Integrand size = 22, antiderivative size = 103 \[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=-\frac {2 \sqrt [4]{5} \sqrt {-1+3 x-x^2} E\left (\left .\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{\sqrt {1-3 x+x^2}}+\frac {2 \sqrt [4]{5} \sqrt {-1+3 x-x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{\sqrt {1-3 x+x^2}} \]
-2*5^(1/4)*EllipticE(1/5*(3-2*x)^(1/2)*5^(3/4),I)*(-x^2+3*x-1)^(1/2)/(x^2- 3*x+1)^(1/2)+2*5^(1/4)*EllipticF(1/5*(3-2*x)^(1/2)*5^(3/4),I)*(-x^2+3*x-1) ^(1/2)/(x^2-3*x+1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=-\frac {2 (3-2 x)^{3/2} \sqrt {-1+3 x-x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {1}{5} (3-2 x)^2\right )}{3 \sqrt {5} \sqrt {1-3 x+x^2}} \]
(-2*(3 - 2*x)^(3/2)*Sqrt[-1 + 3*x - x^2]*Hypergeometric2F1[1/2, 3/4, 7/4, (3 - 2*x)^2/5])/(3*Sqrt[5]*Sqrt[1 - 3*x + x^2])
Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1115, 27, 1114, 27, 836, 27, 762, 1388, 327}
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.
\(\displaystyle \int \frac {\sqrt {3-2 x}}{\sqrt {x^2-3 x+1}} \, dx\) |
\(\Big \downarrow \) 1115 |
\(\displaystyle \frac {\sqrt {-x^2+3 x-1} \int \frac {\sqrt {5} \sqrt {3-2 x}}{\sqrt {-x^2+3 x-1}}dx}{\sqrt {5} \sqrt {x^2-3 x+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {-x^2+3 x-1} \int \frac {\sqrt {3-2 x}}{\sqrt {-x^2+3 x-1}}dx}{\sqrt {x^2-3 x+1}}\) |
\(\Big \downarrow \) 1114 |
\(\displaystyle -\frac {2 \sqrt {-x^2+3 x-1} \int \frac {\sqrt {5} (3-2 x)}{\sqrt {5-(3-2 x)^2}}d\sqrt {3-2 x}}{\sqrt {5} \sqrt {x^2-3 x+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 \sqrt {-x^2+3 x-1} \int \frac {3-2 x}{\sqrt {5-(3-2 x)^2}}d\sqrt {3-2 x}}{\sqrt {x^2-3 x+1}}\) |
\(\Big \downarrow \) 836 |
\(\displaystyle -\frac {2 \sqrt {-x^2+3 x-1} \left (\sqrt {5} \int \frac {-2 x+\sqrt {5}+3}{\sqrt {5} \sqrt {5-(3-2 x)^2}}d\sqrt {3-2 x}-\sqrt {5} \int \frac {1}{\sqrt {5-(3-2 x)^2}}d\sqrt {3-2 x}\right )}{\sqrt {x^2-3 x+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 \sqrt {-x^2+3 x-1} \left (\int \frac {-2 x+\sqrt {5}+3}{\sqrt {5-(3-2 x)^2}}d\sqrt {3-2 x}-\sqrt {5} \int \frac {1}{\sqrt {5-(3-2 x)^2}}d\sqrt {3-2 x}\right )}{\sqrt {x^2-3 x+1}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle -\frac {2 \sqrt {-x^2+3 x-1} \left (\int \frac {-2 x+\sqrt {5}+3}{\sqrt {5-(3-2 x)^2}}d\sqrt {3-2 x}-\sqrt [4]{5} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )\right )}{\sqrt {x^2-3 x+1}}\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle -\frac {2 \sqrt {-x^2+3 x-1} \left (\int \frac {\sqrt {-2 x+\sqrt {5}+3}}{\sqrt {2 x+\sqrt {5}-3}}d\sqrt {3-2 x}-\sqrt [4]{5} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )\right )}{\sqrt {x^2-3 x+1}}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle -\frac {2 \sqrt {-x^2+3 x-1} \left (\sqrt [4]{5} E\left (\left .\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )-\sqrt [4]{5} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )\right )}{\sqrt {x^2-3 x+1}}\) |
(-2*Sqrt[-1 + 3*x - x^2]*(5^(1/4)*EllipticE[ArcSin[Sqrt[3 - 2*x]/5^(1/4)], -1] - 5^(1/4)*EllipticF[ArcSin[Sqrt[3 - 2*x]/5^(1/4)], -1]))/Sqrt[1 - 3*x + x^2]
3.14.77.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) )], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Simp[-q^(-1) Int[1/Sqrt[a + b*x^4], x], x] + Simp[1/q Int[(1 + q*x^2)/S qrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]
Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symb ol] :> Simp[(4/e)*Sqrt[-c/(b^2 - 4*a*c)] Subst[Int[x^2/Sqrt[Simp[1 - b^2* (x^4/(d^2*(b^2 - 4*a*c))), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c , d, e}, x] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]
Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sym bol] :> Simp[Sqrt[(-c)*((a + b*x + c*x^2)/(b^2 - 4*a*c))]/Sqrt[a + b*x + c* x^2] Int[(d + e*x)^m/Sqrt[(-a)*(c/(b^2 - 4*a*c)) - b*c*(x/(b^2 - 4*a*c)) - c^2*(x^2/(b^2 - 4*a*c))], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c* d - b*e, 0] && EqQ[m^2, 1/4]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Time = 2.45 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02
method | result | size |
default | \(-\frac {\sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}\, \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}\, \sqrt {\left (-3+2 x \right ) \sqrt {5}}\, \sqrt {\left (2 x -3+\sqrt {5}\right ) \sqrt {5}}\, \sqrt {5}\, E\left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}}{10}, \sqrt {2}\right )}{5 \left (2 x^{3}-9 x^{2}+11 x -3\right )}\) | \(105\) |
elliptic | \(\frac {\sqrt {-\left (-3+2 x \right ) \left (x^{2}-3 x +1\right )}\, \left (-\frac {6 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, F\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{25 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}+\frac {4 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \left (\frac {\sqrt {5}\, E\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{2}+\frac {3 F\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{2}\right )}{25 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}\right )}{\sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}}\) | \(228\) |
-1/5*(3-2*x)^(1/2)*(x^2-3*x+1)^(1/2)*((-2*x+3+5^(1/2))*5^(1/2))^(1/2)*((-3 +2*x)*5^(1/2))^(1/2)*((2*x-3+5^(1/2))*5^(1/2))^(1/2)*5^(1/2)*EllipticE(1/1 0*2^(1/2)*5^(1/2)*((-2*x+3+5^(1/2))*5^(1/2))^(1/2),2^(1/2))/(2*x^3-9*x^2+1 1*x-3)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.14 \[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=-2 \, \sqrt {-2} {\rm weierstrassZeta}\left (5, 0, {\rm weierstrassPInverse}\left (5, 0, x - \frac {3}{2}\right )\right ) \]
\[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=\int \frac {\sqrt {3 - 2 x}}{\sqrt {x^{2} - 3 x + 1}}\, dx \]
\[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=\int { \frac {\sqrt {-2 \, x + 3}}{\sqrt {x^{2} - 3 \, x + 1}} \,d x } \]
\[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=\int { \frac {\sqrt {-2 \, x + 3}}{\sqrt {x^{2} - 3 \, x + 1}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=\int \frac {\sqrt {3-2\,x}}{\sqrt {x^2-3\,x+1}} \,d x \]