\(\int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx\) [1375]

Optimal result

Integrand size = 22, antiderivative size = 79 \[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=-\frac {4 \sqrt {1-3 x+x^2}}{15 (3-2 x)^{3/2}}-\frac {2 \sqrt {-1+3 x-x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{15 \sqrt [4]{5} \sqrt {1-3 x+x^2}} \]

[Out]

-2/75*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)-4/15*(x^2-3*x+1)^(1/
2)/(3-2*x)^(3/2)

Rubi [A] (verified)

Time = 0.02 (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 = {707, 705, 703, 227} \[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=-\frac {2 \sqrt {-x^2+3 x-1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{15 \sqrt [4]{5} \sqrt {x^2-3 x+1}}-\frac {4 \sqrt {x^2-3 x+1}}{15 (3-2 x)^{3/2}} \]

[In]

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

[Out]

(-4*Sqrt[1 - 3*x + x^2])/(15*(3 - 2*x)^(3/2)) - (2*Sqrt[-1 + 3*x - x^2]*EllipticF[ArcSin[Sqrt[3 - 2*x]/5^(1/4)
], -1])/(15*5^(1/4)*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 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])

Rubi steps \begin{align*} \text {integral}& = -\frac {4 \sqrt {1-3 x+x^2}}{15 (3-2 x)^{3/2}}+\frac {1}{15} \int \frac {1}{\sqrt {3-2 x} \sqrt {1-3 x+x^2}} \, dx \\ & = -\frac {4 \sqrt {1-3 x+x^2}}{15 (3-2 x)^{3/2}}+\frac {\sqrt {-1+3 x-x^2} \int \frac {1}{\sqrt {3-2 x} \sqrt {-\frac {1}{5}+\frac {3 x}{5}-\frac {x^2}{5}}} \, dx}{15 \sqrt {5} \sqrt {1-3 x+x^2}} \\ & = -\frac {4 \sqrt {1-3 x+x^2}}{15 (3-2 x)^{3/2}}-\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 )}{15 \sqrt {5} \sqrt {1-3 x+x^2}} \\ & = -\frac {4 \sqrt {1-3 x+x^2}}{15 (3-2 x)^{3/2}}-\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 )}{15 \sqrt [4]{5} \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 = 0.82 \[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=\frac {2 \sqrt {-1+3 x-x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},\frac {1}{5} (3-2 x)^2\right )}{3 \sqrt {5} (3-2 x)^{3/2} \sqrt {1-3 x+x^2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(141\) vs. \(2(62)=124\).

Time = 2.73 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.80

method	result	size
elliptic	\(\frac {\sqrt {-\left (-3+2 x \right ) \left (x^{2}-3 x +1\right )}\, \left (-\frac {\sqrt {-2 x^{3}+9 x^{2}-11 x +3}}{15 \left (x -\frac {3}{2}\right )^{2}}-\frac {2 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, F\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{375 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}\right )}{\sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}}\)	\(142\)
default	\(\frac {\left (2 \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}\, \sqrt {\left (-3+2 x \right ) \sqrt {5}}\, \sqrt {\left (2 x -3+\sqrt {5}\right ) \sqrt {5}}\, F\left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}}{10}, \sqrt {2}\right ) x -3 \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}\, \sqrt {\left (-3+2 x \right ) \sqrt {5}}\, \sqrt {\left (2 x -3+\sqrt {5}\right ) \sqrt {5}}\, F\left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}}{10}, \sqrt {2}\right )-20 x^{2}+60 x -20\right ) \sqrt {3-2 x}}{75 \sqrt {x^{2}-3 x +1}\, \left (-3+2 x \right )^{2}}\)	\(172\)

[In]

int(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-(-3+2*x)*(x^2-3*x+1))^(1/2)/(3-2*x)^(1/2)/(x^2-3*x+1)^(1/2)*(-1/15*(-2*x^3+9*x^2-11*x+3)^(1/2)/(x-3/2)^2-2/3
75*(-5*(x-3/2-1/2*5^(1/2))*5^(1/2))^(1/2)*10^(1/2)*((x-3/2)*5^(1/2))^(1/2)*((x-3/2+1/2*5^(1/2))*5^(1/2))^(1/2)
/(-2*x^3+9*x^2-11*x+3)^(1/2)*EllipticF(1/5*(-5*(x-3/2-1/2*5^(1/2))*5^(1/2))^(1/2),2^(1/2)))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=-\frac {\sqrt {-2} {\left (4 \, x^{2} - 12 \, x + 9\right )} {\rm weierstrassPInverse}\left (5, 0, x - \frac {3}{2}\right ) + 4 \, \sqrt {x^{2} - 3 \, x + 1} \sqrt {-2 \, x + 3}}{15 \, {\left (4 \, x^{2} - 12 \, x + 9\right )}} \]

[In]

integrate(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x, algorithm="fricas")

[Out]

-1/15*(sqrt(-2)*(4*x^2 - 12*x + 9)*weierstrassPInverse(5, 0, x - 3/2) + 4*sqrt(x^2 - 3*x + 1)*sqrt(-2*x + 3))/
(4*x^2 - 12*x + 9)

Sympy [F]

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

[In]

integrate(1/(3-2*x)**(5/2)/(x**2-3*x+1)**(1/2),x)

[Out]

Integral(1/((3 - 2*x)**(5/2)*sqrt(x**2 - 3*x + 1)), x)

Maxima [F]

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

[In]

integrate(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^2 - 3*x + 1)*(-2*x + 3)^(5/2)), x)

Giac [F]

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

[In]

integrate(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^2 - 3*x + 1)*(-2*x + 3)^(5/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=\int \frac {1}{{\left (3-2\,x\right )}^{5/2}\,\sqrt {x^2-3\,x+1}} \,d x \]

[In]

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

[Out]

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