\(\int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx\) [1377]

Optimal result

Integrand size = 22, antiderivative size = 103 \[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=-\frac {2 \sqrt [4]{5} \sqrt {-1+3 x-x^2} E\left (\left .\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{\sqrt {1-3 x+x^2}}+\frac {2 \sqrt [4]{5} \sqrt {-1+3 x-x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{\sqrt {1-3 x+x^2}} \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {705, 704, 313, 227, 1195, 21, 435} \[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=\frac {2 \sqrt [4]{5} \sqrt {-x^2+3 x-1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{\sqrt {x^2-3 x+1}}-\frac {2 \sqrt [4]{5} \sqrt {-x^2+3 x-1} E\left (\left .\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{\sqrt {x^2-3 x+1}} \]

[In]

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

[Out]

(-2*5^(1/4)*Sqrt[-1 + 3*x - x^2]*EllipticE[ArcSin[Sqrt[3 - 2*x]/5^(1/4)], -1])/Sqrt[1 - 3*x + x^2] + (2*5^(1/4
)*Sqrt[-1 + 3*x - x^2]*EllipticF[ArcSin[Sqrt[3 - 2*x]/5^(1/4)], -1])/Sqrt[1 - 3*x + x^2]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 313

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Dist[-q^(-1), Int[1/Sqrt[a + b*x^4]
, x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[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
]

Rule 704

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(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] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]

Rule 705

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[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] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && EqQ[m^2, 1/4]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[Sqrt[-c], Int
[(d + e*x^2)/(Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-1+3 x-x^2} \int \frac {\sqrt {3-2 x}}{\sqrt {-\frac {1}{5}+\frac {3 x}{5}-\frac {x^2}{5}}} \, dx}{\sqrt {5} \sqrt {1-3 x+x^2}} \\ & = -\frac {\left (2 \sqrt {-1+3 x-x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^4}{5}}} \, dx,x,\sqrt {3-2 x}\right )}{\sqrt {5} \sqrt {1-3 x+x^2}} \\ & = \frac {\left (2 \sqrt {-1+3 x-x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^4}{5}}} \, dx,x,\sqrt {3-2 x}\right )}{\sqrt {1-3 x+x^2}}-\frac {\left (2 \sqrt {-1+3 x-x^2}\right ) \text {Subst}\left (\int \frac {1+\frac {x^2}{\sqrt {5}}}{\sqrt {1-\frac {x^4}{5}}} \, dx,x,\sqrt {3-2 x}\right )}{\sqrt {1-3 x+x^2}} \\ & = \frac {2 \sqrt [4]{5} \sqrt {-1+3 x-x^2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{\sqrt {1-3 x+x^2}}-\frac {\left (2 \sqrt {-1+3 x-x^2}\right ) \text {Subst}\left (\int \frac {1+\frac {x^2}{\sqrt {5}}}{\sqrt {\frac {1}{\sqrt {5}}-\frac {x^2}{5}} \sqrt {\frac {1}{\sqrt {5}}+\frac {x^2}{5}}} \, dx,x,\sqrt {3-2 x}\right )}{\sqrt {5} \sqrt {1-3 x+x^2}} \\ & = \frac {2 \sqrt [4]{5} \sqrt {-1+3 x-x^2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{\sqrt {1-3 x+x^2}}-\frac {\left (2 \sqrt {-1+3 x-x^2}\right ) \text {Subst}\left (\int \frac {\sqrt {\frac {1}{\sqrt {5}}+\frac {x^2}{5}}}{\sqrt {\frac {1}{\sqrt {5}}-\frac {x^2}{5}}} \, dx,x,\sqrt {3-2 x}\right )}{\sqrt {1-3 x+x^2}} \\ & = -\frac {2 \sqrt [4]{5} \sqrt {-1+3 x-x^2} E\left (\left .\sin ^{-1}\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} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{\sqrt {1-3 x+x^2}} \\ \end{align*}

Mathematica [C] (verified)

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 \[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=-\frac {2 (3-2 x)^{3/2} \sqrt {-1+3 x-x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {1}{5} (3-2 x)^2\right )}{3 \sqrt {5} \sqrt {1-3 x+x^2}} \]

[In]

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

[Out]

(-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])

Maple [A] (verified)

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	\(\frac {\sqrt {-\left (-3+2 x \right ) \left (x^{2}-3 x +1\right )}\, \left (-\frac {6 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, F\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{25 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}+\frac {4 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \left (\frac {\sqrt {5}\, E\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{2}+\frac {3 F\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{2}\right )}{25 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}\right )}{\sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}}\)	\(228\)

[In]

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

[Out]

-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(1/10*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
+11*x-3)

Fricas [C] (verification not implemented)

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 \[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=-2 \, \sqrt {-2} {\rm weierstrassZeta}\left (5, 0, {\rm weierstrassPInverse}\left (5, 0, x - \frac {3}{2}\right )\right ) \]

[In]

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

[Out]

-2*sqrt(-2)*weierstrassZeta(5, 0, weierstrassPInverse(5, 0, x - 3/2))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx=\int \frac {\sqrt {3-2\,x}}{\sqrt {x^2-3\,x+1}} \,d x \]

[In]

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

[Out]

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