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

Optimal. Leaf size=21 \[ \log \left (\frac {x^2 (-2+\log (16))}{\left (-5+e^{e^{x^2}}+x\right )^2}\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 0.59, 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 {-10+e^{e^{x^2}} \left (2-4 e^{x^2} x^2\right )}{-5 x+e^{e^{x^2}} x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

2*Log[x] - 2*Defer[Int][(-5 + E^E^x^2 + x)^(-1), x] - 4*Defer[Int][(E^(E^x^2 + x^2)*x)/(-5 + E^E^x^2 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (-5+e^{e^{x^2}}\right )}{x \left (-5+e^{e^{x^2}}+x\right )}-\frac {4 e^{e^{x^2}+x^2} x}{-5+e^{e^{x^2}}+x}\right ) \, dx\\ &=2 \int \frac {-5+e^{e^{x^2}}}{x \left (-5+e^{e^{x^2}}+x\right )} \, dx-4 \int \frac {e^{e^{x^2}+x^2} x}{-5+e^{e^{x^2}}+x} \, dx\\ &=2 \int \left (\frac {1}{x}-\frac {1}{-5+e^{e^{x^2}}+x}\right ) \, dx-4 \int \frac {e^{e^{x^2}+x^2} x}{-5+e^{e^{x^2}}+x} \, dx\\ &=2 \log (x)-2 \int \frac {1}{-5+e^{e^{x^2}}+x} \, dx-4 \int \frac {e^{e^{x^2}+x^2} x}{-5+e^{e^{x^2}}+x} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 22, normalized size = 1.05 \begin {gather*} -2 \log \left (5-e^{e^{x^2}}-x\right )+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*Log[5 - E^E^x^2 - x] + 2*Log[x]

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 16, normalized size = 0.76 \begin {gather*} -2 \, \log \left (x + e^{\left (e^{\left (x^{2}\right )}\right )} - 5\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.14, size = 35, normalized size = 1.67 \begin {gather*} 2 \, x^{2} - 2 \, \log \left (x e^{\left (x^{2}\right )} + e^{\left (x^{2} + e^{\left (x^{2}\right )}\right )} - 5 \, e^{\left (x^{2}\right )}\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 17, normalized size = 0.81

method	result	size
norman	\(2 \ln \relax (x )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x^{2}}}+x -5\right )\)	\(17\)
risch	\(2 \ln \relax (x )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x^{2}}}+x -5\right )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 16, normalized size = 0.76 \begin {gather*} -2 \, \log \left (x + e^{\left (e^{\left (x^{2}\right )}\right )} - 5\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.57, size = 16, normalized size = 0.76 \begin {gather*} 2\,\ln \relax (x)-2\,\ln \left (x+{\mathrm {e}}^{{\mathrm {e}}^{x^2}}-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.27, size = 17, normalized size = 0.81 \begin {gather*} 2 \log {\relax (x )} - 2 \log {\left (x + e^{e^{x^{2}}} - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________