\(\int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} (2+e^{e^{x^2}} (5 x^2+10 e^{x^2} x^4))}{x^2} \, dx\) [2498]

Optimal result

Integrand size = 60, antiderivative size = 33 \[ \int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} \left (2+e^{e^{x^2}} \left (5 x^2+10 e^{x^2} x^4\right )\right )}{x^2} \, dx=3-2 x-\log \left (\frac {1}{2} e^{e^{\frac {2}{x}-5 e^{e^{x^2}} x}} x\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} \left (2+e^{e^{x^2}} \left (5 x^2+10 e^{x^2} x^4\right )\right )}{x^2} \, dx=\int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} \left (2+e^{e^{x^2}} \left (5 x^2+10 e^{x^2} x^4\right )\right )}{x^2} \, dx \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} \left (2+e^{e^{x^2}} \left (5 x^2+10 e^{x^2} x^4\right )\right )}{x^2} \, dx=-e^{\frac {2}{x}-5 e^{e^{x^2}} x}-2 x-\log (x) \]

[In]

Integrate[(-x - 2*x^2 + E^((2 - 5*E^E^x^2*x^2)/x)*(2 + E^E^x^2*(5*x^2 + 10*E^x^2*x^4)))/x^2,x]

[Out]

-E^(2/x - 5*E^E^x^2*x) - 2*x - Log[x]

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88

method	result	size
risch	\(-2 x -\ln \left (x \right )-{\mathrm e}^{-\frac {5 x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}}-2}{x}}\)	\(29\)
parallelrisch	\(-2 x -\ln \left (x \right )-{\mathrm e}^{-\frac {5 x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}}-2}{x}}\)	\(29\)

[In]

int((((10*x^4*exp(x^2)+5*x^2)*exp(exp(x^2))+2)*exp((-5*x^2*exp(exp(x^2))+2)/x)-2*x^2-x)/x^2,x,method=_RETURNVE
RBOSE)

[Out]

-2*x-ln(x)-exp(-(5*x^2*exp(exp(x^2))-2)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} \left (2+e^{e^{x^2}} \left (5 x^2+10 e^{x^2} x^4\right )\right )}{x^2} \, dx=-2 \, x - e^{\left (-\frac {5 \, x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} - 2}{x}\right )} - \log \left (x\right ) \]

[In]

integrate((((10*x^4*exp(x^2)+5*x^2)*exp(exp(x^2))+2)*exp((-5*x^2*exp(exp(x^2))+2)/x)-2*x^2-x)/x^2,x, algorithm
="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} \left (2+e^{e^{x^2}} \left (5 x^2+10 e^{x^2} x^4\right )\right )}{x^2} \, dx=- 2 x - e^{\frac {- 5 x^{2} e^{e^{x^{2}}} + 2}{x}} - \log {\left (x \right )} \]

[In]

integrate((((10*x**4*exp(x**2)+5*x**2)*exp(exp(x**2))+2)*exp((-5*x**2*exp(exp(x**2))+2)/x)-2*x**2-x)/x**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} \left (2+e^{e^{x^2}} \left (5 x^2+10 e^{x^2} x^4\right )\right )}{x^2} \, dx=-2 \, x - e^{\left (-5 \, x e^{\left (e^{\left (x^{2}\right )}\right )} + \frac {2}{x}\right )} - \log \left (x\right ) \]

[In]

integrate((((10*x^4*exp(x^2)+5*x^2)*exp(exp(x^2))+2)*exp((-5*x^2*exp(exp(x^2))+2)/x)-2*x^2-x)/x^2,x, algorithm
="maxima")

[Out]

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

Giac [F]

\[ \int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} \left (2+e^{e^{x^2}} \left (5 x^2+10 e^{x^2} x^4\right )\right )}{x^2} \, dx=\int { -\frac {2 \, x^{2} - {\left (5 \, {\left (2 \, x^{4} e^{\left (x^{2}\right )} + x^{2}\right )} e^{\left (e^{\left (x^{2}\right )}\right )} + 2\right )} e^{\left (-\frac {5 \, x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} - 2}{x}\right )} + x}{x^{2}} \,d x } \]

[In]

integrate((((10*x^4*exp(x^2)+5*x^2)*exp(exp(x^2))+2)*exp((-5*x^2*exp(exp(x^2))+2)/x)-2*x^2-x)/x^2,x, algorithm
="giac")

[Out]

integrate(-(2*x^2 - (5*(2*x^4*e^(x^2) + x^2)*e^(e^(x^2)) + 2)*e^(-(5*x^2*e^(e^(x^2)) - 2)/x) + x)/x^2, x)

Mupad [B] (verification not implemented)

Time = 10.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {-x-2 x^2+e^{\frac {2-5 e^{e^{x^2}} x^2}{x}} \left (2+e^{e^{x^2}} \left (5 x^2+10 e^{x^2} x^4\right )\right )}{x^2} \, dx=-2\,x-\ln \left (x\right )-{\mathrm {e}}^{2/x}\,{\mathrm {e}}^{-5\,x\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}} \]

[In]

int(-(x - exp(-(5*x^2*exp(exp(x^2)) - 2)/x)*(exp(exp(x^2))*(10*x^4*exp(x^2) + 5*x^2) + 2) + 2*x^2)/x^2,x)

[Out]

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