\(\int \frac {1}{(6-5 x+x^2)^{11/4}} \, dx\) [73]

Optimal result

Integrand size = 12, antiderivative size = 90 \[ \int \frac {1}{\left (6-5 x+x^2\right )^{11/4}} \, dx=\frac {4 (5-2 x)}{7 \left (6-5 x+x^2\right )^{7/4}}-\frac {80 (5-2 x)}{21 \left (6-5 x+x^2\right )^{3/4}}-\frac {160 \sqrt {2} \left (-6+5 x-x^2\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin (5-2 x),2\right )}{21 \left (6-5 x+x^2\right )^{3/4}} \] Output:

4/7*(5-2*x)/(x^2-5*x+6)^(7/4)-80/21*(5-2*x)/(x^2-5*x+6)^(3/4)+160/21*(-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 = 8.70 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (6-5 x+x^2\right )^{11/4}} \, dx=\frac {4 \left (-585+734 x-300 x^2+40 x^3+40 \sqrt {2} \left (-6+5 x-x^2\right )^{7/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin (5-2 x),2\right )\right )}{21 \left (6-5 x+x^2\right )^{7/4}} \] Input:

Integrate[(6 - 5*x + x^2)^(-11/4),x]
 

Output:

(4*(-585 + 734*x - 300*x^2 + 40*x^3 + 40*Sqrt[2]*(-6 + 5*x - x^2)^(7/4)*El 
lipticF[ArcSin[5 - 2*x]/2, 2]))/(21*(6 - 5*x + x^2)^(7/4))
 

Rubi [A] (verified)

Time = 0.38 (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 = {1089, 1089, 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.

\(\displaystyle \int \frac {1}{\left (x^2-5 x+6\right )^{11/4}} \, dx\)

\(\Big \downarrow \) 1089

\(\displaystyle \frac {4 (5-2 x)}{7 \left (x^2-5 x+6\right )^{7/4}}-\frac {20}{7} \int \frac {1}{\left (x^2-5 x+6\right )^{7/4}}dx\)

\(\Big \downarrow \) 1089

\(\displaystyle \frac {4 (5-2 x)}{7 \left (x^2-5 x+6\right )^{7/4}}-\frac {20}{7} \left (\frac {4 (5-2 x)}{3 \left (x^2-5 x+6\right )^{3/4}}-\frac {4}{3} \int \frac {1}{\left (x^2-5 x+6\right )^{3/4}}dx\right )\)

\(\Big \downarrow \) 1094

\(\displaystyle \frac {4 (5-2 x)}{7 \left (x^2-5 x+6\right )^{7/4}}-\frac {20}{7} \left (\frac {16 \sqrt {(2 x-5)^2} \int \frac {1}{\sqrt {4 \left (x^2-5 x+6\right )+1}}d\sqrt [4]{x^2-5 x+6}}{3 (5-2 x)}+\frac {4 (5-2 x)}{3 \left (x^2-5 x+6\right )^{3/4}}\right )\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {4 (5-2 x)}{7 \left (x^2-5 x+6\right )^{7/4}}-\frac {20}{7} \left (\frac {4 \sqrt {2} \sqrt {(2 x-5)^2} \left (2 \sqrt {x^2-5 x+6}+1\right ) \sqrt {\frac {4 \left (x^2-5 x+6\right )+1}{\left (2 \sqrt {x^2-5 x+6}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {2} \sqrt [4]{x^2-5 x+6}\right ),\frac {1}{2}\right )}{3 \sqrt {4 \left (x^2-5 x+6\right )+1} (5-2 x)}+\frac {4 (5-2 x)}{3 \left (x^2-5 x+6\right )^{3/4}}\right )\)

Input:

Int[(6 - 5*x + x^2)^(-11/4),x]
 

Output:

(4*(5 - 2*x))/(7*(6 - 5*x + x^2)^(7/4)) - (20*((4*(5 - 2*x))/(3*(6 - 5*x + 
 x^2)^(3/4)) + (4*Sqrt[2]*Sqrt[(-5 + 2*x)^2]*(1 + 2*Sqrt[6 - 5*x + x^2])*S 
qrt[(1 + 4*(6 - 5*x + x^2))/(1 + 2*Sqrt[6 - 5*x + x^2])^2]*EllipticF[2*Arc 
Tan[Sqrt[2]*(6 - 5*x + x^2)^(1/4)], 1/2])/(3*(5 - 2*x)*Sqrt[1 + 4*(6 - 5*x 
 + x^2)])))/7
 

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 1089

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

\[\int \frac {1}{\left (x^{2}-5 x +6\right )^{\frac {11}{4}}}d x\]

Input:

int(1/(x^2-5*x+6)^(11/4),x)
 

Output:

int(1/(x^2-5*x+6)^(11/4),x)
 

Fricas [F]

\[ \int \frac {1}{\left (6-5 x+x^2\right )^{11/4}} \, dx=\int { \frac {1}{{\left (x^{2} - 5 \, x + 6\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate(1/(x^2-5*x+6)^(11/4),x, algorithm="fricas")
 

Output:

integral((x^2 - 5*x + 6)^(1/4)/(x^6 - 15*x^5 + 93*x^4 - 305*x^3 + 558*x^2 
- 540*x + 216), x)
 

Sympy [F]

\[ \int \frac {1}{\left (6-5 x+x^2\right )^{11/4}} \, dx=\int \frac {1}{\left (x^{2} - 5 x + 6\right )^{\frac {11}{4}}}\, dx \] Input:

integrate(1/(x**2-5*x+6)**(11/4),x)
 

Output:

Integral((x**2 - 5*x + 6)**(-11/4), x)
 

Maxima [F]

\[ \int \frac {1}{\left (6-5 x+x^2\right )^{11/4}} \, dx=\int { \frac {1}{{\left (x^{2} - 5 \, x + 6\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate(1/(x^2-5*x+6)^(11/4),x, algorithm="maxima")
 

Output:

integrate((x^2 - 5*x + 6)^(-11/4), x)
 

Giac [F]

\[ \int \frac {1}{\left (6-5 x+x^2\right )^{11/4}} \, dx=\int { \frac {1}{{\left (x^{2} - 5 \, x + 6\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate(1/(x^2-5*x+6)^(11/4),x, algorithm="giac")
 

Output:

integrate((x^2 - 5*x + 6)^(-11/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (6-5 x+x^2\right )^{11/4}} \, dx=\int \frac {1}{{\left (x^2-5\,x+6\right )}^{11/4}} \,d x \] Input:

int(1/(x^2 - 5*x + 6)^(11/4),x)
 

Output:

int(1/(x^2 - 5*x + 6)^(11/4), x)
 

Reduce [F]

\[ \int \frac {1}{\left (6-5 x+x^2\right )^{11/4}} \, dx=\int \frac {1}{\left (x^{2}-5 x +6\right )^{\frac {3}{4}} x^{4}-10 \left (x^{2}-5 x +6\right )^{\frac {3}{4}} x^{3}+37 \left (x^{2}-5 x +6\right )^{\frac {3}{4}} x^{2}-60 \left (x^{2}-5 x +6\right )^{\frac {3}{4}} x +36 \left (x^{2}-5 x +6\right )^{\frac {3}{4}}}d x \] Input:

int(1/(x^2-5*x+6)^(11/4),x)
 

Output:

int(1/((x**2 - 5*x + 6)**(3/4)*x**4 - 10*(x**2 - 5*x + 6)**(3/4)*x**3 + 37 
*(x**2 - 5*x + 6)**(3/4)*x**2 - 60*(x**2 - 5*x + 6)**(3/4)*x + 36*(x**2 - 
5*x + 6)**(3/4)),x)