3.104.28 \(\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} (-10 e x-10 e^{2-x} x) \log (x)) \, dx\) [10328]

Optimal. Leaf size=29 \[ x^{1+\frac {e^{-2 e^{1+5 \left (-e^{1-x}+x\right )}}}{x^2}} \]

[Out]

exp(ln(x)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2)*x

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Rubi [F]
time = 5.71, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \left (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} \left (-10 e x-10 e^{2-x} x\right ) \log (x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x^(-2 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2))*(1 + E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2 - 2*Log[x] + E^
(-5*E^(1 - x) + 5*x)*(-10*E*x - 10*E^(2 - x)*x)*Log[x]))/E^(2*E^(1 - 5*E^(1 - x) + 5*x)),x]

[Out]

Defer[Int][x^(-2 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2))/E^(2*E^(1 - 5*E^(1 - x) + 5*x)), x] - 2*Log[x]*Def
er[Int][x^(-2 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2))/E^(2*E^(1 - 5*E^(1 - x) + 5*x)), x] - 10*Log[x]*Defer
[Int][E^(2 - 5*E^(1 - x) - 2*E^(1 - 5*E^(1 - x) + 5*x) + 4*x)*x^(-1 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2))
, x] - 10*Log[x]*Defer[Int][E^(1 - 5*E^(1 - x) - 2*E^(1 - 5*E^(1 - x) + 5*x) + 5*x)*x^(-1 + 1/(E^(2*E^(1 - 5*E
^(1 - x) + 5*x))*x^2)), x] + Defer[Int][x^(1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2)), x] + 2*Defer[Int][Defer[I
nt][x^(-2 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2))/E^(2*E^(1 - 5*E^(1 - x) + 5*x)), x]/x, x] + 10*Defer[Int]
[Defer[Int][E^(2 - 5*E^(1 - x) - 2*E^(1 - 5*E^(1 - x) + 5*x) + 4*x)*x^(-1 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))
*x^2)), x]/x, x] + 10*Defer[Int][Defer[Int][E^(1 - 5*E^(1 - x) - 2*E^(1 - 5*E^(1 - x) + 5*x) + 5*x)*x^(-1 + 1/
(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2)), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}}+x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}}-2 e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \log (x)-10 e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} \left (e+e^x\right ) x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \log (x)\right ) \, dx\\ &=-\left (2 \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \log (x) \, dx\right )-10 \int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} \left (e+e^x\right ) x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \log (x) \, dx+\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx\\ &=2 \int \frac {\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx+10 \int \frac {\int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx-(2 \log (x)) \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx\\ &=2 \int \frac {\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx+10 \int \left (\frac {\int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x}+\frac {\int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x}\right ) \, dx-(2 \log (x)) \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx\\ &=2 \int \frac {\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx+10 \int \frac {\int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx+10 \int \frac {\int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx-(2 \log (x)) \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.16, size = 28, normalized size = 0.97 \begin {gather*} x^{1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(-2 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2))*(1 + E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2 - 2*Log[x
] + E^(-5*E^(1 - x) + 5*x)*(-10*E*x - 10*E^(2 - x)*x)*Log[x]))/E^(2*E^(1 - 5*E^(1 - x) + 5*x)),x]

[Out]

x^(1 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2))

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Maple [A]
time = 0.09, size = 26, normalized size = 0.90

method	result	size
risch	\(x^{\frac {{\mathrm e}^{-2 \,{\mathrm e}^{1-5 \,{\mathrm e}^{1-x}+5 x}}}{x^{2}}} x\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(exp(1)*exp(-5*exp(1-x)+5*x))^2+(-10*x*exp(1)*exp(1-x)-10*x*exp(1))*ln(x)*exp(-5*exp(1-x)+5*x)-2*l
n(x)+1)*exp(ln(x)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2,x,method=_RET
URNVERBOSE)

[Out]

x^(1/x^2*exp(-2*exp(1-5*exp(1-x)+5*x)))*x

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*exp(exp(1)*exp(-5*exp(1-x)+5*x))^2+(-10*x*exp(1)*exp(1-x)-10*x*exp(1))*log(x)*exp(-5*exp(1-x)+5
*x)-2*log(x)+1)*exp(log(x)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2,x, a
lgorithm="maxima")

[Out]

integrate((x^2*e^(2*e^(5*x - 5*e^(-x + 1) + 1)) - 10*(x*e + x*e^(-x + 2))*e^(5*x - 5*e^(-x + 1))*log(x) - 2*lo
g(x) + 1)*x^(e^(-2*e^(5*x - 5*e^(-x + 1) + 1))/x^2 - 2)*e^(-2*e^(5*x - 5*e^(-x + 1) + 1)), x)

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Fricas [A]
time = 0.38, size = 32, normalized size = 1.10 \begin {gather*} x x^{\frac {e^{\left (-2 \, e^{\left ({\left ({\left (5 \, x + 1\right )} e - 5 \, e^{\left (-x + 2\right )}\right )} e^{\left (-1\right )}\right )}\right )}}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*exp(exp(1)*exp(-5*exp(1-x)+5*x))^2+(-10*x*exp(1)*exp(1-x)-10*x*exp(1))*log(x)*exp(-5*exp(1-x)+5
*x)-2*log(x)+1)*exp(log(x)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2,x, a
lgorithm="fricas")

[Out]

x*x^(e^(-2*e^(((5*x + 1)*e - 5*e^(-x + 2))*e^(-1)))/x^2)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*exp(exp(1)*exp(-5*exp(1-x)+5*x))**2+(-10*x*exp(1)*exp(1-x)-10*x*exp(1))*ln(x)*exp(-5*exp(1-x)+
5*x)-2*ln(x)+1)*exp(ln(x)/x**2/exp(exp(1)*exp(-5*exp(1-x)+5*x))**2)/x**2/exp(exp(1)*exp(-5*exp(1-x)+5*x))**2,x
)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3004 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*exp(exp(1)*exp(-5*exp(1-x)+5*x))^2+(-10*x*exp(1)*exp(1-x)-10*x*exp(1))*log(x)*exp(-5*exp(1-x)+5
*x)-2*log(x)+1)*exp(log(x)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2,x, a
lgorithm="giac")

[Out]

integrate((x^2*e^(2*e^(5*x - 5*e^(-x + 1) + 1)) - 10*(x*e + x*e^(-x + 2))*e^(5*x - 5*e^(-x + 1))*log(x) - 2*lo
g(x) + 1)*x^(e^(-2*e^(5*x - 5*e^(-x + 1) + 1))/x^2)*e^(-2*e^(5*x - 5*e^(-x + 1) + 1))/x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{-2\,\mathrm {e}\,{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}}\,\ln \left (x\right )}{x^2}}\,{\mathrm {e}}^{-2\,\mathrm {e}\,{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}}\,\left (2\,\ln \left (x\right )-x^2\,{\mathrm {e}}^{2\,\mathrm {e}\,{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}}+{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}\,\ln \left (x\right )\,\left (10\,x\,\mathrm {e}+10\,x\,\mathrm {e}\,{\mathrm {e}}^{1-x}\right )-1\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(-2*exp(1)*exp(5*x - 5*exp(1 - x)))*log(x))/x^2)*exp(-2*exp(1)*exp(5*x - 5*exp(1 - x)))*(2*log(x
) - x^2*exp(2*exp(1)*exp(5*x - 5*exp(1 - x))) + exp(5*x - 5*exp(1 - x))*log(x)*(10*x*exp(1) + 10*x*exp(1)*exp(
1 - x)) - 1))/x^2,x)

[Out]

int(-(exp((exp(-2*exp(1)*exp(5*x - 5*exp(1 - x)))*log(x))/x^2)*exp(-2*exp(1)*exp(5*x - 5*exp(1 - x)))*(2*log(x
) - x^2*exp(2*exp(1)*exp(5*x - 5*exp(1 - x))) + exp(5*x - 5*exp(1 - x))*log(x)*(10*x*exp(1) + 10*x*exp(1)*exp(
1 - x)) - 1))/x^2, x)

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