\(\int \frac {-x-e^{5+e^5 x} x}{x} \, dx\) [8851]

Optimal result

Integrand size = 20, antiderivative size = 23 \[ \int \frac {-x-e^{5+e^5 x} x}{x} \, dx=3-e^{e^5 x}-\left (4+e^2\right )^2-x \]

[Out]

3-exp(exp(5+ln(x)))-(exp(1)^2+4)^2-x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {14, 2225} \[ \int \frac {-x-e^{5+e^5 x} x}{x} \, dx=-x-e^{e^5 x} \]

[In]

Int[(-x - E^(5 + E^5*x)*x)/x,x]

[Out]

-E^(E^5*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 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]

Rubi steps \begin{align*} \text {integral}& = \int \left (-1-e^{5+e^5 x}\right ) \, dx \\ & = -x-\int e^{5+e^5 x} \, dx \\ & = -e^{e^5 x}-x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {-x-e^{5+e^5 x} x}{x} \, dx=-e^{e^5 x}-x \]

[In]

Integrate[(-x - E^(5 + E^5*x)*x)/x,x]

[Out]

-E^(E^5*x) - x

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52

method	result	size
norman	\(-x -{\mathrm e}^{x \,{\mathrm e}^{5}}\)	\(12\)
risch	\(-x -{\mathrm e}^{x \,{\mathrm e}^{5}}\)	\(12\)
default	\(-x -{\mathrm e}^{{\mathrm e}^{5+\ln \left (x \right )}}\)	\(13\)
parts	\(-x -{\mathrm e}^{{\mathrm e}^{5+\ln \left (x \right )}}\)	\(13\)
parallelrisch	\(-\frac {x^{3}+{\mathrm e}^{{\mathrm e}^{5+\ln \left (x \right )}} x^{2}}{x^{2}}\)	\(20\)

[In]

int((-exp(5+ln(x))*exp(exp(5+ln(x)))-x)/x,x,method=_RETURNVERBOSE)

[Out]

-x-exp(x*exp(5))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-x-e^{5+e^5 x} x}{x} \, dx=-\frac {{\left (x^{2} e^{5} + e^{\left (x e^{5} + \log \left (x\right ) + 5\right )}\right )} e^{\left (-5\right )}}{x} \]

[In]

integrate((-exp(5+log(x))*exp(exp(5+log(x)))-x)/x,x, algorithm="fricas")

[Out]

-(x^2*e^5 + e^(x*e^5 + log(x) + 5))*e^(-5)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.35 \[ \int \frac {-x-e^{5+e^5 x} x}{x} \, dx=- x - e^{x e^{5}} \]

[In]

integrate((-exp(5+ln(x))*exp(exp(5+ln(x)))-x)/x,x)

[Out]

-x - exp(x*exp(5))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.48 \[ \int \frac {-x-e^{5+e^5 x} x}{x} \, dx=-x - e^{\left (x e^{5}\right )} \]

[In]

integrate((-exp(5+log(x))*exp(exp(5+log(x)))-x)/x,x, algorithm="maxima")

[Out]

-x - e^(x*e^5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.48 \[ \int \frac {-x-e^{5+e^5 x} x}{x} \, dx=-x - e^{\left (x e^{5}\right )} \]

[In]

integrate((-exp(5+log(x))*exp(exp(5+log(x)))-x)/x,x, algorithm="giac")

[Out]

-x - e^(x*e^5)

Mupad [B] (verification not implemented)

Time = 11.94 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.48 \[ \int \frac {-x-e^{5+e^5 x} x}{x} \, dx=-x-{\mathrm {e}}^{x\,{\mathrm {e}}^5} \]

[In]

int(-(x + exp(exp(log(x) + 5))*exp(log(x) + 5))/x,x)

[Out]

- x - exp(x*exp(5))