\(\int \frac {e^{\frac {x^2+\log (x)}{x}} ((120-240 x+120 x^2) \log (2)-120 \log (2) \log (x))}{x^4} \, dx\) [8564]

Optimal result

Integrand size = 36, antiderivative size = 17 \[ \int \frac {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx=\frac {120 e^{x+\frac {\log (x)}{x}} \log (2)}{x^2} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx=\int \frac {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx \]

[In]

Int[(E^((x^2 + Log[x])/x)*((120 - 240*x + 120*x^2)*Log[2] - 120*Log[2]*Log[x]))/x^4,x]

[Out]

120*Log[2]*Defer[Int][E^x*x^(-4 + x^(-1)), x] - 120*Log[2]*Log[x]*Defer[Int][E^x*x^(-4 + x^(-1)), x] - 240*Log
[2]*Defer[Int][E^x*x^(-3 + x^(-1)), x] + 120*Log[2]*Defer[Int][E^x*x^(-2 + x^(-1)), x] + 120*Log[2]*Defer[Int]
[Defer[Int][E^x*x^(-4 + x^(-1)), x]/x, x]

Rubi steps \begin{align*} \text {integral}& = \int 120 e^x x^{-4+\frac {1}{x}} \log (2) \left ((-1+x)^2-\log (x)\right ) \, dx \\ & = (120 \log (2)) \int e^x x^{-4+\frac {1}{x}} \left ((-1+x)^2-\log (x)\right ) \, dx \\ & = (120 \log (2)) \int \left (e^x (-1+x)^2 x^{-4+\frac {1}{x}}-e^x x^{-4+\frac {1}{x}} \log (x)\right ) \, dx \\ & = (120 \log (2)) \int e^x (-1+x)^2 x^{-4+\frac {1}{x}} \, dx-(120 \log (2)) \int e^x x^{-4+\frac {1}{x}} \log (x) \, dx \\ & = (120 \log (2)) \int \left (e^x x^{-4+\frac {1}{x}}-2 e^x x^{-3+\frac {1}{x}}+e^x x^{-2+\frac {1}{x}}\right ) \, dx+(120 \log (2)) \int \frac {\int e^x x^{-4+\frac {1}{x}} \, dx}{x} \, dx-(120 \log (2) \log (x)) \int e^x x^{-4+\frac {1}{x}} \, dx \\ & = (120 \log (2)) \int e^x x^{-4+\frac {1}{x}} \, dx+(120 \log (2)) \int e^x x^{-2+\frac {1}{x}} \, dx+(120 \log (2)) \int \frac {\int e^x x^{-4+\frac {1}{x}} \, dx}{x} \, dx-(240 \log (2)) \int e^x x^{-3+\frac {1}{x}} \, dx-(120 \log (2) \log (x)) \int e^x x^{-4+\frac {1}{x}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx=120 e^x x^{-2+\frac {1}{x}} \log (2) \]

[In]

Integrate[(E^((x^2 + Log[x])/x)*((120 - 240*x + 120*x^2)*Log[2] - 120*Log[2]*Log[x]))/x^4,x]

[Out]

120*E^x*x^(-2 + x^(-1))*Log[2]

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {120 \ln \left (2\right ) x^{\frac {1}{x}} {\mathrm e}^{x}}{x^{2}}\)	\(15\)
norman	\(\frac {120 \ln \left (2\right ) {\mathrm e}^{\frac {\ln \left (x \right )+x^{2}}{x}}}{x^{2}}\)	\(19\)
parallelrisch	\(\frac {120 \ln \left (2\right ) {\mathrm e}^{\frac {\ln \left (x \right )+x^{2}}{x}}}{x^{2}}\)	\(19\)

[In]

int((-120*ln(2)*ln(x)+(120*x^2-240*x+120)*ln(2))*exp((ln(x)+x^2)/x)/x^4,x,method=_RETURNVERBOSE)

[Out]

120/x^2*ln(2)*x^(1/x)*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx=\frac {120 \, e^{\left (\frac {x^{2} + \log \left (x\right )}{x}\right )} \log \left (2\right )}{x^{2}} \]

[In]

integrate((-120*log(2)*log(x)+(120*x^2-240*x+120)*log(2))*exp((log(x)+x^2)/x)/x^4,x, algorithm="fricas")

[Out]

120*e^((x^2 + log(x))/x)*log(2)/x^2

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx=\frac {120 e^{\frac {x^{2} + \log {\left (x \right )}}{x}} \log {\left (2 \right )}}{x^{2}} \]

[In]

integrate((-120*ln(2)*ln(x)+(120*x**2-240*x+120)*ln(2))*exp((ln(x)+x**2)/x)/x**4,x)

[Out]

120*exp((x**2 + log(x))/x)*log(2)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx=\frac {120 \, e^{\left (x + \frac {\log \left (x\right )}{x}\right )} \log \left (2\right )}{x^{2}} \]

[In]

integrate((-120*log(2)*log(x)+(120*x^2-240*x+120)*log(2))*exp((log(x)+x^2)/x)/x^4,x, algorithm="maxima")

[Out]

120*e^(x + log(x)/x)*log(2)/x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx=\frac {120 \, e^{\left (\frac {x^{2} + \log \left (x\right )}{x}\right )} \log \left (2\right )}{x^{2}} \]

[In]

integrate((-120*log(2)*log(x)+(120*x^2-240*x+120)*log(2))*exp((log(x)+x^2)/x)/x^4,x, algorithm="giac")

[Out]

120*e^((x^2 + log(x))/x)*log(2)/x^2

Mupad [B] (verification not implemented)

Time = 14.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx=120\,x^{\frac {1}{x}-2}\,{\mathrm {e}}^x\,\ln \left (2\right ) \]

[In]

int((exp((log(x) + x^2)/x)*(log(2)*(120*x^2 - 240*x + 120) - 120*log(2)*log(x)))/x^4,x)

[Out]

120*x^(1/x - 2)*exp(x)*log(2)