3.80.59 \(\int \frac {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} (-24+x^2-2 x^3) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ \frac {-2+x-\frac {x}{-6+3 e^{x-x^2} \log (x)}}{3 x} \]

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Rubi [F]  time = 6.25, 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 {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(24*E^(-2*x + 2*x^2) + E^(-x + x^2)*x + E^(-x + x^2)*(-24 + x^2 - 2*x^3)*Log[x] + 6*Log[x]^2)/(36*E^(-2*x
+ 2*x^2)*x^2 - 36*E^(-x + x^2)*x^2*Log[x] + 9*x^2*Log[x]^2),x]

[Out]

-2/(3*x) + Defer[Int][E^x/(x*(2*E^x^2 - E^x*Log[x])), x]/18 + Defer[Int][(E^(2*x)*Log[x])/(x*(-2*E^x^2 + E^x*L
og[x])^2), x]/18 + Defer[Int][(E^(2*x)*Log[x]^2)/(-2*E^x^2 + E^x*Log[x])^2, x]/18 - Defer[Int][(E^(2*x)*x*Log[
x]^2)/(-2*E^x^2 + E^x*Log[x])^2, x]/9 - Defer[Int][(E^x*Log[x])/(-2*E^x^2 + E^x*Log[x]), x]/18 + Defer[Int][(E
^x*x*Log[x])/(-2*E^x^2 + E^x*Log[x]), x]/9

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)\right )}{9 x^2 \left (2 e^{x^2}-e^x \log (x)\right )^2} \, dx\\ &=\frac {1}{9} \int \frac {e^{2 x} \left (24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)\right )}{x^2 \left (2 e^{x^2}-e^x \log (x)\right )^2} \, dx\\ &=\frac {1}{9} \int \left (\frac {6}{x^2}-\frac {e^{2 x} \log (x) \left (-1-x \log (x)+2 x^2 \log (x)\right )}{2 x \left (-2 e^{x^2}+e^x \log (x)\right )^2}+\frac {e^x \left (-1-x \log (x)+2 x^2 \log (x)\right )}{2 x \left (-2 e^{x^2}+e^x \log (x)\right )}\right ) \, dx\\ &=-\frac {2}{3 x}-\frac {1}{18} \int \frac {e^{2 x} \log (x) \left (-1-x \log (x)+2 x^2 \log (x)\right )}{x \left (-2 e^{x^2}+e^x \log (x)\right )^2} \, dx+\frac {1}{18} \int \frac {e^x \left (-1-x \log (x)+2 x^2 \log (x)\right )}{x \left (-2 e^{x^2}+e^x \log (x)\right )} \, dx\\ &=-\frac {2}{3 x}-\frac {1}{18} \int \left (-\frac {e^{2 x} \log (x)}{x \left (-2 e^{x^2}+e^x \log (x)\right )^2}-\frac {e^{2 x} \log ^2(x)}{\left (-2 e^{x^2}+e^x \log (x)\right )^2}+\frac {2 e^{2 x} x \log ^2(x)}{\left (-2 e^{x^2}+e^x \log (x)\right )^2}\right ) \, dx+\frac {1}{18} \int \left (\frac {e^x}{x \left (2 e^{x^2}-e^x \log (x)\right )}-\frac {e^x \log (x)}{-2 e^{x^2}+e^x \log (x)}+\frac {2 e^x x \log (x)}{-2 e^{x^2}+e^x \log (x)}\right ) \, dx\\ &=-\frac {2}{3 x}+\frac {1}{18} \int \frac {e^x}{x \left (2 e^{x^2}-e^x \log (x)\right )} \, dx+\frac {1}{18} \int \frac {e^{2 x} \log (x)}{x \left (-2 e^{x^2}+e^x \log (x)\right )^2} \, dx+\frac {1}{18} \int \frac {e^{2 x} \log ^2(x)}{\left (-2 e^{x^2}+e^x \log (x)\right )^2} \, dx-\frac {1}{18} \int \frac {e^x \log (x)}{-2 e^{x^2}+e^x \log (x)} \, dx-\frac {1}{9} \int \frac {e^{2 x} x \log ^2(x)}{\left (-2 e^{x^2}+e^x \log (x)\right )^2} \, dx+\frac {1}{9} \int \frac {e^x x \log (x)}{-2 e^{x^2}+e^x \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.69, size = 33, normalized size = 1.10 \begin {gather*} \frac {1}{9} \left (-\frac {6}{x}-\frac {e^{x^2}}{-2 e^{x^2}+e^x \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24*E^(-2*x + 2*x^2) + E^(-x + x^2)*x + E^(-x + x^2)*(-24 + x^2 - 2*x^3)*Log[x] + 6*Log[x]^2)/(36*E^
(-2*x + 2*x^2)*x^2 - 36*E^(-x + x^2)*x^2*Log[x] + 9*x^2*Log[x]^2),x]

[Out]

(-6/x - E^x^2/(-2*E^x^2 + E^x*Log[x]))/9

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fricas [A]  time = 0.76, size = 38, normalized size = 1.27 \begin {gather*} \frac {{\left (x - 12\right )} e^{\left (x^{2} - x\right )} + 6 \, \log \relax (x)}{9 \, {\left (2 \, x e^{\left (x^{2} - x\right )} - x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(x)^2+(-2*x^3+x^2-24)*exp(x^2-x)*log(x)+24*exp(x^2-x)^2+x*exp(x^2-x))/(9*x^2*log(x)^2-36*x^2*e
xp(x^2-x)*log(x)+36*x^2*exp(x^2-x)^2),x, algorithm="fricas")

[Out]

1/9*((x - 12)*e^(x^2 - x) + 6*log(x))/(2*x*e^(x^2 - x) - x*log(x))

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giac [A]  time = 0.19, size = 40, normalized size = 1.33 \begin {gather*} \frac {x \log \relax (x) - 24 \, e^{\left (x^{2} - x\right )} + 12 \, \log \relax (x)}{18 \, {\left (2 \, x e^{\left (x^{2} - x\right )} - x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(x)^2+(-2*x^3+x^2-24)*exp(x^2-x)*log(x)+24*exp(x^2-x)^2+x*exp(x^2-x))/(9*x^2*log(x)^2-36*x^2*e
xp(x^2-x)*log(x)+36*x^2*exp(x^2-x)^2),x, algorithm="giac")

[Out]

1/18*(x*log(x) - 24*e^(x^2 - x) + 12*log(x))/(2*x*e^(x^2 - x) - x*log(x))

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maple [A]  time = 0.04, size = 30, normalized size = 1.00

method	result	size
risch	\(-\frac {2}{3 x}+\frac {{\mathrm e}^{x \left (x -1\right )}}{18 \,{\mathrm e}^{x \left (x -1\right )}-9 \ln \relax (x )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*ln(x)^2+(-2*x^3+x^2-24)*exp(x^2-x)*ln(x)+24*exp(x^2-x)^2+x*exp(x^2-x))/(9*x^2*ln(x)^2-36*x^2*exp(x^2-x)
*ln(x)+36*x^2*exp(x^2-x)^2),x,method=_RETURNVERBOSE)

[Out]

-2/3/x+1/9*exp(x*(x-1))/(2*exp(x*(x-1))-ln(x))

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maxima [A]  time = 0.43, size = 33, normalized size = 1.10 \begin {gather*} -\frac {{\left (x + 12\right )} e^{x} \log \relax (x) - 24 \, e^{\left (x^{2}\right )}}{18 \, {\left (x e^{x} \log \relax (x) - 2 \, x e^{\left (x^{2}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(x)^2+(-2*x^3+x^2-24)*exp(x^2-x)*log(x)+24*exp(x^2-x)^2+x*exp(x^2-x))/(9*x^2*log(x)^2-36*x^2*e
xp(x^2-x)*log(x)+36*x^2*exp(x^2-x)^2),x, algorithm="maxima")

[Out]

-1/18*((x + 12)*e^x*log(x) - 24*e^(x^2))/(x*e^x*log(x) - 2*x*e^(x^2))

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mupad [B]  time = 4.95, size = 26, normalized size = 0.87 \begin {gather*} \frac {\ln \relax (x)}{18\,\left (2\,{\mathrm {e}}^{x^2-x}-\ln \relax (x)\right )}-\frac {2}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((24*exp(2*x^2 - 2*x) + 6*log(x)^2 + x*exp(x^2 - x) - exp(x^2 - x)*log(x)*(2*x^3 - x^2 + 24))/(36*x^2*exp(2
*x^2 - 2*x) + 9*x^2*log(x)^2 - 36*x^2*exp(x^2 - x)*log(x)),x)

[Out]

log(x)/(18*(2*exp(x^2 - x) - log(x))) - 2/(3*x)

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sympy [A]  time = 0.30, size = 20, normalized size = 0.67 \begin {gather*} \frac {\log {\relax (x )}}{36 e^{x^{2} - x} - 18 \log {\relax (x )}} - \frac {2}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*ln(x)**2+(-2*x**3+x**2-24)*exp(x**2-x)*ln(x)+24*exp(x**2-x)**2+x*exp(x**2-x))/(9*x**2*ln(x)**2-36
*x**2*exp(x**2-x)*ln(x)+36*x**2*exp(x**2-x)**2),x)

[Out]

log(x)/(36*exp(x**2 - x) - 18*log(x)) - 2/(3*x)

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