\(\int \frac {(-3 x-2 x^2) \log (x^2)+(4+6 x+2 x^2) \log (2+3 x+x^2)+(-2 x-3 x^2-x^3) \log ^2(2+3 x+x^2)+e^{e^x} (3 x+2 x^2+e^x (-2 x-3 x^2-x^3) \log (2+3 x+x^2))}{(2 x+3 x^2+x^3) \log ^2(2+3 x+x^2)} \, dx\) [5027]

Optimal result

Integrand size = 129, antiderivative size = 32 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-x+\frac {-e^{e^x}+\log \left (x^2\right )}{\log \left (\frac {(2+x) \left (x+x^2\right )}{x}\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=\int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx \]

[In]

Int[((-3*x - 2*x^2)*Log[x^2] + (4 + 6*x + 2*x^2)*Log[2 + 3*x + x^2] + (-2*x - 3*x^2 - x^3)*Log[2 + 3*x + x^2]^
2 + E^E^x*(3*x + 2*x^2 + E^x*(-2*x - 3*x^2 - x^3)*Log[2 + 3*x + x^2]))/((2*x + 3*x^2 + x^3)*Log[2 + 3*x + x^2]
^2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-x-\frac {e^{e^x}}{\log \left (2+3 x+x^2\right )}+\frac {\log \left (x^2\right )}{\log \left (2+3 x+x^2\right )} \]

[In]

Integrate[((-3*x - 2*x^2)*Log[x^2] + (4 + 6*x + 2*x^2)*Log[2 + 3*x + x^2] + (-2*x - 3*x^2 - x^3)*Log[2 + 3*x +
x^2]^2 + E^E^x*(3*x + 2*x^2 + E^x*(-2*x - 3*x^2 - x^3)*Log[2 + 3*x + x^2]))/((2*x + 3*x^2 + x^3)*Log[2 + 3*x
+ x^2]^2),x]

[Out]

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

Maple [A] (verified)

Time = 59.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53

method	result	size
parallelrisch	\(\frac {-6 x \ln \left (x^{2}+3 x +2\right )+6 \ln \left (x^{2}\right )-6 \,{\mathrm e}^{{\mathrm e}^{x}}+13 \ln \left (x^{2}+3 x +2\right )}{6 \ln \left (x^{2}+3 x +2\right )}\)	\(49\)
risch	\(-x +\frac {4 \ln \left (x \right )-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2 \ln \left (x^{2}+3 x +2\right )}-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{\ln \left (x^{2}+3 x +2\right )}\)	\(88\)

[In]

int((((-x^3-3*x^2-2*x)*exp(x)*ln(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3*x^2-2*x)*ln(x^2+3*x+2)^2+(2*x^2+6*x
+4)*ln(x^2+3*x+2)+(-2*x^2-3*x)*ln(x^2))/(x^3+3*x^2+2*x)/ln(x^2+3*x+2)^2,x,method=_RETURNVERBOSE)

[Out]

1/6*(-6*x*ln(x^2+3*x+2)+6*ln(x^2)-6*exp(exp(x))+13*ln(x^2+3*x+2))/ln(x^2+3*x+2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \]

[In]

integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2
*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x)*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=- x - \frac {e^{e^{x}}}{\log {\left (x^{2} + 3 x + 2 \right )}} + \frac {\log {\left (x^{2} \right )}}{\log {\left (x^{2} + 3 x + 2 \right )}} \]

[In]

integrate((((-x**3-3*x**2-2*x)*exp(x)*ln(x**2+3*x+2)+2*x**2+3*x)*exp(exp(x))+(-x**3-3*x**2-2*x)*ln(x**2+3*x+2)
**2+(2*x**2+6*x+4)*ln(x**2+3*x+2)+(-2*x**2-3*x)*ln(x**2))/(x**3+3*x**2+2*x)/ln(x**2+3*x+2)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x + 2\right ) + x \log \left (x + 1\right ) + e^{\left (e^{x}\right )} - 2 \, \log \left (x\right )}{\log \left (x + 2\right ) + \log \left (x + 1\right )} \]

[In]

integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2
*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x)*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \]

[In]

integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2
*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x)*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=\int \frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (3\,x+2\,x^2-{\mathrm {e}}^x\,\ln \left (x^2+3\,x+2\right )\,\left (x^3+3\,x^2+2\,x\right )\right )-\ln \left (x^2\right )\,\left (2\,x^2+3\,x\right )+\ln \left (x^2+3\,x+2\right )\,\left (2\,x^2+6\,x+4\right )-{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )}{{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )} \,d x \]

[In]

int((exp(exp(x))*(3*x + 2*x^2 - exp(x)*log(3*x + x^2 + 2)*(2*x + 3*x^2 + x^3)) - log(x^2)*(3*x + 2*x^2) + log(
3*x + x^2 + 2)*(6*x + 2*x^2 + 4) - log(3*x + x^2 + 2)^2*(2*x + 3*x^2 + x^3))/(log(3*x + x^2 + 2)^2*(2*x + 3*x^
2 + x^3)),x)

[Out]

int((exp(exp(x))*(3*x + 2*x^2 - exp(x)*log(3*x + x^2 + 2)*(2*x + 3*x^2 + x^3)) - log(x^2)*(3*x + 2*x^2) + log(
3*x + x^2 + 2)*(6*x + 2*x^2 + 4) - log(3*x + x^2 + 2)^2*(2*x + 3*x^2 + x^3))/(log(3*x + x^2 + 2)^2*(2*x + 3*x^
2 + x^3)), x)