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

Optimal. Leaf size=23 \[ -4+2 x+\frac {1}{5} \left (5+x+x^{\frac {1}{(-1+x) x}}\right ) \]

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Rubi [F]  time = 1.43, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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}} \, _2F_1\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}} \, _2F_1\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 19, normalized size = 0.83 \begin {gather*} \frac {1}{5} \left (11 x+x^{\frac {1}{(-1+x) x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.53, size = 17, normalized size = 0.74 \begin {gather*} \frac {11}{5} \, x + \frac {1}{5} \, x^{\left (\frac {1}{x^{2} - x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.88, size = 34, normalized size = 1.48 \begin {gather*} \frac {11}{5} \, x + \frac {e^{\left (\frac {x^{2} \log \relax (x) - x \log \relax (x) + \log \relax (x)}{x^{2} - x}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.09, size = 18, normalized size = 0.78

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

Verification of antiderivative is not currently implemented for this CAS.

[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/(x-1))

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maxima [A]  time = 0.44, size = 23, normalized size = 1.00 \begin {gather*} \frac {11}{5} \, x + \frac {1}{5} \, e^{\left (\frac {\log \relax (x)}{x - 1} - \frac {\log \relax (x)}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 5.59, size = 19, normalized size = 0.83 \begin {gather*} \frac {11\,x}{5}+\frac {1}{5\,x^{\frac {1}{x-x^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 0.40, size = 15, normalized size = 0.65 \begin {gather*} \frac {11 x}{5} + \frac {e^{\frac {\log {\relax (x )}}{x^{2} - x}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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