\(\int \frac {-11 x^2-4 x^3+e^x (4 x^2+4 x^3)+(-12 x-3 x^2+e^x (4 x+4 x^2)) \log (x)+(-3+e^x (1+x)) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx\) [8666]

Optimal result

Integrand size = 83, antiderivative size = 20 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=x \left (-3+e^x-\frac {x^2}{2 x+\log (x)}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx \]

[In]

Int[(-11*x^2 - 4*x^3 + E^x*(4*x^2 + 4*x^3) + (-12*x - 3*x^2 + E^x*(4*x + 4*x^2))*Log[x] + (-3 + E^x*(1 + x))*L
og[x]^2)/(4*x^2 + 4*x*Log[x] + Log[x]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \left (-11-4 x+4 e^x (1+x)\right )+x \left (4 e^x (1+x)-3 (4+x)\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{(2 x+\log (x))^2} \, dx \\ & = \int \left (e^x (1+x)-\frac {11 x^2}{(2 x+\log (x))^2}-\frac {4 x^3}{(2 x+\log (x))^2}-\frac {3 x (4+x) \log (x)}{(2 x+\log (x))^2}-\frac {3 \log ^2(x)}{(2 x+\log (x))^2}\right ) \, dx \\ & = -\left (3 \int \frac {x (4+x) \log (x)}{(2 x+\log (x))^2} \, dx\right )-3 \int \frac {\log ^2(x)}{(2 x+\log (x))^2} \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+\int e^x (1+x) \, dx \\ & = e^x (1+x)-3 \int \left (1+\frac {4 x^2}{(2 x+\log (x))^2}-\frac {4 x}{2 x+\log (x)}\right ) \, dx-3 \int \left (-\frac {2 x^2 (4+x)}{(2 x+\log (x))^2}+\frac {x (4+x)}{2 x+\log (x)}\right ) \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-\int e^x \, dx \\ & = -e^x-3 x+e^x (1+x)-3 \int \frac {x (4+x)}{2 x+\log (x)} \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx+6 \int \frac {x^2 (4+x)}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-12 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+12 \int \frac {x}{2 x+\log (x)} \, dx \\ & = -e^x-3 x+e^x (1+x)-3 \int \left (\frac {4 x}{2 x+\log (x)}+\frac {x^2}{2 x+\log (x)}\right ) \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx+6 \int \left (\frac {4 x^2}{(2 x+\log (x))^2}+\frac {x^3}{(2 x+\log (x))^2}\right ) \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-12 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+12 \int \frac {x}{2 x+\log (x)} \, dx \\ & = -e^x-3 x+e^x (1+x)-3 \int \frac {x^2}{2 x+\log (x)} \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx+6 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-12 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+24 \int \frac {x^2}{(2 x+\log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-3 x+e^x x-\frac {x^3}{2 x+\log (x)} \]

[In]

Integrate[(-11*x^2 - 4*x^3 + E^x*(4*x^2 + 4*x^3) + (-12*x - 3*x^2 + E^x*(4*x + 4*x^2))*Log[x] + (-3 + E^x*(1 +
x))*Log[x]^2)/(4*x^2 + 4*x*Log[x] + Log[x]^2),x]

[Out]

-3*x + E^x*x - x^3/(2*x + Log[x])

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

method	result	size
risch	\({\mathrm e}^{x} x -3 x -\frac {x^{3}}{2 x +\ln \left (x \right )}\)	\(22\)
parallelrisch	\(\frac {-x^{3}+2 \,{\mathrm e}^{x} x^{2}+x \,{\mathrm e}^{x} \ln \left (x \right )-6 x^{2}-3 x \ln \left (x \right )}{2 x +\ln \left (x \right )}\)	\(39\)

[In]

int((((1+x)*exp(x)-3)*ln(x)^2+((4*x^2+4*x)*exp(x)-3*x^2-12*x)*ln(x)+(4*x^3+4*x^2)*exp(x)-4*x^3-11*x^2)/(ln(x)^
2+4*x*ln(x)+4*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {x^{3} - 2 \, x^{2} e^{x} + 6 \, x^{2} - {\left (x e^{x} - 3 \, x\right )} \log \left (x\right )}{2 \, x + \log \left (x\right )} \]

[In]

integrate((((1+x)*exp(x)-3)*log(x)^2+((4*x^2+4*x)*exp(x)-3*x^2-12*x)*log(x)+(4*x^3+4*x^2)*exp(x)-4*x^3-11*x^2)
/(log(x)^2+4*x*log(x)+4*x^2),x, algorithm="fricas")

[Out]

-(x^3 - 2*x^2*e^x + 6*x^2 - (x*e^x - 3*x)*log(x))/(2*x + log(x))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=- \frac {x^{3}}{2 x + \log {\left (x \right )}} + x e^{x} - 3 x \]

[In]

integrate((((1+x)*exp(x)-3)*ln(x)**2+((4*x**2+4*x)*exp(x)-3*x**2-12*x)*ln(x)+(4*x**3+4*x**2)*exp(x)-4*x**3-11*
x**2)/(ln(x)**2+4*x*ln(x)+4*x**2),x)

[Out]

-x**3/(2*x + log(x)) + x*exp(x) - 3*x

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {x^{3} + 6 \, x^{2} - {\left (2 \, x^{2} + x \log \left (x\right )\right )} e^{x} + 3 \, x \log \left (x\right )}{2 \, x + \log \left (x\right )} \]

[In]

integrate((((1+x)*exp(x)-3)*log(x)^2+((4*x^2+4*x)*exp(x)-3*x^2-12*x)*log(x)+(4*x^3+4*x^2)*exp(x)-4*x^3-11*x^2)
/(log(x)^2+4*x*log(x)+4*x^2),x, algorithm="maxima")

[Out]

-(x^3 + 6*x^2 - (2*x^2 + x*log(x))*e^x + 3*x*log(x))/(2*x + log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {x^{3} - 2 \, x^{2} e^{x} - x e^{x} \log \left (x\right ) + 6 \, x^{2} + 3 \, x \log \left (x\right )}{2 \, x + \log \left (x\right )} \]

[In]

integrate((((1+x)*exp(x)-3)*log(x)^2+((4*x^2+4*x)*exp(x)-3*x^2-12*x)*log(x)+(4*x^3+4*x^2)*exp(x)-4*x^3-11*x^2)
/(log(x)^2+4*x*log(x)+4*x^2),x, algorithm="giac")

[Out]

-(x^3 - 2*x^2*e^x - x*e^x*log(x) + 6*x^2 + 3*x*log(x))/(2*x + log(x))

Mupad [B] (verification not implemented)

Time = 14.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.25 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\frac {3}{16\,\left (x+\frac {1}{2}\right )}-\frac {9\,x}{4}+x\,{\mathrm {e}}^x+\frac {\frac {3\,x^3\,\ln \left (x\right )}{2\,x+1}-\frac {x\,\left (x^2-4\,x^3\right )}{2\,x+1}}{2\,x+\ln \left (x\right )}-\frac {3\,x^2}{2} \]

[In]

int(-(log(x)*(12*x - exp(x)*(4*x + 4*x^2) + 3*x^2) - log(x)^2*(exp(x)*(x + 1) - 3) - exp(x)*(4*x^2 + 4*x^3) +
11*x^2 + 4*x^3)/(log(x)^2 + 4*x*log(x) + 4*x^2),x)

[Out]

3/(16*(x + 1/2)) - (9*x)/4 + x*exp(x) + ((3*x^3*log(x))/(2*x + 1) - (x*(x^2 - 4*x^3))/(2*x + 1))/(2*x + log(x)
) - (3*x^2)/2