\(\int (6-5 x+x^2)^{5/4} \, dx\) [69]

Optimal result

Integrand size = 12, antiderivative size = 90 \[ \int \left (6-5 x+x^2\right )^{5/4} \, dx=\frac {5}{84} (5-2 x) \sqrt [4]{6-5 x+x^2}-\frac {1}{7} (5-2 x) \left (6-5 x+x^2\right )^{5/4}-\frac {5 \left (-6+5 x-x^2\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin (5-2 x),2\right )}{84 \sqrt {2} \left (6-5 x+x^2\right )^{3/4}} \] Output:

5/84*(5-2*x)*(x^2-5*x+6)^(1/4)-1/7*(5-2*x)*(x^2-5*x+6)^(5/4)+5/168*(-x^2+5 
*x-6)^(3/4)*InverseJacobiAM(1/2*arcsin(-5+2*x),2^(1/2))*2^(1/2)/(x^2-5*x+6 
)^(3/4)
 

Mathematica [A] (verified)

Time = 8.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.82 \[ \int \left (6-5 x+x^2\right )^{5/4} \, dx=\frac {-4020+8558 x-7170 x^2+2956 x^3-600 x^4+48 x^5-5 \sqrt {2} \left (-6+5 x-x^2\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin (5-2 x),2\right )}{168 \left (6-5 x+x^2\right )^{3/4}} \] Input:

Integrate[(6 - 5*x + x^2)^(5/4),x]
 

Output:

(-4020 + 8558*x - 7170*x^2 + 2956*x^3 - 600*x^4 + 48*x^5 - 5*Sqrt[2]*(-6 + 
 5*x - x^2)^(3/4)*EllipticF[ArcSin[5 - 2*x]/2, 2])/(168*(6 - 5*x + x^2)^(3 
/4))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1087, 1087, 1094, 761}

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.

\(\displaystyle \int \left (x^2-5 x+6\right )^{5/4} \, dx\)

\(\Big \downarrow \) 1087

\(\displaystyle -\frac {5}{28} \int \sqrt [4]{x^2-5 x+6}dx-\frac {1}{7} (5-2 x) \left (x^2-5 x+6\right )^{5/4}\)

\(\Big \downarrow \) 1087

\(\displaystyle -\frac {5}{28} \left (-\frac {1}{12} \int \frac {1}{\left (x^2-5 x+6\right )^{3/4}}dx-\frac {1}{3} \sqrt [4]{x^2-5 x+6} (5-2 x)\right )-\frac {1}{7} (5-2 x) \left (x^2-5 x+6\right )^{5/4}\)

\(\Big \downarrow \) 1094

\(\displaystyle -\frac {5}{28} \left (\frac {\sqrt {(2 x-5)^2} \int \frac {1}{\sqrt {4 \left (x^2-5 x+6\right )+1}}d\sqrt [4]{x^2-5 x+6}}{3 (5-2 x)}-\frac {1}{3} (5-2 x) \sqrt [4]{x^2-5 x+6}\right )-\frac {1}{7} (5-2 x) \left (x^2-5 x+6\right )^{5/4}\)

\(\Big \downarrow \) 761

\(\displaystyle -\frac {5}{28} \left (\frac {\sqrt {(2 x-5)^2} \left (2 \sqrt {x^2-5 x+6}+1\right ) \sqrt {\frac {4 \left (x^2-5 x+6\right )+1}{\left (2 \sqrt {x^2-5 x+6}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {2} \sqrt [4]{x^2-5 x+6}\right ),\frac {1}{2}\right )}{6 \sqrt {2} (5-2 x) \sqrt {4 \left (x^2-5 x+6\right )+1}}-\frac {1}{3} (5-2 x) \sqrt [4]{x^2-5 x+6}\right )-\frac {1}{7} (5-2 x) \left (x^2-5 x+6\right )^{5/4}\)

Input:

Int[(6 - 5*x + x^2)^(5/4),x]
 

Output:

-1/7*((5 - 2*x)*(6 - 5*x + x^2)^(5/4)) - (5*(-1/3*((5 - 2*x)*(6 - 5*x + x^ 
2)^(1/4)) + (Sqrt[(-5 + 2*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]*EllipticF[2*ArcTan[Sqrt[2]*( 
6 - 5*x + x^2)^(1/4)], 1/2])/(6*Sqrt[2]*(5 - 2*x)*Sqrt[1 + 4*(6 - 5*x + x^ 
2)])))/28
 

Defintions of rubi rules used

rule 761

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]
 

rule 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1094

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]
 

Maple [F]

\[\int \left (x^{2}-5 x +6\right )^{\frac {5}{4}}d x\]

Input:

int((x^2-5*x+6)^(5/4),x)
 

Output:

int((x^2-5*x+6)^(5/4),x)
 

Fricas [F]

\[ \int \left (6-5 x+x^2\right )^{5/4} \, dx=\int { {\left (x^{2} - 5 \, x + 6\right )}^{\frac {5}{4}} \,d x } \] Input:

integrate((x^2-5*x+6)^(5/4),x, algorithm="fricas")
 

Output:

integral((x^2 - 5*x + 6)^(5/4), x)
 

Sympy [F]

\[ \int \left (6-5 x+x^2\right )^{5/4} \, dx=\int \left (x^{2} - 5 x + 6\right )^{\frac {5}{4}}\, dx \] Input:

integrate((x**2-5*x+6)**(5/4),x)
 

Output:

Integral((x**2 - 5*x + 6)**(5/4), x)
 

Maxima [F]

\[ \int \left (6-5 x+x^2\right )^{5/4} \, dx=\int { {\left (x^{2} - 5 \, x + 6\right )}^{\frac {5}{4}} \,d x } \] Input:

integrate((x^2-5*x+6)^(5/4),x, algorithm="maxima")
 

Output:

integrate((x^2 - 5*x + 6)^(5/4), x)
 

Giac [F]

\[ \int \left (6-5 x+x^2\right )^{5/4} \, dx=\int { {\left (x^{2} - 5 \, x + 6\right )}^{\frac {5}{4}} \,d x } \] Input:

integrate((x^2-5*x+6)^(5/4),x, algorithm="giac")
 

Output:

integrate((x^2 - 5*x + 6)^(5/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (6-5 x+x^2\right )^{5/4} \, dx=\int {\left (x^2-5\,x+6\right )}^{5/4} \,d x \] Input:

int((x^2 - 5*x + 6)^(5/4),x)
 

Output:

int((x^2 - 5*x + 6)^(5/4), x)
 

Reduce [F]

\[ \int \left (6-5 x+x^2\right )^{5/4} \, dx=\frac {2 \left (x^{2}-5 x +6\right )^{\frac {1}{4}} x^{3}}{7}-\frac {15 \left (x^{2}-5 x +6\right )^{\frac {1}{4}} x^{2}}{7}+\frac {31 \left (x^{2}-5 x +6\right )^{\frac {1}{4}} x}{6}-4 \left (x^{2}-5 x +6\right )^{\frac {1}{4}}+\frac {\left (\int \frac {x}{\left (x^{2}-5 x +6\right )^{\frac {3}{4}}}d x \right )}{168} \] Input:

int((x^2-5*x+6)^(5/4),x)
 

Output:

(48*(x**2 - 5*x + 6)**(1/4)*x**3 - 360*(x**2 - 5*x + 6)**(1/4)*x**2 + 868* 
(x**2 - 5*x + 6)**(1/4)*x - 672*(x**2 - 5*x + 6)**(1/4) + int(((x**2 - 5*x 
 + 6)**(1/4)*x)/(x**2 - 5*x + 6),x))/168