\(\int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} (e^{x+x^2} (x+2 e x^2+2 x^3)+e^x (-1+x+2 e x^2+3 x^3))}{x^2 \log (3)} \, dx\) [6580]

Optimal result

Integrand size = 68, antiderivative size = 27 \[ \int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} \left (e^{x+x^2} \left (x+2 e x^2+2 x^3\right )+e^x \left (-1+x+2 e x^2+3 x^3\right )\right )}{x^2 \log (3)} \, dx=-3+\frac {e^{x+(e+x) \left (e^{x^2}+x^2\right )}}{x \log (3)} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} \left (e^{x+x^2} \left (x+2 e x^2+2 x^3\right )+e^x \left (-1+x+2 e x^2+3 x^3\right )\right )}{x^2 \log (3)} \, dx=\int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} \left (e^{x+x^2} \left (x+2 e x^2+2 x^3\right )+e^x \left (-1+x+2 e x^2+3 x^3\right )\right )}{x^2 \log (3)} \, dx \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} \left (e^{x+x^2} \left (x+2 e x^2+2 x^3\right )+e^x \left (-1+x+2 e x^2+3 x^3\right )\right )}{x^2 \log (3)} \, dx=\frac {e^{x+e x^2+x^3+e^{x^2} (e+x)}}{x \log (3)} \]

[In]

Integrate[(E^(E*x^2 + x^3 + E^x^2*(E + x))*(E^(x + x^2)*(x + 2*E*x^2 + 2*x^3) + E^x*(-1 + x + 2*E*x^2 + 3*x^3)
))/(x^2*Log[3]),x]

[Out]

E^(x + E*x^2 + x^3 + E^x^2*(E + x))/(x*Log[3])

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{x} {\mathrm e}^{\left (x +{\mathrm e}\right ) {\mathrm e}^{x^{2}}+x^{2} {\mathrm e}+x^{3}}}{\ln \left (3\right ) x}\)	\(31\)
risch	\(\frac {{\mathrm e}^{x^{2} {\mathrm e}+x^{3}+{\mathrm e} \,{\mathrm e}^{x^{2}}+{\mathrm e}^{x^{2}} x +x}}{x \ln \left (3\right )}\)	\(34\)

[In]

int(((2*x^2*exp(1)+2*x^3+x)*exp(x)*exp(x^2)+(2*x^2*exp(1)+3*x^3+x-1)*exp(x))*exp((x+exp(1))*exp(x^2)+x^2*exp(1
)+x^3)/x^2/ln(3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} \left (e^{x+x^2} \left (x+2 e x^2+2 x^3\right )+e^x \left (-1+x+2 e x^2+3 x^3\right )\right )}{x^2 \log (3)} \, dx=\frac {e^{\left (x^{3} + x^{2} e + {\left (x + e\right )} e^{\left (x^{2}\right )} + x\right )}}{x \log \left (3\right )} \]

[In]

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

[Out]

e^(x^3 + x^2*e + (x + e)*e^(x^2) + x)/(x*log(3))

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} \left (e^{x+x^2} \left (x+2 e x^2+2 x^3\right )+e^x \left (-1+x+2 e x^2+3 x^3\right )\right )}{x^2 \log (3)} \, dx=\frac {e^{x} e^{x^{3} + e x^{2} + \left (x + e\right ) e^{x^{2}}}}{x \log {\left (3 \right )}} \]

[In]

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

[Out]

exp(x)*exp(x**3 + E*x**2 + (x + E)*exp(x**2))/(x*log(3))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} \left (e^{x+x^2} \left (x+2 e x^2+2 x^3\right )+e^x \left (-1+x+2 e x^2+3 x^3\right )\right )}{x^2 \log (3)} \, dx=\frac {e^{\left (x^{3} + x^{2} e + x e^{\left (x^{2}\right )} + x + e^{\left (x^{2} + 1\right )}\right )}}{x \log \left (3\right )} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} \left (e^{x+x^2} \left (x+2 e x^2+2 x^3\right )+e^x \left (-1+x+2 e x^2+3 x^3\right )\right )}{x^2 \log (3)} \, dx=\int { \frac {{\left ({\left (2 \, x^{3} + 2 \, x^{2} e + x\right )} e^{\left (x^{2} + x\right )} + {\left (3 \, x^{3} + 2 \, x^{2} e + x - 1\right )} e^{x}\right )} e^{\left (x^{3} + x^{2} e + {\left (x + e\right )} e^{\left (x^{2}\right )}\right )}}{x^{2} \log \left (3\right )} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.96 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{e x^2+x^3+e^{x^2} (e+x)} \left (e^{x+x^2} \left (x+2 e x^2+2 x^3\right )+e^x \left (-1+x+2 e x^2+3 x^3\right )\right )}{x^2 \log (3)} \, dx=\frac {{\mathrm {e}}^{x^2\,\mathrm {e}}\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}\,\mathrm {e}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{x^2}}\,{\mathrm {e}}^x}{x\,\ln \left (3\right )} \]

[In]

int((exp(x^2*exp(1) + x^3 + exp(x^2)*(x + exp(1)))*(exp(x)*(x + 2*x^2*exp(1) + 3*x^3 - 1) + exp(x^2)*exp(x)*(x
+ 2*x^2*exp(1) + 2*x^3)))/(x^2*log(3)),x)

[Out]

(exp(x^2*exp(1))*exp(x^3)*exp(exp(x^2)*exp(1))*exp(x*exp(x^2))*exp(x))/(x*log(3))