}-x^{2}\right )^{\frac {3}{4}}}{\left (x -1\right ) x^{2}}\)	\(27\)
meijerg	\(-\frac {4 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {1}{2}\right ], x\right )}{\operatorname {signum}\left (x -1\right )^{\frac {1}{4}} \sqrt {x}}-\frac {4 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} x^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [\frac {5}{4}, \frac {3}{2}\right ], \left [\frac {5}{2}\right ], x\right )}{3 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}}}+\frac {2 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} \sqrt {x}\, \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {3}{2}\right ], x\right )}{\operatorname {signum}\left (x -1\right )^{\frac {1}{4}}}\)	\(80\)

[In]

int((2*x^2-x-2)/(x-1)/x/(x^3-x^2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4*(-1+2*x)/((x-1)*x^2)^(1/4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {4 \, {\left (2 \, x - 1\right )}}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}}} \]

[In]

integrate((2*x^2-x-2)/(-1+x)/x/(x^3-x^2)^(1/4),x, algorithm="fricas")

[Out]

4*(2*x - 1)/(x^3 - x^2)^(1/4)

Sympy [F]

\[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\int \frac {2 x^{2} - x - 2}{x \sqrt [4]{x^{2} \left (x - 1\right )} \left (x - 1\right )}\, dx \]

[In]

integrate((2*x**2-x-2)/(-1+x)/x/(x**3-x**2)**(1/4),x)

[Out]

Integral((2*x**2 - x - 2)/(x*(x**2*(x - 1))**(1/4)*(x - 1)), x)

Maxima [F]

\[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} {\left (x - 1\right )} x} \,d x } \]

[In]

integrate((2*x^2-x-2)/(-1+x)/x/(x^3-x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate((2*x^2 - x - 2)/((x^3 - x^2)^(1/4)*(x - 1)*x), x)

Giac [F]

\[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} {\left (x - 1\right )} x} \,d x } \]

[In]

integrate((2*x^2-x-2)/(-1+x)/x/(x^3-x^2)^(1/4),x, algorithm="giac")

[Out]

integrate((2*x^2 - x - 2)/((x^3 - x^2)^(1/4)*(x - 1)*x), x)

Mupad [B] (verification not implemented)

Time = 5.64 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-2-x+2 x^2}{(-1+x) x \sqrt [4]{-x^2+x^3}} \, dx=\frac {8\,x-4}{{\left (x^3-x^2\right )}^{1/4}} \]

[In]

int(-(x - 2*x^2 + 2)/(x*(x^3 - x^2)^(1/4)*(x - 1)),x)

[Out]

(8*x - 4)/(x^3 - x^2)^(1/4)