\(\int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{5 x^2-10 x^3+5 x^4} \, dx\) [8579]

Optimal result

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

[Out]

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

Rubi [F]

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

[In]

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]

[Out]

(11*x)/5 + (x^(-1 - 1/((1 - x)*x))*Hypergeometric2F1[1, -1 - 1/((1 - x)*x), -(1/((1 - x)*x)), x])/(5*(1 + 1/((
1 - x)*x))) - Defer[Int][(x^(-2 + 1/((-1 + x)*x))*Log[x])/(-1 + x)^2, x]/5 - (2*Defer[Int][(x^(-2 + 1/((-1 + x
)*x))*Log[x])/(-1 + x), x])/5

Rubi steps \begin{align*} \text {integral}& = \int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{x^2 \left (5-10 x+5 x^2\right )} \, dx \\ & = \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 (-1+x)^2 x^2} \, dx \\ & = \frac {1}{5} \int \frac {11 x^2-22 x^3+11 x^4+x^{\frac {1}{-x+x^2}} (-1+x+(1-2 x) \log (x))}{(-1+x)^2 x^2} \, dx \\ & = \frac {1}{5} \int \left (11-\frac {x^{-2+\frac {1}{(-1+x) x}} (1-x-\log (x)+2 x \log (x))}{(-1+x)^2}\right ) \, dx \\ & = \frac {11 x}{5}-\frac {1}{5} \int \frac {x^{-2+\frac {1}{(-1+x) x}} (1-x-\log (x)+2 x \log (x))}{(-1+x)^2} \, dx \\ & = \frac {11 x}{5}-\frac {1}{5} \int \left (\frac {x^{-2+\frac {1}{(-1+x) x}}}{1-x}+\frac {x^{-2+\frac {1}{(-1+x) x}} (-1+2 x) \log (x)}{(-1+x)^2}\right ) \, dx \\ & = \frac {11 x}{5}-\frac {1}{5} \int \frac {x^{-2+\frac {1}{(-1+x) x}}}{1-x} \, dx-\frac {1}{5} \int \frac {x^{-2+\frac {1}{(-1+x) x}} (-1+2 x) \log (x)}{(-1+x)^2} \, dx \\ & = \frac {11 x}{5}+\frac {x^{-1-\frac {1}{(1-x) x}} \operatorname {Hypergeometric2F1}\left (1,-1-\frac {1}{(1-x) x},-\frac {1}{(1-x) x},x\right )}{5 \left (1+\frac {1}{(1-x) x}\right )}-\frac {1}{5} \int \left (\frac {x^{-2+\frac {1}{(-1+x) x}} \log (x)}{(-1+x)^2}+\frac {2 x^{-2+\frac {1}{(-1+x) x}} \log (x)}{-1+x}\right ) \, dx \\ & = \frac {11 x}{5}+\frac {x^{-1-\frac {1}{(1-x) x}} \operatorname {Hypergeometric2F1}\left (1,-1-\frac {1}{(1-x) x},-\frac {1}{(1-x) x},x\right )}{5 \left (1+\frac {1}{(1-x) x}\right )}-\frac {1}{5} \int \frac {x^{-2+\frac {1}{(-1+x) x}} \log (x)}{(-1+x)^2} \, dx-\frac {2}{5} \int \frac {x^{-2+\frac {1}{(-1+x) x}} \log (x)}{-1+x} \, dx \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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]

[Out]

(11*x + x^(1/((-1 + x)*x)))/5

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	\(\frac {-\frac {11 x}{5}+\frac {11 x^{3}}{5}-\frac {x \,{\mathrm e}^{\frac {\ln \left (x \right )}{x^{2}-x}}}{5}+\frac {x^{2} {\mathrm e}^{\frac {\ln \left (x \right )}{x^{2}-x}}}{5}}{x \left (-1+x \right )}\)	\(53\)

[In]

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^4-10*x^3+5*x^2),x,method=_RETURNVERBOSE
)

[Out]

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

Fricas [A] (verification not implemented)

none

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

[In]

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="f
ricas")

[Out]

11/5*x + 1/5*x^(1/(x^2 - x))

Sympy [A] (verification not implemented)

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

[In]

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)

[Out]

11*x/5 + exp(log(x)/(x**2 - x))/5

Maxima [A] (verification not implemented)

none

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

[In]

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="m
axima")

[Out]

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

Giac [A] (verification not implemented)

none

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

[In]

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="g
iac")

[Out]

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

Mupad [B] (verification not implemented)

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

[In]

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
)

[Out]

(11*x)/5 + 1/(5*x^(1/(x - x^2)))