\(\int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x))}{x \log ^2(x)} \, dx\) [215]

Optimal result

Integrand size = 55, antiderivative size = 25 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=5+e^{\frac {x^2}{2}+\frac {15 \left (e^3-x\right )}{\log (x)}} \]

[Out]

5+exp(1/2*x^2+15*(-x+exp(3))/ln(x))

Rubi [F]

\[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=\int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\frac {x^2}{2}+\frac {15 \left (e^3-x\right )}{\log (x)}} \]

[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])

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\)	\(22\)
default	\({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\)	\(22\)
norman	\({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\)	\(22\)
risch	\({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\)	\(22\)
parallelrisch	\({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\)	\(22\)

[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))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\left (\frac {x^{2} \log \left (x\right ) - 30 \, x + 30 \, e^{3}}{2 \, \log \left (x\right )}\right )} \]

[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))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\frac {\frac {x^{2} \log {\left (x \right )}}{2} - 15 x + 15 e^{3}}{\log {\left (x \right )}}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\left (\frac {1}{2} \, x^{2} - \frac {15 \, x}{\log \left (x\right )} + \frac {15 \, e^{3}}{\log \left (x\right )}\right )} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\left (\frac {x^{2} \log \left (x\right ) - 30 \, x + 30 \, e^{3}}{2 \, \log \left (x\right )}\right )} \]

[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*log(x) - 30*x + 30*e^3)/log(x))

Mupad [B] (verification not implemented)

Time = 7.77 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx={\mathrm {e}}^{\frac {15\,{\mathrm {e}}^3}{\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {15\,x}{\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {x^2}{2}} \]

[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)