3.1377 \(\int \frac{\sqrt{3-2 x}}{\sqrt{1-3 x+x^2}} \, dx\)

Optimal. Leaf size=103 \[ \frac{2 \sqrt [4]{5} \sqrt{-x^2+3 x-1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{\sqrt{x^2-3 x+1}}-\frac{2 \sqrt [4]{5} \sqrt{-x^2+3 x-1} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{\sqrt{x^2-3 x+1}} \]

[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]

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Rubi [A]  time = 0.0523656, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {691, 690, 307, 221, 1181, 21, 424} \[ \frac{2 \sqrt [4]{5} \sqrt{-x^2+3 x-1} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{\sqrt{x^2-3 x+1}}-\frac{2 \sqrt [4]{5} \sqrt{-x^2+3 x-1} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{\sqrt{x^2-3 x+1}} \]

Antiderivative was successfully verified.

[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 691

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 690

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(4*Sqrt[-(c/(b^2 - 4*a*
c))])/e, 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 307

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 221

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 1181

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]

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 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{3-2 x}}{\sqrt{1-3 x+x^2}} \, dx &=\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.0133973, size = 65, normalized size = 0.63 \[ -\frac{2 (3-2 x)^{3/2} \sqrt{-x^2+3 x-1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};\frac{1}{5} (3-2 x)^2\right )}{3 \sqrt{5} \sqrt{x^2-3 x+1}} \]

Antiderivative was successfully verified.

[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])

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Maple [A]  time = 0.1, size = 105, normalized size = 1. \begin{align*} -{\frac{\sqrt{5}}{10\,{x}^{3}-45\,{x}^{2}+55\,x-15}\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}}{\it EllipticE} \left ({\frac{\sqrt{2}\sqrt{5}}{10}\sqrt{ \left ( -2\,x+3+\sqrt{5} \right ) \sqrt{5}}},\sqrt{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-2 \, x + 3}}{\sqrt{x^{2} - 3 \, x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-2 \, x + 3}}{\sqrt{x^{2} - 3 \, x + 1}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.5848, size = 41, normalized size = 0.4 \begin{align*} \frac{\sqrt{5} i \left (3 - 2 x\right )^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{\left (3 - 2 x\right )^{2}}{5}} \right )}}{10 \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(5)*I*(3 - 2*x)**(3/2)*gamma(3/4)*hyper((1/2, 3/4), (7/4,), (3 - 2*x)**2/5)/(10*gamma(7/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-2 \, x + 3}}{\sqrt{x^{2} - 3 \, x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)