\(\int \frac {1+x^2}{(-1+x^2) \sqrt {-x+x^3}} \, dx\) [167]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 24, antiderivative size = 20 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {-x+x^3}} \, dx=-\frac {2 \sqrt {-x+x^3}}{-1+x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2081, 460} \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {-x+x^3}} \, dx=-\frac {2 x}{\sqrt {x^3-x}} \]

[In]

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

[Out]

(-2*x)/Sqrt[-x + x^3]

Rule 460

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 1), 0] && NeQ[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 {x} \sqrt {-1+x^2}\right ) \int \frac {1+x^2}{\sqrt {x} \left (-1+x^2\right )^{3/2}} \, dx}{\sqrt {-x+x^3}} \\ & = -\frac {2 x}{\sqrt {-x+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {-x+x^3}} \, dx=-\frac {2 x}{\sqrt {x \left (-1+x^2\right )}} \]

[In]

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

[Out]

(-2*x)/Sqrt[x*(-1 + x^2)]

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65

method result size
gosper \(-\frac {2 x}{\sqrt {x^{3}-x}}\) \(13\)
risch \(-\frac {2 x}{\sqrt {x \left (x^{2}-1\right )}}\) \(13\)
elliptic \(-\frac {2 x}{\sqrt {x \left (x^{2}-1\right )}}\) \(13\)
pseudoelliptic \(-\frac {2 x}{\sqrt {x^{3}-x}}\) \(13\)
trager \(-\frac {2 \sqrt {x^{3}-x}}{x^{2}-1}\) \(19\)
default \(-\frac {x^{2}+x}{\sqrt {\left (x -1\right ) \left (x^{2}+x \right )}}+\frac {x^{2}-x}{\sqrt {\left (1+x \right ) \left (x^{2}-x \right )}}\) \(41\)
meijerg \(-\frac {2 \sqrt {-\operatorname {signum}\left (x^{2}-1\right )}\, \sqrt {x}\, \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {5}{4}\right ], x^{2}\right )}{\sqrt {\operatorname {signum}\left (x^{2}-1\right )}}-\frac {2 \sqrt {-\operatorname {signum}\left (x^{2}-1\right )}\, x^{\frac {5}{2}} \operatorname {hypergeom}\left (\left [\frac {5}{4}, \frac {3}{2}\right ], \left [\frac {9}{4}\right ], x^{2}\right )}{5 \sqrt {\operatorname {signum}\left (x^{2}-1\right )}}\) \(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {-x+x^3}} \, dx=-\frac {2 \, \sqrt {x^{3} - x}}{x^{2} - 1} \]

[In]

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

[Out]

-2*sqrt(x^3 - x)/(x^2 - 1)

Sympy [F]

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

[In]

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

[Out]

Integral((x**2 + 1)/(sqrt(x*(x - 1)*(x + 1))*(x - 1)*(x + 1)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {-x+x^3}} \, dx=-\frac {2\,x}{\sqrt {x^3-x}} \]

[In]

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

[Out]

-(2*x)/(x^3 - x)^(1/2)