3.26.79 \(\int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{5 x^2-10 x^3+5 x^4} \, dx\) [2579]

3.26.79.1 Optimal result

Integrand size = 58, antiderivative size = 23 \[ \int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{5 x^2-10 x^3+5 x^4} \, dx=-4+2 x+\frac {1}{5} \left (5+x+x^{\frac {1}{(-1+x) x}}\right ) \]

output

11/5*x-3+1/5*exp(1/x/(-1+x)*ln(x))
 

3.26.79.2 Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{5 x^2-10 x^3+5 x^4} \, dx=\frac {1}{5} \left (11 x+x^{\frac {1}{(-1+x) x}}\right ) \]

input

Integrate[(11*x^2 - 22*x^3 + 11*x^4 + x^(-x + x^2)^(-1)*(-1 + x + (1 - 2*x 
)*Log[x]))/(5*x^2 - 10*x^3 + 5*x^4),x]
 

output

(11*x + x^(1/((-1 + x)*x)))/5
 

3.26.79.3 Rubi [F]

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.

\(\displaystyle \int \frac {11 x^4-22 x^3+11 x^2+x^{\frac {1}{x^2-x}} (x+(1-2 x) \log (x)-1)}{5 x^4-10 x^3+5 x^2} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {11 x^4-22 x^3+11 x^2+x^{\frac {1}{x^2-x}} (x+(1-2 x) \log (x)-1)}{x^2 \left (5 x^2-10 x+5\right )}dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 20 \int \frac {-\left ((-x-(1-2 x) \log (x)+1) x^{\frac {1}{x^2-x}}\right )+11 x^4-22 x^3+11 x^2}{100 (1-x)^2 x^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {-\left ((-x-(1-2 x) \log (x)+1) x^{\frac {1}{x^2-x}}\right )+11 x^4-22 x^3+11 x^2}{(1-x)^2 x^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{5} \int \left (11-\frac {x^{\frac {1}{(x-1) x}-2} (2 \log (x) x-x-\log (x)+1)}{(x-1)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} \left (-\int \frac {x^{\frac {1}{(x-1) x}-2} \log (x)}{(x-1)^2}dx-2 \int \frac {x^{\frac {1}{(x-1) x}-2} \log (x)}{x-1}dx+\frac {x^{-\frac {1}{(1-x) x}-1} \operatorname {Hypergeometric2F1}\left (1,-1-\frac {1}{(1-x) x},-\frac {1}{(1-x) x},x\right )}{\frac {1}{(1-x) x}+1}+11 x\right )\)

input

Int[(11*x^2 - 22*x^3 + 11*x^4 + x^(-x + x^2)^(-1)*(-1 + x + (1 - 2*x)*Log[ 
x]))/(5*x^2 - 10*x^3 + 5*x^4),x]
 

output

$Aborted
 

3.26.79.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 7277

Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> 
 Simp[1/(4^p*c^p)   Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} 
, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] &&  !AlgebraicFu 
nctionQ[u, x]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.26.79.4 Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {11 x}{5}+\frac {x^{\frac {1}{x \left (-1+x \right )}}}{5}\)	\(18\)
parallelrisch	\(\frac {11 x}{5}+\frac {{\mathrm e}^{\frac {\ln \left (x \right )}{x \left (-1+x \right )}}}{5}+\frac {33}{10}\)	\(20\)
norman	\(\frac {-\frac {11 x}{5}+\frac {11 x^{3}}{5}-\frac {x \,{\mathrm e}^{\frac {\ln \left (x \right )}{x^{2}-x}}}{5}+\frac {x^{2} {\mathrm e}^{\frac {\ln \left (x \right )}{x^{2}-x}}}{5}}{x \left (-1+x \right )}\)	\(53\)

input

int((((1-2*x)*ln(x)+x-1)*exp(ln(x)/(x^2-x))+11*x^4-22*x^3+11*x^2)/(5*x^4-1 
0*x^3+5*x^2),x,method=_RETURNVERBOSE)
 

output

11/5*x+1/5*x^(1/x/(-1+x))
 

3.26.79.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{5 x^2-10 x^3+5 x^4} \, dx=\frac {11}{5} \, x + \frac {1}{5} \, x^{\left (\frac {1}{x^{2} - x}\right )} \]

input

integrate((((1-2*x)*log(x)+x-1)*exp(log(x)/(x^2-x))+11*x^4-22*x^3+11*x^2)/ 
(5*x^4-10*x^3+5*x^2),x, algorithm=\
 

output

11/5*x + 1/5*x^(1/(x^2 - x))
 

3.26.79.6 Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{5 x^2-10 x^3+5 x^4} \, dx=\frac {11 x}{5} + \frac {e^{\frac {\log {\left (x \right )}}{x^{2} - x}}}{5} \]

input

integrate((((1-2*x)*ln(x)+x-1)*exp(ln(x)/(x**2-x))+11*x**4-22*x**3+11*x**2 
)/(5*x**4-10*x**3+5*x**2),x)
 

output

11*x/5 + exp(log(x)/(x**2 - x))/5
 

3.26.79.7 Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{5 x^2-10 x^3+5 x^4} \, dx=\frac {11}{5} \, x + \frac {1}{5} \, e^{\left (\frac {\log \left (x\right )}{x - 1} - \frac {\log \left (x\right )}{x}\right )} \]

input

integrate((((1-2*x)*log(x)+x-1)*exp(log(x)/(x^2-x))+11*x^4-22*x^3+11*x^2)/ 
(5*x^4-10*x^3+5*x^2),x, algorithm=\
 

output

11/5*x + 1/5*e^(log(x)/(x - 1) - log(x)/x)
 

3.26.79.8 Giac [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{5 x^2-10 x^3+5 x^4} \, dx=\frac {11}{5} \, x + \frac {e^{\left (\frac {x^{2} \log \left (x\right ) - x \log \left (x\right ) + \log \left (x\right )}{x^{2} - x}\right )}}{5 \, x} \]

input

integrate((((1-2*x)*log(x)+x-1)*exp(log(x)/(x^2-x))+11*x^4-22*x^3+11*x^2)/ 
(5*x^4-10*x^3+5*x^2),x, algorithm=\
 

output

11/5*x + 1/5*e^((x^2*log(x) - x*log(x) + log(x))/(x^2 - x))/x
 

3.26.79.9 Mupad [B] (verification not implemented)

Time = 14.70 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{5 x^2-10 x^3+5 x^4} \, dx=\frac {11\,x}{5}+\frac {1}{5\,x^{\frac {1}{x-x^2}}} \]

input

int(-(exp(-log(x)/(x - x^2))*(log(x)*(2*x - 1) - x + 1) - 11*x^2 + 22*x^3 
- 11*x^4)/(5*x^2 - 10*x^3 + 5*x^4),x)
 

output

(11*x)/5 + 1/(5*x^(1/(x - x^2)))