Integrand size = 16, antiderivative size = 107 \[ \int \sqrt {x^2+x^3-x^4} \, dx=-\frac {(1-2 x) \sqrt {x^2+x^3-x^4}}{8 x}-\frac {\left (1+x-x^2\right ) \sqrt {x^2+x^3-x^4}}{3 x}-\frac {5 \sqrt {x^2+x^3-x^4} \arcsin \left (\frac {1-2 x}{\sqrt {5}}\right )}{16 x \sqrt {1+x-x^2}} \]
-1/8*(1-2*x)*(-x^4+x^3+x^2)^(1/2)/x-1/3*(-x^2+x+1)*(-x^4+x^3+x^2)^(1/2)/x- 5/16*arcsin(1/5*(1-2*x)*5^(1/2))*(-x^4+x^3+x^2)^(1/2)/x/(-x^2+x+1)^(1/2)
Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.77 \[ \int \sqrt {x^2+x^3-x^4} \, dx=\frac {\sqrt {x^2+x^3-x^4} \left (2 \sqrt {-1-x+x^2} \left (-11-2 x+8 x^2\right )+15 \log \left (1-2 x+2 \sqrt {-1-x+x^2}\right )\right )}{48 x \sqrt {-1-x+x^2}} \]
(Sqrt[x^2 + x^3 - x^4]*(2*Sqrt[-1 - x + x^2]*(-11 - 2*x + 8*x^2) + 15*Log[ 1 - 2*x + 2*Sqrt[-1 - x + x^2]]))/(48*x*Sqrt[-1 - x + x^2])
Time = 0.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1950, 1160, 1087, 1090, 223}
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.
\(\displaystyle \int \sqrt {-x^4+x^3+x^2} \, dx\) |
\(\Big \downarrow \) 1950 |
\(\displaystyle \frac {\sqrt {-x^4+x^3+x^2} \int x \sqrt {-x^2+x+1}dx}{x \sqrt {-x^2+x+1}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {\sqrt {-x^4+x^3+x^2} \left (\frac {1}{2} \int \sqrt {-x^2+x+1}dx-\frac {1}{3} \left (-x^2+x+1\right )^{3/2}\right )}{x \sqrt {-x^2+x+1}}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {\sqrt {-x^4+x^3+x^2} \left (\frac {1}{2} \left (\frac {5}{8} \int \frac {1}{\sqrt {-x^2+x+1}}dx-\frac {1}{4} (1-2 x) \sqrt {-x^2+x+1}\right )-\frac {1}{3} \left (-x^2+x+1\right )^{3/2}\right )}{x \sqrt {-x^2+x+1}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {\sqrt {-x^4+x^3+x^2} \left (\frac {1}{2} \left (-\frac {1}{8} \sqrt {5} \int \frac {1}{\sqrt {1-\frac {1}{5} (1-2 x)^2}}d(1-2 x)-\frac {1}{4} \sqrt {-x^2+x+1} (1-2 x)\right )-\frac {1}{3} \left (-x^2+x+1\right )^{3/2}\right )}{x \sqrt {-x^2+x+1}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\sqrt {-x^4+x^3+x^2} \left (\frac {1}{2} \left (-\frac {5}{8} \arcsin \left (\frac {1-2 x}{\sqrt {5}}\right )-\frac {1}{4} \sqrt {-x^2+x+1} (1-2 x)\right )-\frac {1}{3} \left (-x^2+x+1\right )^{3/2}\right )}{x \sqrt {-x^2+x+1}}\) |
(Sqrt[x^2 + x^3 - x^4]*(-1/3*(1 + x - x^2)^(3/2) + (-1/4*((1 - 2*x)*Sqrt[1 + x - x^2]) - (5*ArcSin[(1 - 2*x)/Sqrt[5]])/8)/2))/(x*Sqrt[1 + x - x^2])
3.10.34.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Simp[Sqrt[a*x^q + b*x^n + c*x^(2*n - q)]/(x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]) Int[x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q]
Time = 0.34 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.44
method | result | size |
pseudoelliptic | \(\frac {\left (16 x^{2}-4 x -22\right ) \sqrt {-x^{2} \left (x^{2}-x -1\right )}+15 \arcsin \left (\frac {\sqrt {5}\, \left (2 x -1\right )}{5}\right ) x}{48 x}\) | \(47\) |
trager | \(\frac {\left (8 x^{2}-2 x -11\right ) \sqrt {-x^{4}+x^{3}+x^{2}}}{24 x}+\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {-x^{4}+x^{3}+x^{2}}}{x}\right )}{16}\) | \(80\) |
default | \(-\frac {\sqrt {-x^{4}+x^{3}+x^{2}}\, \left (16 \left (-x^{2}+x +1\right )^{\frac {3}{2}}-12 x \sqrt {-x^{2}+x +1}+6 \sqrt {-x^{2}+x +1}-15 \arcsin \left (\frac {\sqrt {5}\, \left (2 x -1\right )}{5}\right )\right )}{48 x \sqrt {-x^{2}+x +1}}\) | \(81\) |
risch | \(\frac {\left (8 x^{2}-2 x -11\right ) \sqrt {-x^{2} \left (x^{2}-x -1\right )}}{24 x}-\frac {5 \arcsin \left (\frac {2 \sqrt {5}\, \left (x -\frac {1}{2}\right )}{5}\right ) \sqrt {-x^{2} \left (x^{2}-x -1\right )}\, \sqrt {-x^{2}+x +1}}{16 x \left (x^{2}-x -1\right )}\) | \(81\) |
Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.58 \[ \int \sqrt {x^2+x^3-x^4} \, dx=-\frac {15 \, x \arctan \left (-\frac {x - \sqrt {-x^{4} + x^{3} + x^{2}}}{x^{2}}\right ) - \sqrt {-x^{4} + x^{3} + x^{2}} {\left (8 \, x^{2} - 2 \, x - 11\right )} + 11 \, x}{24 \, x} \]
-1/24*(15*x*arctan(-(x - sqrt(-x^4 + x^3 + x^2))/x^2) - sqrt(-x^4 + x^3 + x^2)*(8*x^2 - 2*x - 11) + 11*x)/x
\[ \int \sqrt {x^2+x^3-x^4} \, dx=\int \sqrt {- x^{4} + x^{3} + x^{2}}\, dx \]
\[ \int \sqrt {x^2+x^3-x^4} \, dx=\int { \sqrt {-x^{4} + x^{3} + x^{2}} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.56 \[ \int \sqrt {x^2+x^3-x^4} \, dx=\frac {1}{48} \, {\left (15 \, \arcsin \left (\frac {1}{5} \, \sqrt {5}\right ) + 22\right )} \mathrm {sgn}\left (x\right ) + \frac {5}{16} \, \arcsin \left (\frac {1}{5} \, \sqrt {5} {\left (2 \, x - 1\right )}\right ) \mathrm {sgn}\left (x\right ) + \frac {1}{24} \, {\left (2 \, {\left (4 \, x \mathrm {sgn}\left (x\right ) - \mathrm {sgn}\left (x\right )\right )} x - 11 \, \mathrm {sgn}\left (x\right )\right )} \sqrt {-x^{2} + x + 1} \]
1/48*(15*arcsin(1/5*sqrt(5)) + 22)*sgn(x) + 5/16*arcsin(1/5*sqrt(5)*(2*x - 1))*sgn(x) + 1/24*(2*(4*x*sgn(x) - sgn(x))*x - 11*sgn(x))*sqrt(-x^2 + x + 1)
Timed out. \[ \int \sqrt {x^2+x^3-x^4} \, dx=\int \sqrt {-x^4+x^3+x^2} \,d x \]