\(\int e^{\frac {1}{2} (-e^x+x)} (-6 x^2-x^3+e^x x^3) \, dx\) [4341]

Optimal result

Integrand size = 32, antiderivative size = 20 \[ \int e^{\frac {1}{2} \left (-e^x+x\right )} \left (-6 x^2-x^3+e^x x^3\right ) \, dx=-4-2 e^{\frac {1}{2} \left (-e^x+x\right )} x^3 \]

[Out]

-4-2*x^3*exp(1/2*x-1/2*exp(x))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {2326} \[ \int e^{\frac {1}{2} \left (-e^x+x\right )} \left (-6 x^2-x^3+e^x x^3\right ) \, dx=-\frac {2 e^{\frac {1}{2} \left (x-e^x\right )} \left (x^3-e^x x^3\right )}{1-e^x} \]

[In]

Int[E^((-E^x + x)/2)*(-6*x^2 - x^3 + E^x*x^3),x]

[Out]

(-2*E^((-E^x + x)/2)*(x^3 - E^x*x^3))/(1 - E^x)

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int e^{\frac {1}{2} \left (-e^x+x\right )} \left (-6 x^2-x^3+e^x x^3\right ) \, dx=-2 e^{\frac {1}{2} \left (-e^x+x\right )} x^3 \]

[In]

Integrate[E^((-E^x + x)/2)*(-6*x^2 - x^3 + E^x*x^3),x]

[Out]

-2*E^((-E^x + x)/2)*x^3

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
norman	\(-2 x^{3} {\mathrm e}^{\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}}\)	\(15\)
risch	\(-2 x^{3} {\mathrm e}^{\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}}\)	\(15\)
parallelrisch	\(-2 x^{3} {\mathrm e}^{\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}}\)	\(15\)

[In]

int((exp(x)*x^3-x^3-6*x^2)*exp(1/2*x-1/2*exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int e^{\frac {1}{2} \left (-e^x+x\right )} \left (-6 x^2-x^3+e^x x^3\right ) \, dx=-2 \, x^{3} e^{\left (\frac {1}{2} \, x - \frac {1}{2} \, e^{x}\right )} \]

[In]

integrate((exp(x)*x^3-x^3-6*x^2)*exp(1/2*x-1/2*exp(x)),x, algorithm="fricas")

[Out]

-2*x^3*e^(1/2*x - 1/2*e^x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int e^{\frac {1}{2} \left (-e^x+x\right )} \left (-6 x^2-x^3+e^x x^3\right ) \, dx=- 2 x^{3} e^{\frac {x}{2} - \frac {e^{x}}{2}} \]

[In]

integrate((exp(x)*x**3-x**3-6*x**2)*exp(1/2*x-1/2*exp(x)),x)

[Out]

-2*x**3*exp(x/2 - exp(x)/2)

Maxima [F]

\[ \int e^{\frac {1}{2} \left (-e^x+x\right )} \left (-6 x^2-x^3+e^x x^3\right ) \, dx=\int { {\left (x^{3} e^{x} - x^{3} - 6 \, x^{2}\right )} e^{\left (\frac {1}{2} \, x - \frac {1}{2} \, e^{x}\right )} \,d x } \]

[In]

integrate((exp(x)*x^3-x^3-6*x^2)*exp(1/2*x-1/2*exp(x)),x, algorithm="maxima")

[Out]

integrate((x^3*e^x - x^3 - 6*x^2)*e^(1/2*x - 1/2*e^x), x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int e^{\frac {1}{2} \left (-e^x+x\right )} \left (-6 x^2-x^3+e^x x^3\right ) \, dx=-2 \, x^{3} e^{\left (\frac {1}{2} \, x - \frac {1}{2} \, e^{x}\right )} \]

[In]

integrate((exp(x)*x^3-x^3-6*x^2)*exp(1/2*x-1/2*exp(x)),x, algorithm="giac")

[Out]

-2*x^3*e^(1/2*x - 1/2*e^x)

Mupad [B] (verification not implemented)

Time = 10.31 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int e^{\frac {1}{2} \left (-e^x+x\right )} \left (-6 x^2-x^3+e^x x^3\right ) \, dx=-2\,x^3\,{\mathrm {e}}^{x/2}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{2}} \]

[In]

int(-exp(x/2 - exp(x)/2)*(6*x^2 - x^3*exp(x) + x^3),x)

[Out]

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