\(\int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx\) [212]

Optimal result

Integrand size = 58, antiderivative size = 21 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=-5+2 x+\log \left (\frac {5+e^2}{x^2}+x-\log (x)\right ) \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 x-2 \log (x)+\log \left (5+e^2+x^3-x^2 \log (x)\right ) \]

[In]

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

[Out]

2*x - 2*Log[x] + Log[5 + E^2 + x^3 - x^2*Log[x]]

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
risch	\(2 x +\ln \left (\ln \left (x \right )-\frac {x^{3}+{\mathrm e}^{2}+5}{x^{2}}\right )\)	\(21\)
default	\(-2 \ln \left (x \right )+2 x +\ln \left (-x^{2} \ln \left (x \right )+x^{3}+{\mathrm e}^{2}+5\right )\)	\(26\)
norman	\(-2 \ln \left (x \right )+2 x +\ln \left (-x^{2} \ln \left (x \right )+x^{3}+{\mathrm e}^{2}+5\right )\)	\(26\)
parallelrisch	\(-2 \ln \left (x \right )+2 x +\ln \left (-x^{2} \ln \left (x \right )+x^{3}+{\mathrm e}^{2}+5\right )\)	\(26\)

[In]

int((2*x^3*ln(x)+(2-2*x)*exp(1)^2-2*x^4-x^3+x^2-10*x+10)/(x^3*ln(x)-x*exp(1)^2-x^4-5*x),x,method=_RETURNVERBOS
E)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 \, x + \log \left (-\frac {x^{3} - x^{2} \log \left (x\right ) + e^{2} + 5}{x^{2}}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 x + \log {\left (\log {\left (x \right )} + \frac {- x^{3} - e^{2} - 5}{x^{2}} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 \, x + \log \left (-\frac {x^{3} - x^{2} \log \left (x\right ) + e^{2} + 5}{x^{2}}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 \, x + \log \left (-x^{3} + x^{2} \log \left (x\right ) - e^{2} - 5\right ) - 2 \, \log \left (x\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.42 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=\ln \left ({\mathrm {e}}^2-x^2\,\ln \left (x\right )+x^3+5\right )-\frac {2\,x^2\,\ln \left (x\right )-2\,x^3}{x^2} \]

[In]

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

[Out]

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