\(\int \frac {e^{\frac {e^x+x^2}{x^3 \log (-2+e^{3/2}) \log (4+2 x)}} (-e^x x-x^3+(-2 x^2-x^3+e^x (-6-x+x^2)) \log (4+2 x))}{(2 x^4+x^5) \log (-2+e^{3/2}) \log ^2(4+2 x)} \, dx\) [8821]

Optimal result

Integrand size = 103, antiderivative size = 33 \[ \int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx=e^{\frac {\frac {e^x}{x}+x}{x^2 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \]

[Out]

exp((exp(x)/x+x)/x^2/ln(4+2*x)/ln(exp(3/2)-2))

Rubi [F]

\[ \int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx=\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx \]

[In]

Int[(E^((E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))*(-(E^x*x) - x^3 + (-2*x^2 - x^3 + E^x*(-6 - x + x^2)
)*Log[4 + 2*x]))/((2*x^4 + x^5)*Log[-2 + E^(3/2)]*Log[4 + 2*x]^2),x]

[Out]

-1/2*Defer[Int][E^(x + (E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))/(x^3*Log[4 + 2*x]^2), x]/Log[-2 + E^(
3/2)] + Defer[Int][E^(x + (E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))/(x^2*Log[4 + 2*x]^2), x]/(4*Log[-2
+ E^(3/2)]) - Defer[Int][E^(x + (E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))/(x*Log[4 + 2*x]^2), x]/(8*L
og[-2 + E^(3/2)]) - Defer[Int][E^((E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))/(x*Log[4 + 2*x]^2), x]/(2*
Log[-2 + E^(3/2)]) + Defer[Int][E^(x + (E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))/((2 + x)*Log[4 + 2*x]
^2), x]/(8*Log[-2 + E^(3/2)]) + Defer[Int][E^((E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))/((2 + x)*Log[4
+ 2*x]^2), x]/(2*Log[-2 + E^(3/2)]) + (3*Defer[Int][E^(x + (E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))/
((-2 - x)*Log[4 + 2*x]), x])/(8*Log[-2 + E^(3/2)]) + Defer[Int][E^((E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 +
2*x]))/((-2 - x)*Log[4 + 2*x]), x]/(2*Log[-2 + E^(3/2)]) - (3*Defer[Int][E^(x + (E^x + x^2)/(x^3*Log[-2 + E^(3
/2)]*Log[4 + 2*x]))/(x^4*Log[4 + 2*x]), x])/Log[-2 + E^(3/2)] + Defer[Int][E^(x + (E^x + x^2)/(x^3*Log[-2 + E^
(3/2)]*Log[4 + 2*x]))/(x^3*Log[4 + 2*x]), x]/Log[-2 + E^(3/2)] - Defer[Int][E^((E^x + x^2)/(x^3*Log[-2 + E^(3/
2)]*Log[4 + 2*x]))/(x^2*Log[4 + 2*x]), x]/Log[-2 + E^(3/2)] + (3*Defer[Int][E^(x + (E^x + x^2)/(x^3*Log[-2 + E
^(3/2)]*Log[4 + 2*x]))/((2 + x)*Log[4 + 2*x]), x])/(8*Log[-2 + E^(3/2)]) + Defer[Int][E^((E^x + x^2)/(x^3*Log[
-2 + E^(3/2)]*Log[4 + 2*x]))/((2 + x)*Log[4 + 2*x]), x]/(2*Log[-2 + E^(3/2)])

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )} \\ & = \frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{x^4 (2+x) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )} \\ & = \frac {\int \left (\frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{x^2 (2+x) \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{x^4 (2+x) \log ^2(4+2 x)}\right ) \, dx}{\log \left (-2+e^{3/2}\right )} \\ & = \frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{x^2 (2+x) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{x^4 (2+x) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )} \\ & = \frac {\int \left (\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (x+6 \log (2 (2+x))+x \log (2 (2+x))-x^2 \log (2 (2+x))\right )}{4 x^3 \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (x+6 \log (2 (2+x))+x \log (2 (2+x))-x^2 \log (2 (2+x))\right )}{16 x \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{2 x^4 \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{8 x^2 \log ^2(4+2 x)}+\frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{16 (2+x) \log ^2(4+2 x)}\right ) \, dx}{\log \left (-2+e^{3/2}\right )}+\frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-(2+x) \log (2 (2+x)))}{x^2 (2+x) \log ^2(4+2 x)} \, dx}{\log \left (-2+e^{3/2}\right )} \\ & = \frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (x+6 \log (2 (2+x))+x \log (2 (2+x))-x^2 \log (2 (2+x))\right )}{x \log ^2(4+2 x)} \, dx}{16 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{(2+x) \log ^2(4+2 x)} \, dx}{16 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{x^2 \log ^2(4+2 x)} \, dx}{8 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (x+6 \log (2 (2+x))+x \log (2 (2+x))-x^2 \log (2 (2+x))\right )}{x^3 \log ^2(4+2 x)} \, dx}{4 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {\exp \left (x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}\right ) \left (-x-6 \log (2 (2+x))-x \log (2 (2+x))+x^2 \log (2 (2+x))\right )}{x^4 \log ^2(4+2 x)} \, dx}{2 \log \left (-2+e^{3/2}\right )}+\frac {\int \left (\frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{2 x^2 \log ^2(4+2 x)}+\frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{4 (2+x) \log ^2(4+2 x)}+\frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (x+2 \log (2 (2+x))+x \log (2 (2+x)))}{4 x \log ^2(4+2 x)}\right ) \, dx}{\log \left (-2+e^{3/2}\right )} \\ & = \frac {\int \left (\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{\log ^2(4+2 x)}+\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{\log (4+2 x)}+\frac {6 e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x \log (4+2 x)}-\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} x}{\log (4+2 x)}\right ) \, dx}{16 \log \left (-2+e^{3/2}\right )}+\frac {\int \left (\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} x}{(-2-x) \log ^2(4+2 x)}+\frac {6 e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{(-2-x) \log (4+2 x)}+\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} x}{(-2-x) \log (4+2 x)}+\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} x^2}{(2+x) \log (4+2 x)}\right ) \, dx}{16 \log \left (-2+e^{3/2}\right )}+\frac {\int \left (-\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x \log ^2(4+2 x)}+\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{\log (4+2 x)}-\frac {6 e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x^2 \log (4+2 x)}-\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x \log (4+2 x)}\right ) \, dx}{8 \log \left (-2+e^{3/2}\right )}+\frac {\int \left (\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x^2 \log ^2(4+2 x)}+\frac {6 e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x^3 \log (4+2 x)}+\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x^2 \log (4+2 x)}-\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x \log (4+2 x)}\right ) \, dx}{4 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{(2+x) \log ^2(4+2 x)} \, dx}{4 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (x+2 \log (2 (2+x))+x \log (2 (2+x)))}{x \log ^2(4+2 x)} \, dx}{4 \log \left (-2+e^{3/2}\right )}+\frac {\int \left (-\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x^3 \log ^2(4+2 x)}-\frac {6 e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x^4 \log (4+2 x)}-\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x^3 \log (4+2 x)}+\frac {e^{x+\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}}}{x^2 \log (4+2 x)}\right ) \, dx}{2 \log \left (-2+e^{3/2}\right )}+\frac {\int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} (-x-2 \log (2 (2+x))-x \log (2 (2+x)))}{x^2 \log ^2(4+2 x)} \, dx}{2 \log \left (-2+e^{3/2}\right )} \\ & = \text {Too large to display} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx=e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (2 (2+x))}} \]

[In]

Integrate[(E^((E^x + x^2)/(x^3*Log[-2 + E^(3/2)]*Log[4 + 2*x]))*(-(E^x*x) - x^3 + (-2*x^2 - x^3 + E^x*(-6 - x
+ x^2))*Log[4 + 2*x]))/((2*x^4 + x^5)*Log[-2 + E^(3/2)]*Log[4 + 2*x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82

\[{\mathrm e}^{\frac {x^{2}+{\mathrm e}^{x}}{x^{3} \ln \left (4+2 x \right ) \ln \left ({\mathrm e}^{\frac {3}{2}}-2\right )}}\]

[In]

int((((x^2-x-6)*exp(x)-x^3-2*x^2)*ln(4+2*x)-exp(x)*x-x^3)*exp((x^2+exp(x))/x^3/ln(4+2*x)/ln(exp(3/2)-2))/(x^5+
2*x^4)/ln(4+2*x)^2/ln(exp(3/2)-2),x)

[Out]

exp((x^2+exp(x))/x^3/ln(4+2*x)/ln(exp(3/2)-2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx=e^{\left (\frac {x^{2} + e^{x}}{x^{3} \log \left (2 \, x + 4\right ) \log \left (e^{\frac {3}{2}} - 2\right )}\right )} \]

[In]

integrate((((x^2-x-6)*exp(x)-x^3-2*x^2)*log(4+2*x)-exp(x)*x-x^3)*exp((x^2+exp(x))/x^3/log(4+2*x)/log(exp(3/2)-
2))/(x^5+2*x^4)/log(4+2*x)^2/log(exp(3/2)-2),x, algorithm="fricas")

[Out]

e^((x^2 + e^x)/(x^3*log(2*x + 4)*log(e^(3/2) - 2)))

Sympy [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx=e^{\frac {x^{2} + e^{x}}{x^{3} \log {\left (-2 + e^{\frac {3}{2}} \right )} \log {\left (2 x + 4 \right )}}} \]

[In]

integrate((((x**2-x-6)*exp(x)-x**3-2*x**2)*ln(4+2*x)-exp(x)*x-x**3)*exp((x**2+exp(x))/x**3/ln(4+2*x)/ln(exp(3/
2)-2))/(x**5+2*x**4)/ln(4+2*x)**2/ln(exp(3/2)-2),x)

[Out]

exp((x**2 + exp(x))/(x**3*log(-2 + exp(3/2))*log(2*x + 4)))

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((((x^2-x-6)*exp(x)-x^3-2*x^2)*log(4+2*x)-exp(x)*x-x^3)*exp((x^2+exp(x))/x^3/log(4+2*x)/log(exp(3/2)-
2))/(x^5+2*x^4)/log(4+2*x)^2/log(exp(3/2)-2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

Giac [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx=e^{\left (\frac {1}{x \log \left (2 \, x + 4\right ) \log \left (e^{\frac {3}{2}} - 2\right )} + \frac {e^{x}}{x^{3} \log \left (2 \, x + 4\right ) \log \left (e^{\frac {3}{2}} - 2\right )}\right )} \]

[In]

integrate((((x^2-x-6)*exp(x)-x^3-2*x^2)*log(4+2*x)-exp(x)*x-x^3)*exp((x^2+exp(x))/x^3/log(4+2*x)/log(exp(3/2)-
2))/(x^5+2*x^4)/log(4+2*x)^2/log(exp(3/2)-2),x, algorithm="giac")

[Out]

e^(1/(x*log(2*x + 4)*log(e^(3/2) - 2)) + e^x/(x^3*log(2*x + 4)*log(e^(3/2) - 2)))

Mupad [B] (verification not implemented)

Time = 14.64 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {e^x+x^2}{x^3 \log \left (-2+e^{3/2}\right ) \log (4+2 x)}} \left (-e^x x-x^3+\left (-2 x^2-x^3+e^x \left (-6-x+x^2\right )\right ) \log (4+2 x)\right )}{\left (2 x^4+x^5\right ) \log \left (-2+e^{3/2}\right ) \log ^2(4+2 x)} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^x+x^2}{x^3\,\ln \left ({\mathrm {e}}^{3/2}-2\right )\,\ln \left (2\,x+4\right )}} \]

[In]

int(-(exp((exp(x) + x^2)/(x^3*log(exp(3/2) - 2)*log(2*x + 4)))*(x*exp(x) + log(2*x + 4)*(exp(x)*(x - x^2 + 6)
+ 2*x^2 + x^3) + x^3))/(log(exp(3/2) - 2)*log(2*x + 4)^2*(2*x^4 + x^5)),x)

[Out]

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