\(\int \frac {2+4 x+e^{-2 x} (-x+2 x^2)}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx\) [6204]

Optimal result

Integrand size = 47, antiderivative size = 25 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log \left (-5-2 x+\frac {1}{2} e^{3+x-3 (1+x)} x-\log (x)\right ) \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

Log[5 + 2*x + Log[x]] - 4*Defer[Int][1/((5 + 2*x + Log[x])*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])),
x] + 10*Defer[Int][x/((5 + 2*x + Log[x])*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])), x] + 4*Defer[Int
][x^2/((5 + 2*x + Log[x])*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])), x] - Defer[Int][Log[x]/((5 + 2*x
+ Log[x])*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])), x] + 2*Defer[Int][(x*Log[x])/((5 + 2*x + Log[x]
)*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])), x]

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

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log \left (10+\left (4-e^{-2 x}\right ) x+2 \log (x)\right ) \]

[In]

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

[Out]

Log[10 + (4 - E^(-2*x))*x + 2*Log[x]]

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64

method	result	size
risch	\(\ln \left (\ln \left (x \right )+2 x +5-\frac {{\mathrm e}^{-2 x} x}{2}\right )\)	\(16\)
parallelrisch	\(\ln \left ({\mathrm e}^{x} {\mathrm e}^{-3 x} x -2 \ln \left (x \right )-4 x -10\right )\)	\(19\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log \left (-x e^{\left (-2 \, x\right )} + 4 \, x + 2 \, \log \left (x\right ) + 10\right ) \]

[In]

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

[Out]

log(-x*e^(-2*x) + 4*x + 2*log(x) + 10)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log {\left (x \right )} + \log {\left (e^{- 2 x} + \frac {- 4 x - 2 \log {\left (x \right )} - 10}{x} \right )} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (17) = 34\).

Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=-2 \, x + \log \left (2 \, x + \log \left (x\right ) + 5\right ) + \log \left (\frac {2 \, {\left (2 \, x + \log \left (x\right ) + 5\right )} e^{\left (2 \, x\right )} - x}{2 \, {\left (2 \, x + \log \left (x\right ) + 5\right )}}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log \left (x e^{\left (-2 \, x\right )} - 4 \, x - 2 \, \log \left (x\right ) - 10\right ) \]

[In]

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

[Out]

log(x*e^(-2*x) - 4*x - 2*log(x) - 10)

Mupad [B] (verification not implemented)

Time = 12.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\ln \left (2\,x+\ln \left (x\right )-\frac {x\,{\mathrm {e}}^{-2\,x}}{2}+5\right ) \]

[In]

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

[Out]

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