3.3.15 \(\int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x))}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ 5+e^{\frac {x^2}{2}+\frac {15 \left (e^3-x\right )}{\log (x)}} \]

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Rubi [F]  time = 1.70, 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 {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((30*E^3 - 30*x + x^2*Log[x])/(2*Log[x]))*(-15*E^3 + 15*x - 15*x*Log[x] + x^2*Log[x]^2))/(x*Log[x]^2),x
]

[Out]

Defer[Int][E^((30*E^3 - 30*x + x^2*Log[x])/(2*Log[x]))*x, x] + 15*Defer[Int][E^((30*E^3 - 30*x + x^2*Log[x])/(
2*Log[x]))/Log[x]^2, x] - 15*Defer[Int][E^((30*E^3 - 30*x + 6*Log[x] + x^2*Log[x])/(2*Log[x]))/(x*Log[x]^2), x
] - 15*Defer[Int][E^((30*E^3 - 30*x + x^2*Log[x])/(2*Log[x]))/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x-\frac {15 e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (e^3-x\right )}{x \log ^2(x)}-\frac {15 e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)}\right ) \, dx\\ &=-\left (15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (e^3-x\right )}{x \log ^2(x)} \, dx\right )-15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)} \, dx+\int e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x \, dx\\ &=-\left (15 \int \left (-\frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log ^2(x)}+\frac {e^{3+\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{x \log ^2(x)}\right ) \, dx\right )-15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)} \, dx+\int e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x \, dx\\ &=15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log ^2(x)} \, dx-15 \int \frac {e^{3+\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{x \log ^2(x)} \, dx-15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)} \, dx+\int e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x \, dx\\ &=15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log ^2(x)} \, dx-15 \int \frac {e^{\frac {30 e^3-30 x+6 \log (x)+x^2 \log (x)}{2 \log (x)}}}{x \log ^2(x)} \, dx-15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)} \, dx+\int e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 23, normalized size = 0.92 \begin {gather*} e^{\frac {x^2}{2}+\frac {15 \left (e^3-x\right )}{\log (x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((30*E^3 - 30*x + x^2*Log[x])/(2*Log[x]))*(-15*E^3 + 15*x - 15*x*Log[x] + x^2*Log[x]^2))/(x*Log[x
]^2),x]

[Out]

E^(x^2/2 + (15*(E^3 - x))/Log[x])

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fricas [A]  time = 0.71, size = 21, normalized size = 0.84 \begin {gather*} e^{\left (\frac {x^{2} \log \relax (x) - 30 \, x + 30 \, e^{3}}{2 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2-15*x*log(x)-15*exp(3)+15*x)*exp(1/2*(x^2*log(x)+30*exp(3)-30*x)/log(x))/x/log(x)^2,x,
algorithm="fricas")

[Out]

e^(1/2*(x^2*log(x) - 30*x + 30*e^3)/log(x))

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giac [A]  time = 0.30, size = 22, normalized size = 0.88 \begin {gather*} e^{\left (\frac {1}{2} \, x^{2} - \frac {15 \, x}{\log \relax (x)} + \frac {15 \, e^{3}}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2-15*x*log(x)-15*exp(3)+15*x)*exp(1/2*(x^2*log(x)+30*exp(3)-30*x)/log(x))/x/log(x)^2,x,
algorithm="giac")

[Out]

e^(1/2*x^2 - 15*x/log(x) + 15*e^3/log(x))

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maple [A]  time = 0.06, size = 22, normalized size = 0.88

method	result	size
norman	\({\mathrm e}^{\frac {x^{2} \ln \relax (x )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \relax (x )}}\)	\(22\)
risch	\({\mathrm e}^{\frac {x^{2} \ln \relax (x )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \relax (x )}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(x)^2-15*x*ln(x)-15*exp(3)+15*x)*exp(1/2*(x^2*ln(x)+30*exp(3)-30*x)/ln(x))/x/ln(x)^2,x,method=_RETU
RNVERBOSE)

[Out]

exp(1/2*(x^2*ln(x)+30*exp(3)-30*x)/ln(x))

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maxima [A]  time = 0.68, size = 22, normalized size = 0.88 \begin {gather*} e^{\left (\frac {1}{2} \, x^{2} - \frac {15 \, x}{\log \relax (x)} + \frac {15 \, e^{3}}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2-15*x*log(x)-15*exp(3)+15*x)*exp(1/2*(x^2*log(x)+30*exp(3)-30*x)/log(x))/x/log(x)^2,x,
algorithm="maxima")

[Out]

e^(1/2*x^2 - 15*x/log(x) + 15*e^3/log(x))

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mupad [B]  time = 0.40, size = 24, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{\frac {15\,{\mathrm {e}}^3}{\ln \relax (x)}}\,{\mathrm {e}}^{-\frac {15\,x}{\ln \relax (x)}}\,{\mathrm {e}}^{\frac {x^2}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((15*exp(3) - 15*x + (x^2*log(x))/2)/log(x))*(15*x - 15*exp(3) + x^2*log(x)^2 - 15*x*log(x)))/(x*log(x
)^2),x)

[Out]

exp((15*exp(3))/log(x))*exp(-(15*x)/log(x))*exp(x^2/2)

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sympy [A]  time = 0.34, size = 20, normalized size = 0.80 \begin {gather*} e^{\frac {\frac {x^{2} \log {\relax (x )}}{2} - 15 x + 15 e^{3}}{\log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(x)**2-15*x*ln(x)-15*exp(3)+15*x)*exp(1/2*(x**2*ln(x)+30*exp(3)-30*x)/ln(x))/x/ln(x)**2,x)

[Out]

exp((x**2*log(x)/2 - 15*x + 15*exp(3))/log(x))

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