\(\int \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}} (-1+e^{1+e^{1-x}+x-\log ^4(2)} (x-e^{1-x} x))}{x^2} \, dx\) [7482]

Optimal result

Integrand size = 57, antiderivative size = 24 \[ \int \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}} \left (-1+e^{1+e^{1-x}+x-\log ^4(2)} \left (x-e^{1-x} x\right )\right )}{x^2} \, dx=\frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}}}{x} \]

[Out]

exp(exp(exp(1-x)-ln(2)^4+x+1))/x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(49\) vs. \(2(24)=48\).

Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {2326} \[ \int \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}} \left (-1+e^{1+e^{1-x}+x-\log ^4(2)} \left (x-e^{1-x} x\right )\right )}{x^2} \, dx=\frac {\left (x-e^{1-x} x\right ) e^{e^{x+e^{1-x}+1-\log ^4(2)}}}{\left (1-e^{1-x}\right ) x^2} \]

[In]

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

[Out]

(E^E^(1 + E^(1 - x) + x - Log[2]^4)*(x - E^(1 - x)*x))/((1 - E^(1 - x))*x^2)

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 {e^{e^{1+e^{1-x}+x-\log ^4(2)}} \left (x-e^{1-x} x\right )}{\left (1-e^{1-x}\right ) x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}} \left (-1+e^{1+e^{1-x}+x-\log ^4(2)} \left (x-e^{1-x} x\right )\right )}{x^2} \, dx=\frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}}}{x} \]

[In]

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

[Out]

E^E^(1 + E^(1 - x) + x - Log[2]^4)/x

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92

method	result	size
norman	\(\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{1-x}-\ln \left (2\right )^{4}+x +1}}}{x}\)	\(22\)
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{1-x}-\ln \left (2\right )^{4}+x +1}}}{x}\)	\(22\)
parallelrisch	\(\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{1-x}-\ln \left (2\right )^{4}+x +1}}}{x}\)	\(22\)

[In]

int(((-x*exp(1-x)+x)*exp(exp(1-x)-ln(2)^4+x+1)-1)*exp(exp(exp(1-x)-ln(2)^4+x+1))/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(exp(exp(1-x)-ln(2)^4+x+1))/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}} \left (-1+e^{1+e^{1-x}+x-\log ^4(2)} \left (x-e^{1-x} x\right )\right )}{x^2} \, dx=\frac {e^{\left (e^{\left (-\log \left (2\right )^{4} + x + e^{\left (-x + 1\right )} + 1\right )}\right )}}{x} \]

[In]

integrate(((-x*exp(1-x)+x)*exp(exp(1-x)-log(2)^4+x+1)-1)*exp(exp(exp(1-x)-log(2)^4+x+1))/x^2,x, algorithm="fri
cas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}} \left (-1+e^{1+e^{1-x}+x-\log ^4(2)} \left (x-e^{1-x} x\right )\right )}{x^2} \, dx=\frac {e^{e^{x + e^{1 - x} - \log {\left (2 \right )}^{4} + 1}}}{x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}} \left (-1+e^{1+e^{1-x}+x-\log ^4(2)} \left (x-e^{1-x} x\right )\right )}{x^2} \, dx=\frac {e^{\left (e^{\left (-\log \left (2\right )^{4} + x + e^{\left (-x + 1\right )} + 1\right )}\right )}}{x} \]

[In]

integrate(((-x*exp(1-x)+x)*exp(exp(1-x)-log(2)^4+x+1)-1)*exp(exp(exp(1-x)-log(2)^4+x+1))/x^2,x, algorithm="max
ima")

[Out]

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

Giac [F]

\[ \int \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}} \left (-1+e^{1+e^{1-x}+x-\log ^4(2)} \left (x-e^{1-x} x\right )\right )}{x^2} \, dx=\int { -\frac {{\left ({\left (x e^{\left (-x + 1\right )} - x\right )} e^{\left (-\log \left (2\right )^{4} + x + e^{\left (-x + 1\right )} + 1\right )} + 1\right )} e^{\left (e^{\left (-\log \left (2\right )^{4} + x + e^{\left (-x + 1\right )} + 1\right )}\right )}}{x^{2}} \,d x } \]

[In]

integrate(((-x*exp(1-x)+x)*exp(exp(1-x)-log(2)^4+x+1)-1)*exp(exp(exp(1-x)-log(2)^4+x+1))/x^2,x, algorithm="gia
c")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}} \left (-1+e^{1+e^{1-x}+x-\log ^4(2)} \left (x-e^{1-x} x\right )\right )}{x^2} \, dx=\frac {{\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^{{\mathrm {e}}^{-x}\,\mathrm {e}}\,{\mathrm {e}}^{-{\ln \left (2\right )}^4}\,{\mathrm {e}}^x}}{x} \]

[In]

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

[Out]

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