\(\int \frac {1+x^2}{(-1+2 x+x^2) \sqrt {-x-x^2+x^3}} \, dx\) [513]

   Optimal result
   Rubi [C] (warning: unable to verify)
   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 = 32, antiderivative size = 40 \[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {3} \sqrt {-x-x^2+x^3}}{-1-x+x^2}\right )}{\sqrt {3}} \]

[Out]

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

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.92 (sec) , antiderivative size = 387, normalized size of antiderivative = 9.68, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2081, 6860, 730, 1112, 948, 174, 552, 551} \[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=\frac {\sqrt {x} \sqrt {-\left (\left (1-\sqrt {5}\right ) x\right )-2} \sqrt {\frac {\left (1+\sqrt {5}\right ) x+2}{\left (1-\sqrt {5}\right ) x+2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} \sqrt {x}}{\sqrt {-\left (\left (1-\sqrt {5}\right ) x\right )-2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{\left (1-\sqrt {5}\right ) x+2}} \sqrt {x^3-x^2-x}}-\frac {\sqrt {3+\sqrt {5}} \sqrt {x} \sqrt {2 x+\sqrt {5}-1} \sqrt {1-\frac {2 x}{1+\sqrt {5}}} \operatorname {EllipticPi}\left (\frac {1}{2} \left (1-\sqrt {2}\right ) \left (1+\sqrt {5}\right ),\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\sqrt {x^3-x^2-x}}-\frac {\sqrt {3+\sqrt {5}} \sqrt {x} \sqrt {2 x+\sqrt {5}-1} \sqrt {1-\frac {2 x}{1+\sqrt {5}}} \operatorname {EllipticPi}\left (\frac {1}{2} \left (1+\sqrt {2}\right ) \left (1+\sqrt {5}\right ),\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\sqrt {x^3-x^2-x}} \]

[In]

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

[Out]

(Sqrt[x]*Sqrt[-2 - (1 - Sqrt[5])*x]*Sqrt[(2 + (1 + Sqrt[5])*x)/(2 + (1 - Sqrt[5])*x)]*EllipticF[ArcSin[(Sqrt[2
]*5^(1/4)*Sqrt[x])/Sqrt[-2 - (1 - Sqrt[5])*x]], (5 - Sqrt[5])/10])/(5^(1/4)*Sqrt[(2 + (1 - Sqrt[5])*x)^(-1)]*S
qrt[-x - x^2 + x^3]) - (Sqrt[3 + Sqrt[5]]*Sqrt[x]*Sqrt[-1 + Sqrt[5] + 2*x]*Sqrt[1 - (2*x)/(1 + Sqrt[5])]*Ellip
ticPi[((1 - Sqrt[2])*(1 + Sqrt[5]))/2, ArcSin[Sqrt[2/(1 + Sqrt[5])]*Sqrt[x]], (-3 - Sqrt[5])/2])/Sqrt[-x - x^2
 + x^3] - (Sqrt[3 + Sqrt[5]]*Sqrt[x]*Sqrt[-1 + Sqrt[5] + 2*x]*Sqrt[1 - (2*x)/(1 + Sqrt[5])]*EllipticPi[((1 + S
qrt[2])*(1 + Sqrt[5]))/2, ArcSin[Sqrt[2/(1 + Sqrt[5])]*Sqrt[x]], (-3 - Sqrt[5])/2])/Sqrt[-x - x^2 + x^3]

Rule 174

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + f*(x^2/d), x]]*Sqrt[Simp[(d
*g - c*h)/d + h*(x^2/d), x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-f/e, -d/c])

Rule 552

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 +
(d/c)*x^2]/Sqrt[c + d*x^2], Int[1/((a + b*x^2)*Sqrt[1 + (d/c)*x^2]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c,
 d, e, f}, x] &&  !GtQ[c, 0]

Rule 730

Int[(x_)^(m_)/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[x^(2*m + 1)/Sqrt[a + b*x^
2 + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[m^2, 1/4]

Rule 948

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[Sqrt[b - q + 2*c*x]*(Sqrt[b + q + 2*c*x]/Sqrt[a + b*x + c*x^2]), Int[1/((d +
 e*x)*Sqrt[f + g*x]*Sqrt[b - q + 2*c*x]*Sqrt[b + q + 2*c*x]), x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1112

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*(Sqrt[(2*a + (b + q)*x^2)/q]/(2*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2*a + (b + q)
*x^2)]))*EllipticF[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q)]], (b + q)/(2*q)], x]] /; FreeQ[{a, b, c}, x] && Gt
Q[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]

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]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt {-1-x+x^2}\right ) \int \frac {1+x^2}{\sqrt {x} \sqrt {-1-x+x^2} \left (-1+2 x+x^2\right )} \, dx}{\sqrt {-x-x^2+x^3}} \\ & = \frac {\left (\sqrt {x} \sqrt {-1-x+x^2}\right ) \int \left (\frac {1}{\sqrt {x} \sqrt {-1-x+x^2}}+\frac {2 (1-x)}{\sqrt {x} \sqrt {-1-x+x^2} \left (-1+2 x+x^2\right )}\right ) \, dx}{\sqrt {-x-x^2+x^3}} \\ & = \frac {\left (\sqrt {x} \sqrt {-1-x+x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {-1-x+x^2}} \, dx}{\sqrt {-x-x^2+x^3}}+\frac {\left (2 \sqrt {x} \sqrt {-1-x+x^2}\right ) \int \frac {1-x}{\sqrt {x} \sqrt {-1-x+x^2} \left (-1+2 x+x^2\right )} \, dx}{\sqrt {-x-x^2+x^3}} \\ & = \frac {\left (2 \sqrt {x} \sqrt {-1-x+x^2}\right ) \int \left (\frac {-1+\sqrt {2}}{\sqrt {x} \left (2-2 \sqrt {2}+2 x\right ) \sqrt {-1-x+x^2}}+\frac {-1-\sqrt {2}}{\sqrt {x} \left (2+2 \sqrt {2}+2 x\right ) \sqrt {-1-x+x^2}}\right ) \, dx}{\sqrt {-x-x^2+x^3}}+\frac {\left (2 \sqrt {x} \sqrt {-1-x+x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-1-x^2+x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x-x^2+x^3}} \\ & = \frac {\sqrt {x} \sqrt {-2-\left (1-\sqrt {5}\right ) x} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x}{2+\left (1-\sqrt {5}\right ) x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} \sqrt {x}}{\sqrt {-2-\left (1-\sqrt {5}\right ) x}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x}} \sqrt {-x-x^2+x^3}}+\frac {\left (2 \left (-1-\sqrt {2}\right ) \sqrt {x} \sqrt {-1-x+x^2}\right ) \int \frac {1}{\sqrt {x} \left (2+2 \sqrt {2}+2 x\right ) \sqrt {-1-x+x^2}} \, dx}{\sqrt {-x-x^2+x^3}}+\frac {\left (2 \left (-1+\sqrt {2}\right ) \sqrt {x} \sqrt {-1-x+x^2}\right ) \int \frac {1}{\sqrt {x} \left (2-2 \sqrt {2}+2 x\right ) \sqrt {-1-x+x^2}} \, dx}{\sqrt {-x-x^2+x^3}} \\ & = \frac {\sqrt {x} \sqrt {-2-\left (1-\sqrt {5}\right ) x} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x}{2+\left (1-\sqrt {5}\right ) x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} \sqrt {x}}{\sqrt {-2-\left (1-\sqrt {5}\right ) x}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x}} \sqrt {-x-x^2+x^3}}+\frac {\left (2 \left (-1-\sqrt {2}\right ) \sqrt {x} \sqrt {-1-\sqrt {5}+2 x} \sqrt {-1+\sqrt {5}+2 x}\right ) \int \frac {1}{\sqrt {x} \left (2+2 \sqrt {2}+2 x\right ) \sqrt {-1-\sqrt {5}+2 x} \sqrt {-1+\sqrt {5}+2 x}} \, dx}{\sqrt {-x-x^2+x^3}}+\frac {\left (2 \left (-1+\sqrt {2}\right ) \sqrt {x} \sqrt {-1-\sqrt {5}+2 x} \sqrt {-1+\sqrt {5}+2 x}\right ) \int \frac {1}{\sqrt {x} \left (2-2 \sqrt {2}+2 x\right ) \sqrt {-1-\sqrt {5}+2 x} \sqrt {-1+\sqrt {5}+2 x}} \, dx}{\sqrt {-x-x^2+x^3}} \\ & = \frac {\sqrt {x} \sqrt {-2-\left (1-\sqrt {5}\right ) x} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x}{2+\left (1-\sqrt {5}\right ) x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} \sqrt {x}}{\sqrt {-2-\left (1-\sqrt {5}\right ) x}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x}} \sqrt {-x-x^2+x^3}}-\frac {\left (4 \left (-1-\sqrt {2}\right ) \sqrt {x} \sqrt {-1-\sqrt {5}+2 x} \sqrt {-1+\sqrt {5}+2 x}\right ) \text {Subst}\left (\int \frac {1}{\left (-2 \left (1+\sqrt {2}\right )-2 x^2\right ) \sqrt {-1-\sqrt {5}+2 x^2} \sqrt {-1+\sqrt {5}+2 x^2}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x-x^2+x^3}}-\frac {\left (4 \left (-1+\sqrt {2}\right ) \sqrt {x} \sqrt {-1-\sqrt {5}+2 x} \sqrt {-1+\sqrt {5}+2 x}\right ) \text {Subst}\left (\int \frac {1}{\left (-2 \left (1-\sqrt {2}\right )-2 x^2\right ) \sqrt {-1-\sqrt {5}+2 x^2} \sqrt {-1+\sqrt {5}+2 x^2}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x-x^2+x^3}} \\ & = \frac {\sqrt {x} \sqrt {-2-\left (1-\sqrt {5}\right ) x} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x}{2+\left (1-\sqrt {5}\right ) x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} \sqrt {x}}{\sqrt {-2-\left (1-\sqrt {5}\right ) x}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x}} \sqrt {-x-x^2+x^3}}-\frac {\left (4 \left (-1-\sqrt {2}\right ) \sqrt {x} \sqrt {-1+\sqrt {5}+2 x} \sqrt {1-\frac {2 x}{1+\sqrt {5}}}\right ) \text {Subst}\left (\int \frac {1}{\left (-2 \left (1+\sqrt {2}\right )-2 x^2\right ) \sqrt {-1+\sqrt {5}+2 x^2} \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x-x^2+x^3}}-\frac {\left (4 \left (-1+\sqrt {2}\right ) \sqrt {x} \sqrt {-1+\sqrt {5}+2 x} \sqrt {1-\frac {2 x}{1+\sqrt {5}}}\right ) \text {Subst}\left (\int \frac {1}{\left (-2 \left (1-\sqrt {2}\right )-2 x^2\right ) \sqrt {-1+\sqrt {5}+2 x^2} \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}} \, dx,x,\sqrt {x}\right )}{\sqrt {-x-x^2+x^3}} \\ & = \frac {\sqrt {x} \sqrt {-2-\left (1-\sqrt {5}\right ) x} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x}{2+\left (1-\sqrt {5}\right ) x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} \sqrt {x}}{\sqrt {-2-\left (1-\sqrt {5}\right ) x}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x}} \sqrt {-x-x^2+x^3}}-\frac {\sqrt {3+\sqrt {5}} \sqrt {x} \sqrt {-1+\sqrt {5}+2 x} \sqrt {1-\frac {2 x}{1+\sqrt {5}}} \operatorname {EllipticPi}\left (\frac {1}{2} \left (1-\sqrt {2}\right ) \left (1+\sqrt {5}\right ),\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\sqrt {-x-x^2+x^3}}-\frac {\sqrt {3+\sqrt {5}} \sqrt {x} \sqrt {-1+\sqrt {5}+2 x} \sqrt {1-\frac {2 x}{1+\sqrt {5}}} \operatorname {EllipticPi}\left (\frac {1}{2} \left (1+\sqrt {2}\right ) \left (1+\sqrt {5}\right ),\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\sqrt {-x-x^2+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.55 \[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=-\frac {2 \sqrt {x} \sqrt {-1-x+x^2} \arctan \left (\frac {\sqrt {3} \sqrt {x}}{\sqrt {-1-x+x^2}}\right )}{\sqrt {3} \sqrt {x \left (-1-x+x^2\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.68

method result size
default \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {x \left (x^{2}-x -1\right )}\, \sqrt {3}}{3 x}\right )}{3}\) \(27\)
pseudoelliptic \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {x \left (x^{2}-x -1\right )}\, \sqrt {3}}{3 x}\right )}{3}\) \(27\)
trager \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x +6 \sqrt {x^{3}-x^{2}-x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right )}{x^{2}+2 x -1}\right )}{3}\) \(64\)
elliptic \(\text {Expression too large to display}\) \(1531\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x^{2} - 4 \, x - 1\right )}}{6 \, \sqrt {x^{3} - x^{2} - x}}\right ) \]

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/6*sqrt(3)*(x^2 - 4*x - 1)/sqrt(x^3 - x^2 - x))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 223, normalized size of antiderivative = 5.58 \[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=-\frac {\sqrt {\frac {x}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\,\sqrt {\frac {x+\frac {\sqrt {5}}{2}-\frac {1}{2}}{\frac {\sqrt {5}}{2}-\frac {1}{2}}}\,\left (\sqrt {5}+1\right )\,\sqrt {\frac {\frac {\sqrt {5}}{2}-x+\frac {1}{2}}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\,\left (\Pi \left (\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\sqrt {2}-1};\mathrm {asin}\left (\sqrt {\frac {x}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\right )\middle |-\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\frac {\sqrt {5}}{2}-\frac {1}{2}}\right )-\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\right )\middle |-\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\frac {\sqrt {5}}{2}-\frac {1}{2}}\right )+\Pi \left (-\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\sqrt {2}+1};\mathrm {asin}\left (\sqrt {\frac {x}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\right )\middle |-\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\frac {\sqrt {5}}{2}-\frac {1}{2}}\right )\right )}{\sqrt {x^3-x^2-\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right )\,\left (\frac {\sqrt {5}}{2}+\frac {1}{2}\right )\,x}} \]

[In]

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

[Out]

-((x/(5^(1/2)/2 + 1/2))^(1/2)*((x + 5^(1/2)/2 - 1/2)/(5^(1/2)/2 - 1/2))^(1/2)*(5^(1/2) + 1)*((5^(1/2)/2 - x +
1/2)/(5^(1/2)/2 + 1/2))^(1/2)*(ellipticPi((5^(1/2)/2 + 1/2)/(2^(1/2) - 1), asin((x/(5^(1/2)/2 + 1/2))^(1/2)),
-(5^(1/2)/2 + 1/2)/(5^(1/2)/2 - 1/2)) - ellipticF(asin((x/(5^(1/2)/2 + 1/2))^(1/2)), -(5^(1/2)/2 + 1/2)/(5^(1/
2)/2 - 1/2)) + ellipticPi(-(5^(1/2)/2 + 1/2)/(2^(1/2) + 1), asin((x/(5^(1/2)/2 + 1/2))^(1/2)), -(5^(1/2)/2 + 1
/2)/(5^(1/2)/2 - 1/2))))/(x^3 - x^2 - x*(5^(1/2)/2 - 1/2)*(5^(1/2)/2 + 1/2))^(1/2)