\(\int e^{2 x+2 x^2+x^3+x \log (\frac {x^2}{e^3})} (4+4 x+3 x^2+\log (\frac {x^2}{e^3})) \, dx\) [8865]

Optimal result

Integrand size = 43, antiderivative size = 26 \[ \int e^{2 x+2 x^2+x^3+x \log \left (\frac {x^2}{e^3}\right )} \left (4+4 x+3 x^2+\log \left (\frac {x^2}{e^3}\right )\right ) \, dx=e^{x+x \left (\frac {\left (x+x^2\right )^2}{x^2}+\log \left (\frac {x^2}{e^3}\right )\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6838} \[ \int e^{2 x+2 x^2+x^3+x \log \left (\frac {x^2}{e^3}\right )} \left (4+4 x+3 x^2+\log \left (\frac {x^2}{e^3}\right )\right ) \, dx=e^{x^3+2 x^2-x} \left (x^2\right )^x \]

[In]

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

[Out]

E^(-x + 2*x^2 + x^3)*(x^2)^x

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int e^{2 x+2 x^2+x^3+x \log \left (\frac {x^2}{e^3}\right )} \left (4+4 x+3 x^2+\log \left (\frac {x^2}{e^3}\right )\right ) \, dx=e^{-x+2 x^2+x^3} \left (x^2\right )^x \]

[In]

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

[Out]

E^(-x + 2*x^2 + x^3)*(x^2)^x

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81

method	result	size
risch	\(\left ({\mathrm e}^{-3} x^{2}\right )^{x} {\mathrm e}^{x \left (x^{2}+2 x +2\right )}\)	\(21\)
parallelrisch	\({\mathrm e}^{x \left (x^{2}+\ln \left ({\mathrm e}^{-3} x^{2}\right )+2 x +2\right )}\)	\(21\)
derivativedivides	\({\mathrm e}^{x \ln \left ({\mathrm e}^{-3} x^{2}\right )+x^{3}+2 x^{2}+2 x}\)	\(25\)
default	\({\mathrm e}^{x \ln \left ({\mathrm e}^{-3} x^{2}\right )+x^{3}+2 x^{2}+2 x}\)	\(25\)
norman	\({\mathrm e}^{x \ln \left ({\mathrm e}^{-3} x^{2}\right )+x^{3}+2 x^{2}+2 x}\)	\(25\)

[In]

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

[Out]

(exp(-3)*x^2)^x*exp(x*(x^2+2*x+2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{2 x+2 x^2+x^3+x \log \left (\frac {x^2}{e^3}\right )} \left (4+4 x+3 x^2+\log \left (\frac {x^2}{e^3}\right )\right ) \, dx=e^{\left (x^{3} + 2 \, x^{2} + x \log \left (x^{2} e^{\left (-3\right )}\right ) + 2 \, x\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{2 x+2 x^2+x^3+x \log \left (\frac {x^2}{e^3}\right )} \left (4+4 x+3 x^2+\log \left (\frac {x^2}{e^3}\right )\right ) \, dx=e^{x^{3} + 2 x^{2} + x \log {\left (\frac {x^{2}}{e^{3}} \right )} + 2 x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{2 x+2 x^2+x^3+x \log \left (\frac {x^2}{e^3}\right )} \left (4+4 x+3 x^2+\log \left (\frac {x^2}{e^3}\right )\right ) \, dx=e^{\left (x^{3} + 2 \, x^{2} + x \log \left (x^{2} e^{\left (-3\right )}\right ) + 2 \, x\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int e^{2 x+2 x^2+x^3+x \log \left (\frac {x^2}{e^3}\right )} \left (4+4 x+3 x^2+\log \left (\frac {x^2}{e^3}\right )\right ) \, dx=e^{\left (x^{3} + 2 \, x^{2} + x \log \left (x^{2}\right ) - x\right )} \]

[In]

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

[Out]

e^(x^3 + 2*x^2 + x*log(x^2) - x)

Mupad [B] (verification not implemented)

Time = 12.44 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int e^{2 x+2 x^2+x^3+x \log \left (\frac {x^2}{e^3}\right )} \left (4+4 x+3 x^2+\log \left (\frac {x^2}{e^3}\right )\right ) \, dx={\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{2\,x^2}\,{\left (x^2\right )}^x \]

[In]

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

[Out]

exp(-x)*exp(x^3)*exp(2*x^2)*(x^2)^x