\(\int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+(2 x-8 e^{2 x} x-8 x^2) \log (x)+(-2 e^{2 x}-2 x) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx\) [1812]

Optimal result

Integrand size = 78, antiderivative size = 33 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\frac {11}{3}-e^{2 x}-x+(1-x) x+\frac {x}{2+\frac {\log (x)}{x}} \]

[Out]

11/3+x*(1-x)-x-exp(x)^2+x/(ln(x)/x+2)

Rubi [F]

\[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx \]

[In]

Int[(-x + 2*x^2 - 8*E^(2*x)*x^2 - 8*x^3 + (2*x - 8*E^(2*x)*x - 8*x^2)*Log[x] + (-2*E^(2*x) - 2*x)*Log[x]^2)/(4
*x^2 + 4*x*Log[x] + Log[x]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{(2 x+\log (x))^2} \, dx \\ & = \int \left (-2 e^{2 x}-\frac {x}{(2 x+\log (x))^2}+\frac {2 x^2}{(2 x+\log (x))^2}-\frac {8 x^3}{(2 x+\log (x))^2}+\frac {2 x \log (x)}{(2 x+\log (x))^2}-\frac {8 x^2 \log (x)}{(2 x+\log (x))^2}-\frac {2 x \log ^2(x)}{(2 x+\log (x))^2}\right ) \, dx \\ & = -\left (2 \int e^{2 x} \, dx\right )+2 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+2 \int \frac {x \log (x)}{(2 x+\log (x))^2} \, dx-2 \int \frac {x \log ^2(x)}{(2 x+\log (x))^2} \, dx-8 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-8 \int \frac {x^2 \log (x)}{(2 x+\log (x))^2} \, dx-\int \frac {x}{(2 x+\log (x))^2} \, dx \\ & = -e^{2 x}+2 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+2 \int \left (-\frac {2 x^2}{(2 x+\log (x))^2}+\frac {x}{2 x+\log (x)}\right ) \, dx-2 \int \left (x+\frac {4 x^3}{(2 x+\log (x))^2}-\frac {4 x^2}{2 x+\log (x)}\right ) \, dx-8 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-8 \int \left (-\frac {2 x^3}{(2 x+\log (x))^2}+\frac {x^2}{2 x+\log (x)}\right ) \, dx-\int \frac {x}{(2 x+\log (x))^2} \, dx \\ & = -e^{2 x}-x^2+2 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+2 \int \frac {x}{2 x+\log (x)} \, dx-4 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-2 \left (8 \int \frac {x^3}{(2 x+\log (x))^2} \, dx\right )+16 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-\int \frac {x}{(2 x+\log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-e^{2 x}-x^2+\frac {x^2}{2 x+\log (x)} \]

[In]

Integrate[(-x + 2*x^2 - 8*E^(2*x)*x^2 - 8*x^3 + (2*x - 8*E^(2*x)*x - 8*x^2)*Log[x] + (-2*E^(2*x) - 2*x)*Log[x]
^2)/(4*x^2 + 4*x*Log[x] + Log[x]^2),x]

[Out]

-E^(2*x) - x^2 + x^2/(2*x + Log[x])

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76

method	result	size
risch	\(-x^{2}-{\mathrm e}^{2 x}+\frac {x^{2}}{2 x +\ln \left (x \right )}\)	\(25\)
parallelrisch	\(\frac {-2 x^{3}-x^{2} \ln \left (x \right )-2 x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (x \right )+x^{2}}{2 x +\ln \left (x \right )}\)	\(41\)
default	\(-\frac {x^{2} \ln \left (x \right )-x^{2}+2 x^{3}}{2 x +\ln \left (x \right )}+\frac {-2 x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (x \right )}{2 x +\ln \left (x \right )}\)	\(54\)

[In]

int(((-2*exp(x)^2-2*x)*ln(x)^2+(-8*x*exp(x)^2-8*x^2+2*x)*ln(x)-8*exp(x)^2*x^2-8*x^3+2*x^2-x)/(ln(x)^2+4*x*ln(x
)+4*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {2 \, x^{3} - x^{2} + 2 \, x e^{\left (2 \, x\right )} + {\left (x^{2} + e^{\left (2 \, x\right )}\right )} \log \left (x\right )}{2 \, x + \log \left (x\right )} \]

[In]

integrate(((-2*exp(x)^2-2*x)*log(x)^2+(-8*x*exp(x)^2-8*x^2+2*x)*log(x)-8*exp(x)^2*x^2-8*x^3+2*x^2-x)/(log(x)^2
+4*x*log(x)+4*x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.52 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=- x^{2} + \frac {x^{2}}{2 x + \log {\left (x \right )}} - e^{2 x} \]

[In]

integrate(((-2*exp(x)**2-2*x)*ln(x)**2+(-8*x*exp(x)**2-8*x**2+2*x)*ln(x)-8*exp(x)**2*x**2-8*x**3+2*x**2-x)/(ln
(x)**2+4*x*ln(x)+4*x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {2 \, x^{3} + x^{2} \log \left (x\right ) - x^{2} + {\left (2 \, x + \log \left (x\right )\right )} e^{\left (2 \, x\right )}}{2 \, x + \log \left (x\right )} \]

[In]

integrate(((-2*exp(x)^2-2*x)*log(x)^2+(-8*x*exp(x)^2-8*x^2+2*x)*log(x)-8*exp(x)^2*x^2-8*x^3+2*x^2-x)/(log(x)^2
+4*x*log(x)+4*x^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {2 \, x^{3} + x^{2} \log \left (x\right ) - x^{2} + 2 \, x e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \left (x\right )}{2 \, x + \log \left (x\right )} \]

[In]

integrate(((-2*exp(x)^2-2*x)*log(x)^2+(-8*x*exp(x)^2-8*x^2+2*x)*log(x)-8*exp(x)^2*x^2-8*x^3+2*x^2-x)/(log(x)^2
+4*x*log(x)+4*x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.63 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\frac {x^3}{x\,\ln \left (x\right )+2\,x^2}-x^2-{\mathrm {e}}^{2\,x} \]

[In]

int(-(x + 8*x^2*exp(2*x) + log(x)^2*(2*x + 2*exp(2*x)) + log(x)*(8*x*exp(2*x) - 2*x + 8*x^2) - 2*x^2 + 8*x^3)/
(log(x)^2 + 4*x*log(x) + 4*x^2),x)

[Out]

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