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

Optimal. Leaf size=33 \[ 3-2 x-\log \left (\frac {1}{2} e^{e^{\frac {2}{x}-5 e^{e^{x^2}} x}} x\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 1.47, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.37, size = 28, normalized size = 0.85 \begin {gather*} -e^{\frac {2}{x}-5 e^{e^{x^2}} x}-2 x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 28, normalized size = 0.85 \begin {gather*} -2 \, x - e^{\left (-\frac {5 \, x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} - 2}{x}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.08, size = 29, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [A]  time = 0.54, size = 25, normalized size = 0.76 \begin {gather*} -2 \, x - e^{\left (-5 \, x e^{\left (e^{\left (x^{2}\right )}\right )} + \frac {2}{x}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 1.42, size = 25, normalized size = 0.76 \begin {gather*} -2\,x-\ln \relax (x)-{\mathrm {e}}^{2/x}\,{\mathrm {e}}^{-5\,x\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

________________________________________________________________________________________

sympy [A]  time = 5.84, size = 24, normalized size = 0.73 \begin {gather*} - 2 x - e^{\frac {- 5 x^{2} e^{e^{x^{2}}} + 2}{x}} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________