Integrand size = 22, antiderivative size = 79 \[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=-\frac {4 \sqrt {1-3 x+x^2}}{15 (3-2 x)^{3/2}}-\frac {2 \sqrt {-1+3 x-x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{15 \sqrt [4]{5} \sqrt {1-3 x+x^2}} \] Output:
-4/15*(x^2-3*x+1)^(1/2)/(3-2*x)^(3/2)-2/75*5^(3/4)*(-x^2+3*x-1)^(1/2)*Elli pticF(1/5*(3-2*x)^(1/2)*5^(3/4),I)/(x^2-3*x+1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=\frac {2 \sqrt {-1+3 x-x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},\frac {1}{5} (3-2 x)^2\right )}{3 \sqrt {5} (3-2 x)^{3/2} \sqrt {1-3 x+x^2}} \] Input:
Integrate[1/((3 - 2*x)^(5/2)*Sqrt[1 - 3*x + x^2]),x]
Output:
(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])
Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1117, 1115, 27, 1113, 762}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(3-2 x)^{5/2} \sqrt {x^2-3 x+1}} \, dx\) |
\(\Big \downarrow \) 1117 |
\(\displaystyle \frac {1}{15} \int \frac {1}{\sqrt {3-2 x} \sqrt {x^2-3 x+1}}dx-\frac {4 \sqrt {x^2-3 x+1}}{15 (3-2 x)^{3/2}}\) |
\(\Big \downarrow \) 1115 |
\(\displaystyle \frac {\sqrt {-x^2+3 x-1} \int \frac {\sqrt {5}}{\sqrt {3-2 x} \sqrt {-x^2+3 x-1}}dx}{15 \sqrt {5} \sqrt {x^2-3 x+1}}-\frac {4 \sqrt {x^2-3 x+1}}{15 (3-2 x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {-x^2+3 x-1} \int \frac {1}{\sqrt {3-2 x} \sqrt {-x^2+3 x-1}}dx}{15 \sqrt {x^2-3 x+1}}-\frac {4 \sqrt {x^2-3 x+1}}{15 (3-2 x)^{3/2}}\) |
\(\Big \downarrow \) 1113 |
\(\displaystyle -\frac {2 \sqrt {-x^2+3 x-1} \int \frac {1}{\sqrt {1-\frac {1}{5} (3-2 x)^2}}d\sqrt {3-2 x}}{15 \sqrt {5} \sqrt {x^2-3 x+1}}-\frac {4 \sqrt {x^2-3 x+1}}{15 (3-2 x)^{3/2}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle -\frac {2 \sqrt {-x^2+3 x-1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{15 \sqrt [4]{5} \sqrt {x^2-3 x+1}}-\frac {4 \sqrt {x^2-3 x+1}}{15 (3-2 x)^{3/2}}\) |
Input:
Int[1/((3 - 2*x)^(5/2)*Sqrt[1 - 3*x + x^2]),x]
Output:
(-4*Sqrt[1 - 3*x + x^2])/(15*(3 - 2*x)^(3/2)) - (2*Sqrt[-1 + 3*x - x^2]*El lipticF[ArcSin[Sqrt[3 - 2*x]/5^(1/4)], -1])/(15*5^(1/4)*Sqrt[1 - 3*x + x^2 ])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_ Symbol] :> Simp[(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] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]
Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sym bol] :> Simp[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] && EqQ[2*c* d - b*e, 0] && EqQ[m^2, 1/4]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> 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] + Simp[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] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & & (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) /2])
Leaf count of result is larger than twice the leaf count of optimal. \(141\) vs. \(2(62)=124\).
Time = 0.85 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.80
method | result | size |
elliptic | \(\frac {\sqrt {-\left (2 x -3\right ) \left (x^{2}-3 x +1\right )}\, \left (-\frac {\sqrt {-2 x^{3}+9 x^{2}-11 x +3}}{15 \left (x -\frac {3}{2}\right )^{2}}-\frac {2 \sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \sqrt {10}\, \sqrt {\left (x -\frac {3}{2}\right ) \sqrt {5}}\, \sqrt {\left (x -\frac {3}{2}+\frac {\sqrt {5}}{2}\right ) \sqrt {5}}\, \operatorname {EllipticF}\left (\frac {\sqrt {-5 \left (x -\frac {3}{2}-\frac {\sqrt {5}}{2}\right ) \sqrt {5}}}{5}, \sqrt {2}\right )}{375 \sqrt {-2 x^{3}+9 x^{2}-11 x +3}}\right )}{\sqrt {3-2 x}\, \sqrt {x^{2}-3 x +1}}\) | \(142\) |
default | \(\frac {2 \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}\, \sqrt {\left (2 x -3\right ) \sqrt {5}}\, \sqrt {\left (2 x -3+\sqrt {5}\right ) \sqrt {5}}\, \operatorname {EllipticF}\left (\frac {\sqrt {5}\, \sqrt {2}\, \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}}{10}, \sqrt {2}\right ) x -3 \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}\, \sqrt {\left (2 x -3\right ) \sqrt {5}}\, \sqrt {\left (2 x -3+\sqrt {5}\right ) \sqrt {5}}\, \operatorname {EllipticF}\left (\frac {\sqrt {5}\, \sqrt {2}\, \sqrt {\left (-2 x +3+\sqrt {5}\right ) \sqrt {5}}}{10}, \sqrt {2}\right )-20 x^{2}+60 x -20}{75 \sqrt {x^{2}-3 x +1}\, \left (3-2 x \right )^{\frac {3}{2}}}\) | \(165\) |
Input:
int(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
(-(2*x-3)*(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)^2-2/375*(-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)))
Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=-\frac {\sqrt {-2} {\left (4 \, x^{2} - 12 \, x + 9\right )} {\rm weierstrassPInverse}\left (5, 0, x - \frac {3}{2}\right ) + 4 \, \sqrt {x^{2} - 3 \, x + 1} \sqrt {-2 \, x + 3}}{15 \, {\left (4 \, x^{2} - 12 \, x + 9\right )}} \] Input:
integrate(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x, algorithm="fricas")
Output:
-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)
\[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=\int \frac {1}{\left (3 - 2 x\right )^{\frac {5}{2}} \sqrt {x^{2} - 3 x + 1}}\, dx \] Input:
integrate(1/(3-2*x)**(5/2)/(x**2-3*x+1)**(1/2),x)
Output:
Integral(1/((3 - 2*x)**(5/2)*sqrt(x**2 - 3*x + 1)), x)
\[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=\int { \frac {1}{\sqrt {x^{2} - 3 \, x + 1} {\left (-2 \, x + 3\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(x^2 - 3*x + 1)*(-2*x + 3)^(5/2)), x)
\[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=\int { \frac {1}{\sqrt {x^{2} - 3 \, x + 1} {\left (-2 \, x + 3\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x, algorithm="giac")
Output:
integrate(1/(sqrt(x^2 - 3*x + 1)*(-2*x + 3)^(5/2)), x)
Timed out. \[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=\int \frac {1}{{\left (3-2\,x\right )}^{5/2}\,\sqrt {x^2-3\,x+1}} \,d x \] Input:
int(1/((3 - 2*x)^(5/2)*(x^2 - 3*x + 1)^(1/2)),x)
Output:
int(1/((3 - 2*x)^(5/2)*(x^2 - 3*x + 1)^(1/2)), x)
\[ \int \frac {1}{(3-2 x)^{5/2} \sqrt {1-3 x+x^2}} \, dx=-\left (\int \frac {\sqrt {-2 x +3}\, \sqrt {x^{2}-3 x +1}}{8 x^{5}-60 x^{4}+170 x^{3}-225 x^{2}+135 x -27}d x \right ) \] Input:
int(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x)
Output:
- int((sqrt( - 2*x + 3)*sqrt(x**2 - 3*x + 1))/(8*x**5 - 60*x**4 + 170*x** 3 - 225*x**2 + 135*x - 27),x)