\(\int \frac {(3-2 x)^{3/2}}{\sqrt {1-3 x+x^2}} \, dx\) [1373]

Optimal result

Integrand size = 22, antiderivative size = 79 \[ \int \frac {(3-2 x)^{3/2}}{\sqrt {1-3 x+x^2}} \, dx=-\frac {4}{3} \sqrt {3-2 x} \sqrt {1-3 x+x^2}-\frac {2\ 5^{3/4} \sqrt {-1+3 x-x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{3 \sqrt {1-3 x+x^2}} \]

[Out]

-2/3*5^(3/4)*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)-4/3*(3-2*x)^(1/2)*(x^
2-3*x+1)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {706, 705, 703, 227} \[ \int \frac {(3-2 x)^{3/2}}{\sqrt {1-3 x+x^2}} \, dx=-\frac {2\ 5^{3/4} \sqrt {-x^2+3 x-1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{3 \sqrt {x^2-3 x+1}}-\frac {4}{3} \sqrt {3-2 x} \sqrt {x^2-3 x+1} \]

[In]

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

[Out]

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

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 703

Int[1/(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[1/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 706

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*d*(d + e*x)^(m - 1
)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Dist[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rubi steps \begin{align*} \text {integral}& = -\frac {4}{3} \sqrt {3-2 x} \sqrt {1-3 x+x^2}+\frac {5}{3} \int \frac {1}{\sqrt {3-2 x} \sqrt {1-3 x+x^2}} \, dx \\ & = -\frac {4}{3} \sqrt {3-2 x} \sqrt {1-3 x+x^2}+\frac {\left (\sqrt {5} \sqrt {-1+3 x-x^2}\right ) \int \frac {1}{\sqrt {3-2 x} \sqrt {-\frac {1}{5}+\frac {3 x}{5}-\frac {x^2}{5}}} \, dx}{3 \sqrt {1-3 x+x^2}} \\ & = -\frac {4}{3} \sqrt {3-2 x} \sqrt {1-3 x+x^2}-\frac {\left (2 \sqrt {5} \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 )}{3 \sqrt {1-3 x+x^2}} \\ & = -\frac {4}{3} \sqrt {3-2 x} \sqrt {1-3 x+x^2}-\frac {2\ 5^{3/4} \sqrt {-1+3 x-x^2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{3 \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 = 10.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.96 \[ \int \frac {(3-2 x)^{3/2}}{\sqrt {1-3 x+x^2}} \, dx=-\frac {2 \sqrt {3-2 x} \left (2-6 x+2 x^2+\sqrt {5} \sqrt {-1+3 x-x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {1}{5} (3-2 x)^2\right )\right )}{3 \sqrt {1-3 x+x^2}} \]

[In]

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

[Out]

(-2*Sqrt[3 - 2*x]*(2 - 6*x + 2*x^2 + Sqrt[5]*Sqrt[-1 + 3*x - x^2]*Hypergeometric2F1[1/4, 1/2, 5/4, (3 - 2*x)^2
/5]))/(3*Sqrt[1 - 3*x + x^2])

Maple [A] (verified)

Time = 2.43 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.49

method	result	size
default	\(\frac {\sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}\, \left (\sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}\, \sqrt {\left (-3+2 x \right ) \sqrt {5}}\, \sqrt {\left (2 x -3+\sqrt {5}\right ) \sqrt {5}}\, F\left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}}{10}, \sqrt {2}\right )-8 x^{3}+36 x^{2}-44 x +12\right )}{6 x^{3}-27 x^{2}+33 x -9}\)	\(118\)
elliptic	\(\frac {\sqrt {-\left (-3+2 x \right ) \left (x^{2}-3 x +1\right )}\, \left (-\frac {4 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}{3}-\frac {2 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, F\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{15 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}\right )}{\sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}}\)	\(137\)
risch	\(\frac {4 \left (-3+2 x \right ) \sqrt {x^{2}-3 x +1}\, \sqrt {\left (3-2 x \right ) \left (x^{2}-3 x +1\right )}}{3 \sqrt {-\left (-3+2 x \right ) \left (x^{2}-3 x +1\right )}\, \sqrt {3-2 x}}-\frac {2 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, F\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right ) \sqrt {\left (3-2 x \right ) \left (x^{2}-3 x +1\right )}}{15 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}\, \sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}}\)	\(173\)

[In]

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

[Out]

1/3*(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)*EllipticF(1/10*2^(1/2)*5^(1/2)*((-2*x+3+5^(1/2))*5^(1/2))^(1/2),2^(1/2))-8*x^3+36*x^2-44*x+12
)/(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 = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.39 \[ \int \frac {(3-2 x)^{3/2}}{\sqrt {1-3 x+x^2}} \, dx=-\frac {5}{3} \, \sqrt {-2} {\rm weierstrassPInverse}\left (5, 0, x - \frac {3}{2}\right ) - \frac {4}{3} \, \sqrt {x^{2} - 3 \, x + 1} \sqrt {-2 \, x + 3} \]

[In]

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

[Out]

-5/3*sqrt(-2)*weierstrassPInverse(5, 0, x - 3/2) - 4/3*sqrt(x^2 - 3*x + 1)*sqrt(-2*x + 3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(3-2 x)^{3/2}}{\sqrt {1-3 x+x^2}} \, dx=\int { \frac {{\left (-2 \, x + 3\right )}^{\frac {3}{2}}}{\sqrt {x^{2} - 3 \, x + 1}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(3-2 x)^{3/2}}{\sqrt {1-3 x+x^2}} \, dx=\int { \frac {{\left (-2 \, x + 3\right )}^{\frac {3}{2}}}{\sqrt {x^{2} - 3 \, x + 1}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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