\(\int e^{9-5 x-3 x^2+2 x^3+x^4+x \log (x)} (-4-6 x+6 x^2+4 x^3+\log (x)) \, dx\) [6917]

Optimal result

Integrand size = 42, antiderivative size = 23 \[ \int e^{9-5 x-3 x^2+2 x^3+x^4+x \log (x)} \left (-4-6 x+6 x^2+4 x^3+\log (x)\right ) \, dx=e^{5-x+\left (2-x-x^2\right )^2+x \log (x)} \]

[Out]

exp(5+(-x^2-x+2)^2-x+x*ln(x))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6838} \[ \int e^{9-5 x-3 x^2+2 x^3+x^4+x \log (x)} \left (-4-6 x+6 x^2+4 x^3+\log (x)\right ) \, dx=e^{x^4+2 x^3-3 x^2-5 x+9} x^x \]

[In]

Int[E^(9 - 5*x - 3*x^2 + 2*x^3 + x^4 + x*Log[x])*(-4 - 6*x + 6*x^2 + 4*x^3 + Log[x]),x]

[Out]

E^(9 - 5*x - 3*x^2 + 2*x^3 + x^4)*x^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^{9-5 x-3 x^2+2 x^3+x^4} x^x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int e^{9-5 x-3 x^2+2 x^3+x^4+x \log (x)} \left (-4-6 x+6 x^2+4 x^3+\log (x)\right ) \, dx=e^{9-5 x-3 x^2+2 x^3+x^4} x^x \]

[In]

Integrate[E^(9 - 5*x - 3*x^2 + 2*x^3 + x^4 + x*Log[x])*(-4 - 6*x + 6*x^2 + 4*x^3 + Log[x]),x]

[Out]

E^(9 - 5*x - 3*x^2 + 2*x^3 + x^4)*x^x

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\({\mathrm e}^{x \ln \left (x \right )+x^{4}+2 x^{3}-3 x^{2}-5 x +9}\)	\(24\)
default	\({\mathrm e}^{x \ln \left (x \right )+x^{4}+2 x^{3}-3 x^{2}-5 x +9}\)	\(24\)
norman	\({\mathrm e}^{x \ln \left (x \right )+x^{4}+2 x^{3}-3 x^{2}-5 x +9}\)	\(24\)
risch	\(x^{x} {\mathrm e}^{x^{4}+2 x^{3}-3 x^{2}-5 x +9}\)	\(24\)
parallelrisch	\({\mathrm e}^{x \ln \left (x \right )+x^{4}+2 x^{3}-3 x^{2}-5 x +9}\)	\(24\)

[In]

int((ln(x)+4*x^3+6*x^2-6*x-4)*exp(x*ln(x)+x^4+2*x^3-3*x^2-5*x+9),x,method=_RETURNVERBOSE)

[Out]

exp(x*ln(x)+x^4+2*x^3-3*x^2-5*x+9)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int e^{9-5 x-3 x^2+2 x^3+x^4+x \log (x)} \left (-4-6 x+6 x^2+4 x^3+\log (x)\right ) \, dx=e^{\left (x^{4} + 2 \, x^{3} - 3 \, x^{2} + x \log \left (x\right ) - 5 \, x + 9\right )} \]

[In]

integrate((log(x)+4*x^3+6*x^2-6*x-4)*exp(x*log(x)+x^4+2*x^3-3*x^2-5*x+9),x, algorithm="fricas")

[Out]

e^(x^4 + 2*x^3 - 3*x^2 + x*log(x) - 5*x + 9)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int e^{9-5 x-3 x^2+2 x^3+x^4+x \log (x)} \left (-4-6 x+6 x^2+4 x^3+\log (x)\right ) \, dx=e^{x^{4} + 2 x^{3} - 3 x^{2} + x \log {\left (x \right )} - 5 x + 9} \]

[In]

integrate((ln(x)+4*x**3+6*x**2-6*x-4)*exp(x*ln(x)+x**4+2*x**3-3*x**2-5*x+9),x)

[Out]

exp(x**4 + 2*x**3 - 3*x**2 + x*log(x) - 5*x + 9)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int e^{9-5 x-3 x^2+2 x^3+x^4+x \log (x)} \left (-4-6 x+6 x^2+4 x^3+\log (x)\right ) \, dx=e^{\left (x^{4} + 2 \, x^{3} - 3 \, x^{2} + x \log \left (x\right ) - 5 \, x + 9\right )} \]

[In]

integrate((log(x)+4*x^3+6*x^2-6*x-4)*exp(x*log(x)+x^4+2*x^3-3*x^2-5*x+9),x, algorithm="maxima")

[Out]

e^(x^4 + 2*x^3 - 3*x^2 + x*log(x) - 5*x + 9)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int e^{9-5 x-3 x^2+2 x^3+x^4+x \log (x)} \left (-4-6 x+6 x^2+4 x^3+\log (x)\right ) \, dx=e^{\left (x^{4} + 2 \, x^{3} - 3 \, x^{2} + x \log \left (x\right ) - 5 \, x + 9\right )} \]

[In]

integrate((log(x)+4*x^3+6*x^2-6*x-4)*exp(x*log(x)+x^4+2*x^3-3*x^2-5*x+9),x, algorithm="giac")

[Out]

e^(x^4 + 2*x^3 - 3*x^2 + x*log(x) - 5*x + 9)

Mupad [B] (verification not implemented)

Time = 12.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int e^{9-5 x-3 x^2+2 x^3+x^4+x \log (x)} \left (-4-6 x+6 x^2+4 x^3+\log (x)\right ) \, dx=x^x\,{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^9\,{\mathrm {e}}^{-3\,x^2}\,{\mathrm {e}}^{2\,x^3} \]

[In]

int(exp(x*log(x) - 5*x - 3*x^2 + 2*x^3 + x^4 + 9)*(log(x) - 6*x + 6*x^2 + 4*x^3 - 4),x)

[Out]

x^x*exp(-5*x)*exp(x^4)*exp(9)*exp(-3*x^2)*exp(2*x^3)