3.1.86 \(\int \frac {e^{\frac {18}{2 x^2+x^2 \log (x)}} (-90+4 x^2+(-36+4 x^2) \log (x)+x^2 \log ^2(x))}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {9}{x \left (\frac {x}{2}+\frac {x}{\log (x)}\right ) \log (x)}} x \]

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Rubi [F]  time = 1.00, 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 {e^{\frac {18}{2 x^2+x^2 \log (x)}} \left (-90+4 x^2+\left (-36+4 x^2\right ) \log (x)+x^2 \log ^2(x)\right )}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(18/(2*x^2 + x^2*Log[x]))*(-90 + 4*x^2 + (-36 + 4*x^2)*Log[x] + x^2*Log[x]^2))/(4*x^2 + 4*x^2*Log[x] +
x^2*Log[x]^2),x]

[Out]

Defer[Int][E^(18/(x^2*(2 + Log[x]))), x] - 18*Defer[Int][E^(18/(x^2*(2 + Log[x])))/(x^2*(2 + Log[x])^2), x] -
36*Defer[Int][E^(18/(x^2*(2 + Log[x])))/(x^2*(2 + Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {18}{x^2 (2+\log (x))}} \left (-90+4 x^2+4 \left (-9+x^2\right ) \log (x)+x^2 \log ^2(x)\right )}{x^2 (2+\log (x))^2} \, dx\\ &=\int \left (e^{\frac {18}{x^2 (2+\log (x))}}-\frac {18 e^{\frac {18}{x^2 (2+\log (x))}}}{x^2 (2+\log (x))^2}-\frac {36 e^{\frac {18}{x^2 (2+\log (x))}}}{x^2 (2+\log (x))}\right ) \, dx\\ &=-\left (18 \int \frac {e^{\frac {18}{x^2 (2+\log (x))}}}{x^2 (2+\log (x))^2} \, dx\right )-36 \int \frac {e^{\frac {18}{x^2 (2+\log (x))}}}{x^2 (2+\log (x))} \, dx+\int e^{\frac {18}{x^2 (2+\log (x))}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 15, normalized size = 0.56 \begin {gather*} e^{\frac {18}{x^2 (2+\log (x))}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(18/(2*x^2 + x^2*Log[x]))*(-90 + 4*x^2 + (-36 + 4*x^2)*Log[x] + x^2*Log[x]^2))/(4*x^2 + 4*x^2*Log
[x] + x^2*Log[x]^2),x]

[Out]

E^(18/(x^2*(2 + Log[x])))*x

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fricas [A]  time = 0.63, size = 19, normalized size = 0.70 \begin {gather*} x e^{\left (\frac {18}{x^{2} \log \relax (x) + 2 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+(4*x^2-36)*log(x)+4*x^2-90)*exp(18/(x^2*log(x)+2*x^2))/(x^2*log(x)^2+4*x^2*log(x)+4*x^
2),x, algorithm="fricas")

[Out]

x*e^(18/(x^2*log(x) + 2*x^2))

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giac [A]  time = 0.42, size = 19, normalized size = 0.70 \begin {gather*} x e^{\left (\frac {18}{x^{2} \log \relax (x) + 2 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+(4*x^2-36)*log(x)+4*x^2-90)*exp(18/(x^2*log(x)+2*x^2))/(x^2*log(x)^2+4*x^2*log(x)+4*x^
2),x, algorithm="giac")

[Out]

x*e^(18/(x^2*log(x) + 2*x^2))

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maple [A]  time = 0.04, size = 15, normalized size = 0.56

method	result	size
risch	\(x \,{\mathrm e}^{\frac {18}{x^{2} \left (\ln \relax (x )+2\right )}}\)	\(15\)
norman	\(\frac {x^{2} \ln \relax (x ) {\mathrm e}^{\frac {18}{x^{2} \ln \relax (x )+2 x^{2}}}+2 x^{2} {\mathrm e}^{\frac {18}{x^{2} \ln \relax (x )+2 x^{2}}}}{\left (\ln \relax (x )+2\right ) x}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(x)^2+(4*x^2-36)*ln(x)+4*x^2-90)*exp(18/(x^2*ln(x)+2*x^2))/(x^2*ln(x)^2+4*x^2*ln(x)+4*x^2),x,method
=_RETURNVERBOSE)

[Out]

x*exp(18/x^2/(ln(x)+2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+(4*x^2-36)*log(x)+4*x^2-90)*exp(18/(x^2*log(x)+2*x^2))/(x^2*log(x)^2+4*x^2*log(x)+4*x^
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

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mupad [B]  time = 0.37, size = 14, normalized size = 0.52 \begin {gather*} x\,{\mathrm {e}}^{\frac {18}{x^2\,\left (\ln \relax (x)+2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(18/(x^2*log(x) + 2*x^2))*(x^2*log(x)^2 + 4*x^2 + log(x)*(4*x^2 - 36) - 90))/(4*x^2*log(x) + x^2*log(x
)^2 + 4*x^2),x)

[Out]

x*exp(18/(x^2*(log(x) + 2)))

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sympy [A]  time = 3.96, size = 15, normalized size = 0.56 \begin {gather*} x e^{\frac {18}{x^{2} \log {\relax (x )} + 2 x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(x)**2+(4*x**2-36)*ln(x)+4*x**2-90)*exp(18/(x**2*ln(x)+2*x**2))/(x**2*ln(x)**2+4*x**2*ln(x)+
4*x**2),x)

[Out]

x*exp(18/(x**2*log(x) + 2*x**2))

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