Integrand size = 15, antiderivative size = 45 \[ \int \frac {1}{\sqrt [4]{(2-x) (3-x)}} \, dx=-\frac {\sqrt {2} \sqrt [4]{-6+5 x-x^2} E\left (\left .\frac {1}{2} \arcsin (5-2 x)\right |2\right )}{\sqrt [4]{6-5 x+x^2}} \] Output:
(-x^2+5*x-6)^(1/4)*EllipticE(sin(1/2*arcsin(-5+2*x)),2^(1/2))*2^(1/2)/(x^2 -5*x+6)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\sqrt [4]{(2-x) (3-x)}} \, dx=-\frac {(5-2 x) \sqrt [4]{-6+5 x-x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},(5-2 x)^2\right )}{\sqrt {2} \sqrt [4]{6-5 x+x^2}} \] Input:
Integrate[((2 - x)*(3 - x))^(-1/4),x]
Output:
-(((5 - 2*x)*(-6 + 5*x - x^2)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (5 - 2*x)^2])/(Sqrt[2]*(6 - 5*x + x^2)^(1/4)))
Leaf count is larger than twice the leaf count of optimal. \(273\) vs. \(2(45)=90\).
Time = 0.49 (sec) , antiderivative size = 273, normalized size of antiderivative = 6.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2048, 1094, 834, 761, 1510}
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.
\(\displaystyle \int \frac {1}{\sqrt [4]{(2-x) (3-x)}} \, dx\) |
\(\Big \downarrow \) 2048 |
\(\displaystyle \int \frac {1}{\sqrt [4]{x^2-5 x+6}}dx\) |
\(\Big \downarrow \) 1094 |
\(\displaystyle -\frac {4 \sqrt {(2 x-5)^2} \int \frac {\sqrt {x^2-5 x+6}}{\sqrt {4 \left (x^2-5 x+6\right )+1}}d\sqrt [4]{x^2-5 x+6}}{5-2 x}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle -\frac {4 \sqrt {(2 x-5)^2} \left (\frac {1}{2} \int \frac {1}{\sqrt {4 \left (x^2-5 x+6\right )+1}}d\sqrt [4]{x^2-5 x+6}-\frac {1}{2} \int \frac {1-2 \sqrt {x^2-5 x+6}}{\sqrt {4 \left (x^2-5 x+6\right )+1}}d\sqrt [4]{x^2-5 x+6}\right )}{5-2 x}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {4 \sqrt {(2 x-5)^2} \left (\frac {\left (2 \sqrt {x^2-5 x+6}+1\right ) \sqrt {\frac {4 \left (x^2-5 x+6\right )+1}{\left (2 \sqrt {x^2-5 x+6}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {2} \sqrt [4]{x^2-5 x+6}\right ),\frac {1}{2}\right )}{4 \sqrt {2} \sqrt {4 \left (x^2-5 x+6\right )+1}}-\frac {1}{2} \int \frac {1-2 \sqrt {x^2-5 x+6}}{\sqrt {4 \left (x^2-5 x+6\right )+1}}d\sqrt [4]{x^2-5 x+6}\right )}{5-2 x}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle -\frac {4 \sqrt {(2 x-5)^2} \left (\frac {\left (2 \sqrt {x^2-5 x+6}+1\right ) \sqrt {\frac {4 \left (x^2-5 x+6\right )+1}{\left (2 \sqrt {x^2-5 x+6}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {2} \sqrt [4]{x^2-5 x+6}\right ),\frac {1}{2}\right )}{4 \sqrt {2} \sqrt {4 \left (x^2-5 x+6\right )+1}}+\frac {1}{2} \left (\frac {\sqrt [4]{x^2-5 x+6} \sqrt {4 \left (x^2-5 x+6\right )+1}}{2 \sqrt {x^2-5 x+6}+1}-\frac {\left (2 \sqrt {x^2-5 x+6}+1\right ) \sqrt {\frac {4 \left (x^2-5 x+6\right )+1}{\left (2 \sqrt {x^2-5 x+6}+1\right )^2}} E\left (2 \arctan \left (\sqrt {2} \sqrt [4]{x^2-5 x+6}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt {4 \left (x^2-5 x+6\right )+1}}\right )\right )}{5-2 x}\) |
Input:
Int[((2 - x)*(3 - x))^(-1/4),x]
Output:
(-4*Sqrt[(-5 + 2*x)^2]*((((6 - 5*x + x^2)^(1/4)*Sqrt[1 + 4*(6 - 5*x + x^2) ])/(1 + 2*Sqrt[6 - 5*x + x^2]) - ((1 + 2*Sqrt[6 - 5*x + x^2])*Sqrt[(1 + 4* (6 - 5*x + x^2))/(1 + 2*Sqrt[6 - 5*x + x^2])^2]*EllipticE[2*ArcTan[Sqrt[2] *(6 - 5*x + x^2)^(1/4)], 1/2])/(Sqrt[2]*Sqrt[1 + 4*(6 - 5*x + x^2)]))/2 + ((1 + 2*Sqrt[6 - 5*x + x^2])*Sqrt[(1 + 4*(6 - 5*x + x^2))/(1 + 2*Sqrt[6 - 5*x + x^2])^2]*EllipticF[2*ArcTan[Sqrt[2]*(6 - 5*x + x^2)^(1/4)], 1/2])/(4 *Sqrt[2]*Sqrt[1 + 4*(6 - 5*x + x^2)])))/(5 - 2*x)
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[4*(Sqrt[(b + 2*c*x)^2]/(b + 2*c*x)) Subst[Int[x^(4*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4 *c*x^4], x], x, (a + b*x + c*x^2)^(1/4)], x] /; FreeQ[{a, b, c}, x] && Inte gerQ[4*p]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_) , x_Symbol] :> Int[u*(a*c*e + (b*c + a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; F reeQ[{a, b, c, d, e, n, p}, x]
\[\int \frac {1}{\left (\left (2-x \right ) \left (3-x \right )\right )^{\frac {1}{4}}}d x\]
Input:
int(1/((2-x)*(3-x))^(1/4),x)
Output:
int(1/((2-x)*(3-x))^(1/4),x)
\[ \int \frac {1}{\sqrt [4]{(2-x) (3-x)}} \, dx=\int { \frac {1}{\left ({\left (x - 2\right )} {\left (x - 3\right )}\right )^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/((2-x)*(3-x))^(1/4),x, algorithm="fricas")
Output:
integral((x^2 - 5*x + 6)^(-1/4), x)
\[ \int \frac {1}{\sqrt [4]{(2-x) (3-x)}} \, dx=\int \frac {1}{\sqrt [4]{\left (2 - x\right ) \left (3 - x\right )}}\, dx \] Input:
integrate(1/((2-x)*(3-x))**(1/4),x)
Output:
Integral(((2 - x)*(3 - x))**(-1/4), x)
\[ \int \frac {1}{\sqrt [4]{(2-x) (3-x)}} \, dx=\int { \frac {1}{\left ({\left (x - 2\right )} {\left (x - 3\right )}\right )^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/((2-x)*(3-x))^(1/4),x, algorithm="maxima")
Output:
integrate(((x - 2)*(x - 3))^(-1/4), x)
\[ \int \frac {1}{\sqrt [4]{(2-x) (3-x)}} \, dx=\int { \frac {1}{\left ({\left (x - 2\right )} {\left (x - 3\right )}\right )^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/((2-x)*(3-x))^(1/4),x, algorithm="giac")
Output:
integrate(((x - 2)*(x - 3))^(-1/4), x)
Timed out. \[ \int \frac {1}{\sqrt [4]{(2-x) (3-x)}} \, dx=\int \frac {1}{{\left (\left (x-2\right )\,\left (x-3\right )\right )}^{1/4}} \,d x \] Input:
int(1/((x - 2)*(x - 3))^(1/4),x)
Output:
int(1/((x - 2)*(x - 3))^(1/4), x)
\[ \int \frac {1}{\sqrt [4]{(2-x) (3-x)}} \, dx=\int \frac {1}{\left (x^{2}-5 x +6\right )^{\frac {1}{4}}}d x \] Input:
int(1/((2-x)*(3-x))^(1/4),x)
Output:
int(1/(x**2 - 5*x + 6)**(1/4),x)