\(\int \frac {1}{(3-2 x)^{3/2} \sqrt {1-3 x+x^2}} \, dx\) [1378]

Optimal result

Integrand size = 22, antiderivative size = 128 \[ \int \frac {1}{(3-2 x)^{3/2} \sqrt {1-3 x+x^2}} \, dx=-\frac {4 \sqrt {1-3 x+x^2}}{5 \sqrt {3-2 x}}+\frac {2 \sqrt {-1+3 x-x^2} E\left (\left .\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{5^{3/4} \sqrt {1-3 x+x^2}}-\frac {2 \sqrt {-1+3 x-x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{5^{3/4} \sqrt {1-3 x+x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {707, 705, 704, 313, 227, 1195, 21, 435} \[ \int \frac {1}{(3-2 x)^{3/2} \sqrt {1-3 x+x^2}} \, dx=-\frac {2 \sqrt {-x^2+3 x-1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{5^{3/4} \sqrt {x^2-3 x+1}}+\frac {2 \sqrt {-x^2+3 x-1} E\left (\left .\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{5^{3/4} \sqrt {x^2-3 x+1}}-\frac {4 \sqrt {x^2-3 x+1}}{5 \sqrt {3-2 x}} \]

[In]

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

[Out]

(-4*Sqrt[1 - 3*x + x^2])/(5*Sqrt[3 - 2*x]) + (2*Sqrt[-1 + 3*x - x^2]*EllipticE[ArcSin[Sqrt[3 - 2*x]/5^(1/4)],
-1])/(5^(3/4)*Sqrt[1 - 3*x + x^2]) - (2*Sqrt[-1 + 3*x - x^2]*EllipticF[ArcSin[Sqrt[3 - 2*x]/5^(1/4)], -1])/(5^
(3/4)*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 707

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[-2*b*d*(d + e*x)^(m
+ 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Dist[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 -
4*a*c))), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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 {4 \sqrt {1-3 x+x^2}}{5 \sqrt {3-2 x}}-\frac {1}{5} \int \frac {\sqrt {3-2 x}}{\sqrt {1-3 x+x^2}} \, dx \\ & = -\frac {4 \sqrt {1-3 x+x^2}}{5 \sqrt {3-2 x}}-\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}{5 \sqrt {5} \sqrt {1-3 x+x^2}} \\ & = -\frac {4 \sqrt {1-3 x+x^2}}{5 \sqrt {3-2 x}}+\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 )}{5 \sqrt {5} \sqrt {1-3 x+x^2}} \\ & = -\frac {4 \sqrt {1-3 x+x^2}}{5 \sqrt {3-2 x}}-\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 )}{5 \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 )}{5 \sqrt {1-3 x+x^2}} \\ & = -\frac {4 \sqrt {1-3 x+x^2}}{5 \sqrt {3-2 x}}-\frac {2 \sqrt {-1+3 x-x^2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{5^{3/4} \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 )}{5 \sqrt {5} \sqrt {1-3 x+x^2}} \\ & = -\frac {4 \sqrt {1-3 x+x^2}}{5 \sqrt {3-2 x}}-\frac {2 \sqrt {-1+3 x-x^2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{5^{3/4} \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 )}{5 \sqrt {1-3 x+x^2}} \\ & = -\frac {4 \sqrt {1-3 x+x^2}}{5 \sqrt {3-2 x}}+\frac {2 \sqrt {-1+3 x-x^2} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{5^{3/4} \sqrt {1-3 x+x^2}}-\frac {2 \sqrt {-1+3 x-x^2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{5^{3/4} \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 = 63, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(3-2 x)^{3/2} \sqrt {1-3 x+x^2}} \, dx=\frac {2 \sqrt {-1+3 x-x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},\frac {1}{5} (3-2 x)^2\right )}{\sqrt {5} \sqrt {3-2 x} \sqrt {1-3 x+x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.70 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}\, \left (\sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}\, \sqrt {5}\, \sqrt {\left (-3+2 x \right ) \sqrt {5}}\, \sqrt {\left (2 x -3+\sqrt {5}\right ) \sqrt {5}}\, E\left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}}{10}, \sqrt {2}\right )+20 x^{2}-60 x +20\right )}{50 x^{3}-225 x^{2}+275 x -75}\)	\(116\)
elliptic	\(\frac {\sqrt {-\left (-3+2 x \right ) \left (x^{2}-3 x +1\right )}\, \left (\frac {-\frac {4}{5} x^{2}+\frac {12}{5} x -\frac {4}{5}}{\sqrt {\left (x -\frac {3}{2}\right ) \left (-2 x^{2}+6 x -2\right )}}+\frac {6 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, F\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{125 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}-\frac {4 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \left (\frac {\sqrt {5}\, E\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{2}+\frac {3 F\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{2}\right )}{125 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}\right )}{\sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}}\)	\(256\)

[In]

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

[Out]

1/25*(3-2*x)^(1/2)*(x^2-3*x+1)^(1/2)*(((-2*x+3+5^(1/2))*5^(1/2))^(1/2)*5^(1/2)*((-3+2*x)*5^(1/2))^(1/2)*((2*x-
3+5^(1/2))*5^(1/2))^(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))+20*x^2-60*x
+20)/(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 = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.37 \[ \int \frac {1}{(3-2 x)^{3/2} \sqrt {1-3 x+x^2}} \, dx=\frac {2 \, {\left (\sqrt {-2} {\left (2 \, x - 3\right )} {\rm weierstrassZeta}\left (5, 0, {\rm weierstrassPInverse}\left (5, 0, x - \frac {3}{2}\right )\right ) + 2 \, \sqrt {x^{2} - 3 \, x + 1} \sqrt {-2 \, x + 3}\right )}}{5 \, {\left (2 \, x - 3\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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