\(\int \frac {e^{4/3} (-x+3 x^2-2 x^3+x^4)+e^{4/3} (-2 x+4 x^2-2 x^3) \log (x)+e^{4/3} (-2 x+x^2) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+(10-20 x+20 x^2-10 x^3) \log (x)+(5-10 x+5 x^2) \log ^2(x)} \, dx\) [1204]

Optimal result

Integrand size = 119, antiderivative size = 25 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {e^{4/3} x^2}{5 \left (-1+x+\frac {1}{x-\log (x)}\right )} \]

[Out]

1/5*x^2/(1/(x-ln(x))+x-1)*exp(2/3)^2

Rubi [F]

\[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx \]

[In]

Int[(E^(4/3)*(-x + 3*x^2 - 2*x^3 + x^4) + E^(4/3)*(-2*x + 4*x^2 - 2*x^3)*Log[x] + E^(4/3)*(-2*x + x^2)*Log[x]^
2)/(5 - 10*x + 15*x^2 - 10*x^3 + 5*x^4 + (10 - 20*x + 20*x^2 - 10*x^3)*Log[x] + (5 - 10*x + 5*x^2)*Log[x]^2),x
]

[Out]

-1/5*(E^(4/3)*x^2)/(1 - x) - (E^(4/3)*Defer[Int][(1 - x + x^2 + Log[x] - x*Log[x])^(-2), x])/5 - (E^(4/3)*Defe
r[Int][1/((-1 + x)^2*(1 - x + x^2 + Log[x] - x*Log[x])^2), x])/5 - (2*E^(4/3)*Defer[Int][1/((-1 + x)*(1 - x +
x^2 + Log[x] - x*Log[x])^2), x])/5 - (E^(4/3)*Defer[Int][x/(1 - x + x^2 + Log[x] - x*Log[x])^2, x])/5 + (E^(4/
3)*Defer[Int][x^2/(1 - x + x^2 + Log[x] - x*Log[x])^2, x])/5 + (2*E^(4/3)*Defer[Int][1/((-1 + x)^2*(1 - x + x^
2 + Log[x] - x*Log[x])), x])/5 + (2*E^(4/3)*Defer[Int][1/((-1 + x)*(1 - x + x^2 + Log[x] - x*Log[x])), x])/5

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {1}{5} e^{4/3} \left (\frac {1}{-1+x}+x+\frac {x^2}{(-1+x) \left (-1+x-x^2-\log (x)+x \log (x)\right )}\right ) \]

[In]

Integrate[(E^(4/3)*(-x + 3*x^2 - 2*x^3 + x^4) + E^(4/3)*(-2*x + 4*x^2 - 2*x^3)*Log[x] + E^(4/3)*(-2*x + x^2)*L
og[x]^2)/(5 - 10*x + 15*x^2 - 10*x^3 + 5*x^4 + (10 - 20*x + 20*x^2 - 10*x^3)*Log[x] + (5 - 10*x + 5*x^2)*Log[x
]^2),x]

[Out]

(E^(4/3)*((-1 + x)^(-1) + x + x^2/((-1 + x)*(-1 + x - x^2 - Log[x] + x*Log[x]))))/5

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48

method	result	size
default	\(\frac {{\mathrm e}^{\frac {4}{3}} \left (x^{2} \ln \left (x \right )-x^{3}\right )}{5 x \ln \left (x \right )-5 x^{2}-5 \ln \left (x \right )+5 x -5}\)	\(37\)
norman	\(\frac {\frac {{\mathrm e}^{\frac {4}{3}} x^{3}}{5}-\frac {{\mathrm e}^{\frac {4}{3}} x^{2} \ln \left (x \right )}{5}}{x^{2}-x \ln \left (x \right )-x +\ln \left (x \right )+1}\)	\(40\)
parallelrisch	\(\frac {{\mathrm e}^{\frac {4}{3}} x^{3}-{\mathrm e}^{\frac {4}{3}} x^{2} \ln \left (x \right )}{5 x^{2}-5 x \ln \left (x \right )-5 x +5 \ln \left (x \right )+5}\)	\(40\)
risch	\(\frac {{\mathrm e}^{\frac {4}{3}} \left (x^{2}-x +1\right )}{5 x -5}-\frac {{\mathrm e}^{\frac {4}{3}} x^{2}}{5 \left (-1+x \right ) \left (x^{2}-x \ln \left (x \right )-x +\ln \left (x \right )+1\right )}\)	\(48\)

[In]

int(((x^2-2*x)*exp(2/3)^2*ln(x)^2+(-2*x^3+4*x^2-2*x)*exp(2/3)^2*ln(x)+(x^4-2*x^3+3*x^2-x)*exp(2/3)^2)/((5*x^2-
10*x+5)*ln(x)^2+(-10*x^3+20*x^2-20*x+10)*ln(x)+5*x^4-10*x^3+15*x^2-10*x+5),x,method=_RETURNVERBOSE)

[Out]

1/5*exp(2/3)^2*(x^2*ln(x)-x^3)/(x*ln(x)-x^2-ln(x)+x-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=-\frac {{\left (x^{2} - x + 1\right )} e^{\frac {4}{3}} \log \left (x\right ) - {\left (x^{3} - x^{2} + x - 1\right )} e^{\frac {4}{3}}}{5 \, {\left (x^{2} - {\left (x - 1\right )} \log \left (x\right ) - x + 1\right )}} \]

[In]

integrate(((x^2-2*x)*exp(2/3)^2*log(x)^2+(-2*x^3+4*x^2-2*x)*exp(2/3)^2*log(x)+(x^4-2*x^3+3*x^2-x)*exp(2/3)^2)/
((5*x^2-10*x+5)*log(x)^2+(-10*x^3+20*x^2-20*x+10)*log(x)+5*x^4-10*x^3+15*x^2-10*x+5),x, algorithm="fricas")

[Out]

-1/5*((x^2 - x + 1)*e^(4/3)*log(x) - (x^3 - x^2 + x - 1)*e^(4/3))/(x^2 - (x - 1)*log(x) - x + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (19) = 38\).

Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {x^{2} e^{\frac {4}{3}}}{- 5 x^{3} + 10 x^{2} - 10 x + \left (5 x^{2} - 10 x + 5\right ) \log {\left (x \right )} + 5} + \frac {x e^{\frac {4}{3}}}{5} + \frac {e^{\frac {4}{3}}}{5 x - 5} \]

[In]

integrate(((x**2-2*x)*exp(2/3)**2*ln(x)**2+(-2*x**3+4*x**2-2*x)*exp(2/3)**2*ln(x)+(x**4-2*x**3+3*x**2-x)*exp(2
/3)**2)/((5*x**2-10*x+5)*ln(x)**2+(-10*x**3+20*x**2-20*x+10)*ln(x)+5*x**4-10*x**3+15*x**2-10*x+5),x)

[Out]

x**2*exp(4/3)/(-5*x**3 + 10*x**2 - 10*x + (5*x**2 - 10*x + 5)*log(x) + 5) + x*exp(4/3)/5 + exp(4/3)/(5*x - 5)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (20) = 40\).

Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.36 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {x^{3} e^{\frac {4}{3}} - x^{2} e^{\frac {4}{3}} + x e^{\frac {4}{3}} - {\left (x^{2} e^{\frac {4}{3}} - x e^{\frac {4}{3}} + e^{\frac {4}{3}}\right )} \log \left (x\right ) - e^{\frac {4}{3}}}{5 \, {\left (x^{2} - {\left (x - 1\right )} \log \left (x\right ) - x + 1\right )}} \]

[In]

integrate(((x^2-2*x)*exp(2/3)^2*log(x)^2+(-2*x^3+4*x^2-2*x)*exp(2/3)^2*log(x)+(x^4-2*x^3+3*x^2-x)*exp(2/3)^2)/
((5*x^2-10*x+5)*log(x)^2+(-10*x^3+20*x^2-20*x+10)*log(x)+5*x^4-10*x^3+15*x^2-10*x+5),x, algorithm="maxima")

[Out]

1/5*(x^3*e^(4/3) - x^2*e^(4/3) + x*e^(4/3) - (x^2*e^(4/3) - x*e^(4/3) + e^(4/3))*log(x) - e^(4/3))/(x^2 - (x -
1)*log(x) - x + 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.48 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {x^{3} e^{\frac {4}{3}} - x^{2} e^{\frac {4}{3}} \log \left (x\right ) - x^{2} e^{\frac {4}{3}} + x e^{\frac {4}{3}} \log \left (x\right ) + x e^{\frac {4}{3}} - e^{\frac {4}{3}} \log \left (x\right ) - e^{\frac {4}{3}}}{5 \, {\left (x^{2} - x \log \left (x\right ) - x + \log \left (x\right ) + 1\right )}} \]

[In]

integrate(((x^2-2*x)*exp(2/3)^2*log(x)^2+(-2*x^3+4*x^2-2*x)*exp(2/3)^2*log(x)+(x^4-2*x^3+3*x^2-x)*exp(2/3)^2)/
((5*x^2-10*x+5)*log(x)^2+(-10*x^3+20*x^2-20*x+10)*log(x)+5*x^4-10*x^3+15*x^2-10*x+5),x, algorithm="giac")

[Out]

1/5*(x^3*e^(4/3) - x^2*e^(4/3)*log(x) - x^2*e^(4/3) + x*e^(4/3)*log(x) + x*e^(4/3) - e^(4/3)*log(x) - e^(4/3))
/(x^2 - x*log(x) - x + log(x) + 1)

Mupad [B] (verification not implemented)

Time = 8.63 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=-\frac {{\mathrm {e}}^{4/3}\,\left (\ln \left (x\right )-x+x^2\,\ln \left (x\right )-x\,\ln \left (x\right )+x^2-x^3+1\right )}{5\,\left (\ln \left (x\right )-x-x\,\ln \left (x\right )+x^2+1\right )} \]

[In]

int(-(exp(4/3)*(x - 3*x^2 + 2*x^3 - x^4) + exp(4/3)*log(x)*(2*x - 4*x^2 + 2*x^3) + exp(4/3)*log(x)^2*(2*x - x^
2))/(log(x)^2*(5*x^2 - 10*x + 5) - 10*x + 15*x^2 - 10*x^3 + 5*x^4 - log(x)*(20*x - 20*x^2 + 10*x^3 - 10) + 5),
x)

[Out]

-(exp(4/3)*(log(x) - x + x^2*log(x) - x*log(x) + x^2 - x^3 + 1))/(5*(log(x) - x - x*log(x) + x^2 + 1))