\(\int \frac {e^{-x} (e^x+e^{e^{-x} (e^{15} x^2+e^{15+x} x^2)} (2 e^{15+x} x^3+e^{15} (2 x^3-x^4))-e^x \log (x))}{x^2} \, dx\) [7285]

Optimal result

Integrand size = 72, antiderivative size = 24 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=-1+e^{e^{15} x \left (x+e^{-x} x\right )}+\frac {\log (x)}{x} \]

[Out]

exp(exp(15)*x*(x/exp(x)+x))+ln(x)/x-1

Rubi [F]

\[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx \]

[In]

Int[(E^x + E^((E^15*x^2 + E^(15 + x)*x^2)/E^x)*(2*E^(15 + x)*x^3 + E^15*(2*x^3 - x^4)) - E^x*Log[x])/(E^x*x^2)
,x]

[Out]

Log[x]/x + 2*Defer[Int][E^(15 + E^(15 - x)*(1 + E^x)*x^2)*x, x] + 2*Defer[Int][E^(15 - x + E^(15 - x)*(1 + E^x
)*x^2)*x, x] - Defer[Int][E^(15 - x + E^(15 - x)*(1 + E^x)*x^2)*x^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1+e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} \left (2+2 e^x-x\right ) x^3-\log (x)}{x^2} \, dx \\ & = \int \left (e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} \left (2+2 e^x-x\right ) x+\frac {1-\log (x)}{x^2}\right ) \, dx \\ & = \int e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} \left (2+2 e^x-x\right ) x \, dx+\int \frac {1-\log (x)}{x^2} \, dx \\ & = \frac {\log (x)}{x}+\int \left (2 e^{15+e^{15-x} \left (1+e^x\right ) x^2} x-e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} (-2+x) x\right ) \, dx \\ & = \frac {\log (x)}{x}+2 \int e^{15+e^{15-x} \left (1+e^x\right ) x^2} x \, dx-\int e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} (-2+x) x \, dx \\ & = \frac {\log (x)}{x}+2 \int e^{15+e^{15-x} \left (1+e^x\right ) x^2} x \, dx-\int \left (-2 e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} x+e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} x^2\right ) \, dx \\ & = \frac {\log (x)}{x}+2 \int e^{15+e^{15-x} \left (1+e^x\right ) x^2} x \, dx+2 \int e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} x \, dx-\int e^{15-x+e^{15-x} \left (1+e^x\right ) x^2} x^2 \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.72 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=e^{e^{15-x} \left (1+e^x\right ) x^2}+\frac {\log (x)}{x} \]

[In]

Integrate[(E^x + E^((E^15*x^2 + E^(15 + x)*x^2)/E^x)*(2*E^(15 + x)*x^3 + E^15*(2*x^3 - x^4)) - E^x*Log[x])/(E^
x*x^2),x]

[Out]

E^(E^(15 - x)*(1 + E^x)*x^2) + Log[x]/x

Maple [A] (verified)

Time = 4.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {\ln \left (x \right )}{x}+{\mathrm e}^{x^{2} \left ({\mathrm e}^{x +15}+{\mathrm e}^{15}\right ) {\mathrm e}^{-x}}\)	\(24\)
parallelrisch	\(\frac {{\mathrm e}^{x^{2} {\mathrm e}^{15} \left ({\mathrm e}^{x}+1\right ) {\mathrm e}^{-x}} x +\ln \left (x \right )}{x}\)	\(25\)
default	\(\frac {\ln \left (x \right )}{x}+{\mathrm e}^{x^{2} \left ({\mathrm e}^{15} {\mathrm e}^{x}+{\mathrm e}^{15}\right ) {\mathrm e}^{-x}}\)	\(29\)

[In]

int(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*ln(x)+exp
(x))/exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\frac {x e^{\left ({\left (x^{2} e^{30} + x^{2} e^{\left (x + 30\right )}\right )} e^{\left (-x - 15\right )}\right )} + \log \left (x\right )}{x} \]

[In]

integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*log
(x)+exp(x))/exp(x)/x^2,x, algorithm="fricas")

[Out]

(x*e^((x^2*e^30 + x^2*e^(x + 30))*e^(-x - 15)) + log(x))/x

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=e^{\left (x^{2} e^{15} e^{x} + x^{2} e^{15}\right ) e^{- x}} + \frac {\log {\left (x \right )}}{x} \]

[In]

integrate(((2*x**3*exp(15)*exp(x)+(-x**4+2*x**3)*exp(15))*exp((x**2*exp(15)*exp(x)+x**2*exp(15))/exp(x))-exp(x
)*ln(x)+exp(x))/exp(x)/x**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\frac {x e^{\left (x^{2} e^{15} + x^{2} e^{\left (-x + 15\right )}\right )} + \log \left (x\right ) + 1}{x} - \frac {1}{x} \]

[In]

integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*log
(x)+exp(x))/exp(x)/x^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\frac {\log \left (x\right )}{x} + e^{\left (x^{2} e^{15} + x^{2} e^{\left (-x + 15\right )}\right )} \]

[In]

integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*log
(x)+exp(x))/exp(x)/x^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (e^x+e^{e^{-x} \left (e^{15} x^2+e^{15+x} x^2\right )} \left (2 e^{15+x} x^3+e^{15} \left (2 x^3-x^4\right )\right )-e^x \log (x)\right )}{x^2} \, dx=\frac {\ln \left (x\right )}{x}+{\mathrm {e}}^{x^2\,{\mathrm {e}}^{15}}\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{15}} \]

[In]

int((exp(-x)*(exp(x) - exp(x)*log(x) + exp(exp(-x)*(x^2*exp(15) + x^2*exp(15)*exp(x)))*(exp(15)*(2*x^3 - x^4)
+ 2*x^3*exp(15)*exp(x))))/x^2,x)

[Out]

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