Integrand size = 16, antiderivative size = 96 \[ \int \sqrt {x^2+x^3-x^4} \, dx=-\frac {1}{12} \sqrt {x^2+x^3-x^4}-\frac {11 \sqrt {x^2+x^3-x^4}}{24 x}+\frac {1}{3} x \sqrt {x^2+x^3-x^4}-\frac {5}{16} \arctan \left (\frac {(1-2 x) x}{2 \sqrt {x^2+x^3-x^4}}\right ) \] Output:
-1/12*(-x^4+x^3+x^2)^(1/2)-11/24*(-x^4+x^3+x^2)^(1/2)/x+1/3*x*(-x^4+x^3+x^ 2)^(1/2)-5/16*arctan(1/2*(1-2*x)*x/(-x^4+x^3+x^2)^(1/2))
Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \sqrt {x^2+x^3-x^4} \, dx=\frac {\sqrt {x^2+x^3-x^4} \left (2 \sqrt {-1-x+x^2} \left (-11-2 x+8 x^2\right )+15 \log \left (1-2 x+2 \sqrt {-1-x+x^2}\right )\right )}{48 x \sqrt {-1-x+x^2}} \] Input:
Integrate[Sqrt[x^2 + x^3 - x^4],x]
Output:
(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 = 0.37 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.95, 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 |
\(\displaystyle \frac {\sqrt {-x^4+x^3+x^2} \left (\frac {1}{2} \left (-\frac {5}{8} \arcsin \left (\frac {1-2 x}{\sqrt {5}}\right )-\frac {1}{4} \sqrt {-x^2+x+1} (1-2 x)\right )-\frac {1}{3} \left (-x^2+x+1\right )^{3/2}\right )}{x \sqrt {-x^2+x+1}}\) |
Input:
Int[Sqrt[x^2 + x^3 - x^4],x]
Output:
(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])
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 = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.49
method | result | size |
pseudoelliptic | \(\frac {\left (16 x^{2}-4 x -22\right ) \sqrt {-x^{2} \left (x^{2}-x -1\right )}+15 \arcsin \left (\frac {\left (2 x -1\right ) \sqrt {5}}{5}\right ) x}{48 x}\) | \(47\) |
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 | \(-\frac {\sqrt {-x^{4}+x^{3}+x^{2}}\, \left (16 \left (-x^{2}+x +1\right )^{\frac {3}{2}}-12 x \sqrt {-x^{2}+x +1}+6 \sqrt {-x^{2}+x +1}-15 \arcsin \left (\frac {\left (2 x -1\right ) \sqrt {5}}{5}\right )\right )}{48 x \sqrt {-x^{2}+x +1}}\) | \(81\) |
risch | \(\frac {\left (8 x^{2}-2 x -11\right ) \sqrt {-x^{2} \left (x^{2}-x -1\right )}}{24 x}-\frac {5 \arcsin \left (\frac {2 \sqrt {5}\, \left (x -\frac {1}{2}\right )}{5}\right ) \sqrt {-x^{2} \left (x^{2}-x -1\right )}\, \sqrt {-x^{2}+x +1}}{16 x \left (x^{2}-x -1\right )}\) | \(81\) |
Input:
int((-x^4+x^3+x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/48*((16*x^2-4*x-22)*(-x^2*(x^2-x-1))^(1/2)+15*arcsin(1/5*(2*x-1)*5^(1/2) )*x)/x
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \sqrt {x^2+x^3-x^4} \, dx=-\frac {15 \, x \arctan \left (-\frac {x - \sqrt {-x^{4} + x^{3} + x^{2}}}{x^{2}}\right ) - \sqrt {-x^{4} + x^{3} + x^{2}} {\left (8 \, x^{2} - 2 \, x - 11\right )} + 11 \, x}{24 \, x} \] Input:
integrate((-x^4+x^3+x^2)^(1/2),x, algorithm="fricas")
Output:
-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
\[ \int \sqrt {x^2+x^3-x^4} \, dx=\int \sqrt {- x^{4} + x^{3} + x^{2}}\, dx \] Input:
integrate((-x**4+x**3+x**2)**(1/2),x)
Output:
Integral(sqrt(-x**4 + x**3 + x**2), x)
\[ \int \sqrt {x^2+x^3-x^4} \, dx=\int { \sqrt {-x^{4} + x^{3} + x^{2}} \,d x } \] Input:
integrate((-x^4+x^3+x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(-x^4 + x^3 + x^2), x)
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.62 \[ \int \sqrt {x^2+x^3-x^4} \, dx=\frac {1}{48} \, {\left (15 \, \arcsin \left (\frac {1}{5} \, \sqrt {5}\right ) + 22\right )} \mathrm {sgn}\left (x\right ) + \frac {5}{16} \, \arcsin \left (\frac {1}{5} \, \sqrt {5} {\left (2 \, x - 1\right )}\right ) \mathrm {sgn}\left (x\right ) + \frac {1}{24} \, {\left (2 \, {\left (4 \, x \mathrm {sgn}\left (x\right ) - \mathrm {sgn}\left (x\right )\right )} x - 11 \, \mathrm {sgn}\left (x\right )\right )} \sqrt {-x^{2} + x + 1} \] Input:
integrate((-x^4+x^3+x^2)^(1/2),x, algorithm="giac")
Output:
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. \[ \int \sqrt {x^2+x^3-x^4} \, dx=\int \sqrt {-x^4+x^3+x^2} \,d x \] Input:
int((x^2 + x^3 - x^4)^(1/2),x)
Output:
int((x^2 + x^3 - x^4)^(1/2), x)
Time = 0.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.57 \[ \int \sqrt {x^2+x^3-x^4} \, dx=\frac {5 \mathit {asin} \left (\frac {2 x -1}{\sqrt {5}}\right )}{16}+\frac {\sqrt {-x^{2}+x +1}\, x^{2}}{3}-\frac {\sqrt {-x^{2}+x +1}\, x}{12}-\frac {11 \sqrt {-x^{2}+x +1}}{24}+\frac {5 \sqrt {5}}{24} \] Input:
int((-x^4+x^3+x^2)^(1/2),x)
Output:
(15*asin((2*x - 1)/sqrt(5)) + 16*sqrt( - x**2 + x + 1)*x**2 - 4*sqrt( - x* *2 + x + 1)*x - 22*sqrt( - x**2 + x + 1) + 10*sqrt(5))/48