\(\int \frac {e^{\frac {-2 e^{1+x}-4 x+e (8+16 x)}{x}} (-8 e+e^{1+x} (2-2 x))-x^2}{x^2} \, dx\) [10061]

Optimal result

Integrand size = 50, antiderivative size = 25 \[ \int \frac {e^{\frac {-2 e^{1+x}-4 x+e (8+16 x)}{x}} \left (-8 e+e^{1+x} (2-2 x)\right )-x^2}{x^2} \, dx=e^{-4-2 e \left (-8-\frac {4}{x}+\frac {e^x}{x}\right )}-x \]

[Out]

exp(-2*exp(1)*(exp(x)/x-4/x-8)-4)-x

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {14, 6838} \[ \int \frac {e^{\frac {-2 e^{1+x}-4 x+e (8+16 x)}{x}} \left (-8 e+e^{1+x} (2-2 x)\right )-x^2}{x^2} \, dx=e^{-\frac {2 e^{x+1}}{x}+\frac {8 e}{x}-4 (1-4 e)}-x \]

[In]

Int[(E^((-2*E^(1 + x) - 4*x + E*(8 + 16*x))/x)*(-8*E + E^(1 + x)*(2 - 2*x)) - x^2)/x^2,x]

[Out]

E^(-4*(1 - 4*E) + (8*E)/x - (2*E^(1 + x))/x) - 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 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 (-1+\frac {2 e^{-3 \left (1-\frac {16 e}{3}\right )+\frac {8 e}{x}-\frac {2 e^{1+x}}{x}} \left (-4+e^x-e^x x\right )}{x^2}\right ) \, dx \\ & = -x+2 \int \frac {e^{-3 \left (1-\frac {16 e}{3}\right )+\frac {8 e}{x}-\frac {2 e^{1+x}}{x}} \left (-4+e^x-e^x x\right )}{x^2} \, dx \\ & = e^{-4 (1-4 e)+\frac {8 e}{x}-\frac {2 e^{1+x}}{x}}-x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {-2 e^{1+x}-4 x+e (8+16 x)}{x}} \left (-8 e+e^{1+x} (2-2 x)\right )-x^2}{x^2} \, dx=e^{-4+16 e+\frac {8 e}{x}-\frac {2 e^{1+x}}{x}}-x \]

[In]

Integrate[(E^((-2*E^(1 + x) - 4*x + E*(8 + 16*x))/x)*(-8*E + E^(1 + x)*(2 - 2*x)) - x^2)/x^2,x]

[Out]

E^(-4 + 16*E + (8*E)/x - (2*E^(1 + x))/x) - x

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(-x +{\mathrm e}^{\frac {-2 \,{\mathrm e} \,{\mathrm e}^{x}+\left (16 x +8\right ) {\mathrm e}-4 x}{x}}\)	\(28\)
risch	\(-x +{\mathrm e}^{\frac {16 x \,{\mathrm e}+8 \,{\mathrm e}-2 \,{\mathrm e}^{1+x}-4 x}{x}}\)	\(30\)
norman	\(\frac {x \,{\mathrm e}^{\frac {-2 \,{\mathrm e} \,{\mathrm e}^{x}+\left (16 x +8\right ) {\mathrm e}-4 x}{x}}-x^{2}}{x}\)	\(36\)

[In]

int((((2-2*x)*exp(1)*exp(x)-8*exp(1))*exp((-2*exp(1)*exp(x)+(16*x+8)*exp(1)-4*x)/x)-x^2)/x^2,x,method=_RETURNV
ERBOSE)

[Out]

-x+exp((-2*exp(1)*exp(x)+(16*x+8)*exp(1)-4*x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {-2 e^{1+x}-4 x+e (8+16 x)}{x}} \left (-8 e+e^{1+x} (2-2 x)\right )-x^2}{x^2} \, dx=-x + e^{\left (\frac {2 \, {\left (4 \, {\left (2 \, x + 1\right )} e - 2 \, x - e^{\left (x + 1\right )}\right )}}{x}\right )} \]

[In]

integrate((((2-2*x)*exp(1)*exp(x)-8*exp(1))*exp((-2*exp(1)*exp(x)+(16*x+8)*exp(1)-4*x)/x)-x^2)/x^2,x, algorith
m="fricas")

[Out]

-x + e^(2*(4*(2*x + 1)*e - 2*x - e^(x + 1))/x)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {-2 e^{1+x}-4 x+e (8+16 x)}{x}} \left (-8 e+e^{1+x} (2-2 x)\right )-x^2}{x^2} \, dx=- x + e^{\frac {- 4 x + e \left (16 x + 8\right ) - 2 e e^{x}}{x}} \]

[In]

integrate((((2-2*x)*exp(1)*exp(x)-8*exp(1))*exp((-2*exp(1)*exp(x)+(16*x+8)*exp(1)-4*x)/x)-x**2)/x**2,x)

[Out]

-x + exp((-4*x + E*(16*x + 8) - 2*E*exp(x))/x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {-2 e^{1+x}-4 x+e (8+16 x)}{x}} \left (-8 e+e^{1+x} (2-2 x)\right )-x^2}{x^2} \, dx=-x + e^{\left (\frac {8 \, e}{x} - \frac {2 \, e^{\left (x + 1\right )}}{x} + 16 \, e - 4\right )} \]

[In]

integrate((((2-2*x)*exp(1)*exp(x)-8*exp(1))*exp((-2*exp(1)*exp(x)+(16*x+8)*exp(1)-4*x)/x)-x^2)/x^2,x, algorith
m="maxima")

[Out]

-x + e^(8*e/x - 2*e^(x + 1)/x + 16*e - 4)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {-2 e^{1+x}-4 x+e (8+16 x)}{x}} \left (-8 e+e^{1+x} (2-2 x)\right )-x^2}{x^2} \, dx=-x + e^{\left (\frac {8 \, e}{x} - \frac {2 \, e^{\left (x + 1\right )}}{x} + 16 \, e - 4\right )} \]

[In]

integrate((((2-2*x)*exp(1)*exp(x)-8*exp(1))*exp((-2*exp(1)*exp(x)+(16*x+8)*exp(1)-4*x)/x)-x^2)/x^2,x, algorith
m="giac")

[Out]

-x + e^(8*e/x - 2*e^(x + 1)/x + 16*e - 4)

Mupad [B] (verification not implemented)

Time = 17.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {-2 e^{1+x}-4 x+e (8+16 x)}{x}} \left (-8 e+e^{1+x} (2-2 x)\right )-x^2}{x^2} \, dx={\mathrm {e}}^{\frac {8\,\mathrm {e}}{x}}\,{\mathrm {e}}^{16\,\mathrm {e}}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{-\frac {2\,\mathrm {e}\,{\mathrm {e}}^x}{x}}-x \]

[In]

int(-(exp(-(4*x + 2*exp(1)*exp(x) - exp(1)*(16*x + 8))/x)*(8*exp(1) + exp(1)*exp(x)*(2*x - 2)) + x^2)/x^2,x)

[Out]

exp((8*exp(1))/x)*exp(16*exp(1))*exp(-4)*exp(-(2*exp(1)*exp(x))/x) - x