\(\int \frac {-2 x^2-e^x x^2+e^{\frac {-3-2 x+3 e^4 x+e^x x^2}{x}} (-3+e^x (-x^2-x^3))}{x^2} \, dx\) [3687]

Optimal result

Integrand size = 60, antiderivative size = 31 \[ \int \frac {-2 x^2-e^x x^2+e^{\frac {-3-2 x+3 e^4 x+e^x x^2}{x}} \left (-3+e^x \left (-x^2-x^3\right )\right )}{x^2} \, dx=1-e^x-e^{-2+3 e^4-\frac {3}{x}+e^x x}-2 x \]

[Out]

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14, 2225, 6820, 6838} \[ \int \frac {-2 x^2-e^x x^2+e^{\frac {-3-2 x+3 e^4 x+e^x x^2}{x}} \left (-3+e^x \left (-x^2-x^3\right )\right )}{x^2} \, dx=-2 x-e^x-e^{e^x x-\frac {3}{x}+3 e^4-2} \]

[In]

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

[Out]

-E^x - E^(-2 + 3*E^4 - 3/x + E^x*x) - 2*x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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}& = \int \left (-2-e^x+\frac {e^{-2 \left (1-\frac {3 e^4}{2}\right )-\frac {3}{x}+e^x x} \left (-3-e^x x^2-e^x x^3\right )}{x^2}\right ) \, dx \\ & = -2 x-\int e^x \, dx+\int \frac {e^{-2 \left (1-\frac {3 e^4}{2}\right )-\frac {3}{x}+e^x x} \left (-3-e^x x^2-e^x x^3\right )}{x^2} \, dx \\ & = -e^x-2 x+\int \frac {e^{-2 \left (1-\frac {3 e^4}{2}\right )-\frac {3}{x}+e^x x} \left (-3-e^x x^2 (1+x)\right )}{x^2} \, dx \\ & = -e^x-e^{-2+3 e^4-\frac {3}{x}+e^x x}-2 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-2 x^2-e^x x^2+e^{\frac {-3-2 x+3 e^4 x+e^x x^2}{x}} \left (-3+e^x \left (-x^2-x^3\right )\right )}{x^2} \, dx=-e^x-e^{-2+3 e^4-\frac {3}{x}+e^x x}-2 x \]

[In]

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

[Out]

-E^x - E^(-2 + 3*E^4 - 3/x + E^x*x) - 2*x

Maple [A] (verified)

Time = 2.98 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03

method	result	size
risch	\(-2 x -{\mathrm e}^{x}-{\mathrm e}^{\frac {{\mathrm e}^{x} x^{2}+3 x \,{\mathrm e}^{4}-2 x -3}{x}}\)	\(32\)
parallelrisch	\(-2 x -{\mathrm e}^{x}-{\mathrm e}^{\frac {{\mathrm e}^{x} x^{2}+3 x \,{\mathrm e}^{4}-2 x -3}{x}}\)	\(32\)
norman	\(\frac {-2 x^{2}-{\mathrm e}^{x} x -{\mathrm e}^{\frac {{\mathrm e}^{x} x^{2}+3 x \,{\mathrm e}^{4}-2 x -3}{x}} x}{x}\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x^2-e^x x^2+e^{\frac {-3-2 x+3 e^4 x+e^x x^2}{x}} \left (-3+e^x \left (-x^2-x^3\right )\right )}{x^2} \, dx=-2 \, x - e^{x} - e^{\left (\frac {x^{2} e^{x} + 3 \, x e^{4} - 2 \, x - 3}{x}\right )} \]

[In]

integrate((((-x^3-x^2)*exp(x)-3)*exp((exp(x)*x^2+3*x*exp(4)-2*x-3)/x)-exp(x)*x^2-2*x^2)/x^2,x, algorithm="fric
as")

[Out]

-2*x - e^x - e^((x^2*e^x + 3*x*e^4 - 2*x - 3)/x)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {-2 x^2-e^x x^2+e^{\frac {-3-2 x+3 e^4 x+e^x x^2}{x}} \left (-3+e^x \left (-x^2-x^3\right )\right )}{x^2} \, dx=- 2 x - e^{x} - e^{\frac {x^{2} e^{x} - 2 x + 3 x e^{4} - 3}{x}} \]

[In]

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

[Out]

-2*x - exp(x) - exp((x**2*exp(x) - 2*x + 3*x*exp(4) - 3)/x)

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {-2 x^2-e^x x^2+e^{\frac {-3-2 x+3 e^4 x+e^x x^2}{x}} \left (-3+e^x \left (-x^2-x^3\right )\right )}{x^2} \, dx=-2 \, x - e^{\left (x e^{x} - \frac {3}{x} + 3 \, e^{4} - 2\right )} - e^{x} \]

[In]

integrate((((-x^3-x^2)*exp(x)-3)*exp((exp(x)*x^2+3*x*exp(4)-2*x-3)/x)-exp(x)*x^2-2*x^2)/x^2,x, algorithm="maxi
ma")

[Out]

-2*x - e^(x*e^x - 3/x + 3*e^4 - 2) - e^x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {-2 x^2-e^x x^2+e^{\frac {-3-2 x+3 e^4 x+e^x x^2}{x}} \left (-3+e^x \left (-x^2-x^3\right )\right )}{x^2} \, dx=-2 \, x - e^{\left (x e^{x} - \frac {3}{x} + 3 \, e^{4} - 2\right )} - e^{x} \]

[In]

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

[Out]

-2*x - e^(x*e^x - 3/x + 3*e^4 - 2) - e^x

Mupad [B] (verification not implemented)

Time = 8.60 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-2 x^2-e^x x^2+e^{\frac {-3-2 x+3 e^4 x+e^x x^2}{x}} \left (-3+e^x \left (-x^2-x^3\right )\right )}{x^2} \, dx=-2\,x-{\mathrm {e}}^x-{\mathrm {e}}^{x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{3\,{\mathrm {e}}^4}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{-\frac {3}{x}} \]

[In]

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

[Out]

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