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

Optimal. Leaf size=24 \[ \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}}}{x} \]

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Rubi [B]  time = 0.21, antiderivative size = 49, normalized size of antiderivative = 2.04, number of steps used = 1, number of rules used = 1, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {2288} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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 2288

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 24, normalized size = 1.00 \begin {gather*} \frac {e^{e^{1+e^{1-x}+x-\log ^4(2)}}}{x} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.78, size = 21, normalized size = 0.88 \begin {gather*} \frac {e^{\left (e^{\left (-\log \relax (2)^{4} + x + e^{\left (-x + 1\right )} + 1\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (x e^{\left (-x + 1\right )} - x\right )} e^{\left (-\log \relax (2)^{4} + x + e^{\left (-x + 1\right )} + 1\right )} + 1\right )} e^{\left (e^{\left (-\log \relax (2)^{4} + x + e^{\left (-x + 1\right )} + 1\right )}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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maple [A]  time = 0.07, size = 22, normalized size = 0.92

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.51, size = 21, normalized size = 0.88 \begin {gather*} \frac {e^{\left (e^{\left (-\log \relax (2)^{4} + x + e^{\left (-x + 1\right )} + 1\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.80, size = 25, normalized size = 1.04 \begin {gather*} \frac {{\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^{{\mathrm {e}}^{-x}\,\mathrm {e}}\,{\mathrm {e}}^{-{\ln \relax (2)}^4}\,{\mathrm {e}}^x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.24, size = 17, normalized size = 0.71 \begin {gather*} \frac {e^{e^{x + e^{1 - x} - \log {\relax (2 )}^{4} + 1}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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