Integrand size = 32, antiderivative size = 40 \[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {3} \sqrt {-x-x^2+x^3}}{-1-x+x^2}\right )}{\sqrt {3}} \]
Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.55 \[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=-\frac {2 \sqrt {x} \sqrt {-1-x+x^2} \arctan \left (\frac {\sqrt {3} \sqrt {x}}{\sqrt {-1-x+x^2}}\right )}{\sqrt {3} \sqrt {x \left (-1-x+x^2\right )}} \]
(-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)])
Time = 0.45 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.60, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2467, 25, 2035, 2537, 216}
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.
\(\displaystyle \int \frac {x^2+1}{\left (x^2+2 x-1\right ) \sqrt {x^3-x^2-x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt {x^2-x-1} \int -\frac {x^2+1}{\sqrt {x} \left (-x^2-2 x+1\right ) \sqrt {x^2-x-1}}dx}{\sqrt {x^3-x^2-x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt {x^2-x-1} \int \frac {x^2+1}{\sqrt {x} \left (-x^2-2 x+1\right ) \sqrt {x^2-x-1}}dx}{\sqrt {x^3-x^2-x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^2-x-1} \int \frac {x^2+1}{\left (-x^2-2 x+1\right ) \sqrt {x^2-x-1}}d\sqrt {x}}{\sqrt {x^3-x^2-x}}\) |
\(\Big \downarrow \) 2537 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^2-x-1} \int \frac {1}{3 x+1}d\frac {\sqrt {x}}{\sqrt {x^2-x-1}}}{\sqrt {x^3-x^2-x}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^2-x-1} \arctan \left (\frac {\sqrt {3} \sqrt {x}}{\sqrt {x^2-x-1}}\right )}{\sqrt {3} \sqrt {x^3-x^2-x}}\) |
(-2*Sqrt[x]*Sqrt[-1 - x + x^2]*ArcTan[(Sqrt[3]*Sqrt[x])/Sqrt[-1 - x + x^2] ])/(Sqrt[3]*Sqrt[-x - x^2 + x^3])
3.6.13.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
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]
Int[((u_)*((A_) + (B_.)*(x_)^4))/Sqrt[v_], x_Symbol] :> With[{a = Coeff[v, x, 0], b = Coeff[v, x, 2], c = Coeff[v, x, 4], d = Coeff[1/u, x, 0], e = Co eff[1/u, x, 2], f = Coeff[1/u, x, 4]}, Simp[A Subst[Int[1/(d - (b*d - a*e )*x^2), x], x, x/Sqrt[v]], x] /; EqQ[a*B + A*c, 0] && EqQ[c*d - a*f, 0]] /; FreeQ[{A, B}, x] && PolyQ[v, x^2, 2] && PolyQ[1/u, x^2, 2]
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 | \(\text {Expression too large to display}\) | \(1531\) |
Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x^{2} - 4 \, x - 1\right )}}{6 \, \sqrt {x^{3} - x^{2} - x}}\right ) \]
\[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=\int \frac {x^{2} + 1}{\sqrt {x \left (x^{2} - x - 1\right )} \left (x^{2} + 2 x - 1\right )}\, dx \]
\[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=\int { \frac {x^{2} + 1}{\sqrt {x^{3} - x^{2} - x} {\left (x^{2} + 2 \, x - 1\right )}} \,d x } \]
\[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=\int { \frac {x^{2} + 1}{\sqrt {x^{3} - x^{2} - x} {\left (x^{2} + 2 \, x - 1\right )}} \,d x } \]
Time = 0.17 (sec) , antiderivative size = 223, normalized size of antiderivative = 5.58 \[ \int \frac {1+x^2}{\left (-1+2 x+x^2\right ) \sqrt {-x-x^2+x^3}} \, dx=-\frac {\sqrt {\frac {x}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\,\sqrt {\frac {x+\frac {\sqrt {5}}{2}-\frac {1}{2}}{\frac {\sqrt {5}}{2}-\frac {1}{2}}}\,\left (\sqrt {5}+1\right )\,\sqrt {\frac {\frac {\sqrt {5}}{2}-x+\frac {1}{2}}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\,\left (\Pi \left (\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\sqrt {2}-1};\mathrm {asin}\left (\sqrt {\frac {x}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\right )\middle |-\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\frac {\sqrt {5}}{2}-\frac {1}{2}}\right )-\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\right )\middle |-\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\frac {\sqrt {5}}{2}-\frac {1}{2}}\right )+\Pi \left (-\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\sqrt {2}+1};\mathrm {asin}\left (\sqrt {\frac {x}{\frac {\sqrt {5}}{2}+\frac {1}{2}}}\right )\middle |-\frac {\frac {\sqrt {5}}{2}+\frac {1}{2}}{\frac {\sqrt {5}}{2}-\frac {1}{2}}\right )\right )}{\sqrt {x^3-x^2-\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right )\,\left (\frac {\sqrt {5}}{2}+\frac {1}{2}\right )\,x}} \]
-((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)*(ellipti cPi((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)