3.90.97 \(\int \frac {e^{-e^x-2 x} (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} (e^{3 x} x^2+e^{2 x} (-x+x^2)))}{x^3} \, dx\)

Optimal. Leaf size=34 \[ 25 e^{-2 x}+\frac {e^{e^x+x}-\frac {5 e^{-e^x}}{x}}{x}+\log (3) \]

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Rubi [F]  time = 1.52, 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 \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-E^x - 2*x)*(10*E^(2*x) + 5*E^(3*x)*x - 50*E^E^x*x^3 + E^(2*E^x + x)*(E^(3*x)*x^2 + E^(2*x)*(-x + x^2)
)))/x^3,x]

[Out]

25/E^(2*x) + 10*Defer[Int][1/(E^E^x*x^3), x] + 5*Defer[Int][E^(-E^x + x)/x^2, x] - Defer[Int][E^(E^x + x)/x^2,
x] + Defer[Int][E^(E^x + x)/x, x] + Defer[Int][E^(E^x + 2*x)/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-50 e^{-2 x}+\frac {10 e^{-e^x}}{x^3}+\frac {e^{-e^x-2 x+2 \left (e^x+2 x\right )}}{x}+\frac {e^{-e^x+x} \left (5-e^{2 e^x}+e^{2 e^x} x\right )}{x^2}\right ) \, dx\\ &=10 \int \frac {e^{-e^x}}{x^3} \, dx-50 \int e^{-2 x} \, dx+\int \frac {e^{-e^x-2 x+2 \left (e^x+2 x\right )}}{x} \, dx+\int \frac {e^{-e^x+x} \left (5-e^{2 e^x}+e^{2 e^x} x\right )}{x^2} \, dx\\ &=25 e^{-2 x}+10 \int \frac {e^{-e^x}}{x^3} \, dx+\int \frac {e^{-e^x+x} \left (5+e^{2 e^x} (-1+x)\right )}{x^2} \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx\\ &=25 e^{-2 x}+10 \int \frac {e^{-e^x}}{x^3} \, dx+\int \left (\frac {5 e^{-e^x+x}}{x^2}+\frac {e^{e^x+x} (-1+x)}{x^2}\right ) \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx\\ &=25 e^{-2 x}+5 \int \frac {e^{-e^x+x}}{x^2} \, dx+10 \int \frac {e^{-e^x}}{x^3} \, dx+\int \frac {e^{e^x+x} (-1+x)}{x^2} \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx\\ &=25 e^{-2 x}+5 \int \frac {e^{-e^x+x}}{x^2} \, dx+10 \int \frac {e^{-e^x}}{x^3} \, dx+\int \left (-\frac {e^{e^x+x}}{x^2}+\frac {e^{e^x+x}}{x}\right ) \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx\\ &=25 e^{-2 x}+5 \int \frac {e^{-e^x+x}}{x^2} \, dx+10 \int \frac {e^{-e^x}}{x^3} \, dx-\int \frac {e^{e^x+x}}{x^2} \, dx+\int \frac {e^{e^x+x}}{x} \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 31, normalized size = 0.91 \begin {gather*} 25 e^{-2 x}-\frac {5 e^{-e^x}}{x^2}+\frac {e^{e^x+x}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-E^x - 2*x)*(10*E^(2*x) + 5*E^(3*x)*x - 50*E^E^x*x^3 + E^(2*E^x + x)*(E^(3*x)*x^2 + E^(2*x)*(-x
+ x^2))))/x^3,x]

[Out]

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

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fricas [A]  time = 0.55, size = 39, normalized size = 1.15 \begin {gather*} \frac {{\left (25 \, x^{2} e^{\left (e^{x}\right )} + x e^{\left (3 \, x + 2 \, e^{x}\right )} - 5 \, e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x - e^{x}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(x)^3+(x^2-x)*exp(x)^2)*exp(exp(x))*exp(exp(x)+x)-50*x^3*exp(exp(x))+5*x*exp(x)^3+10*exp(x)
^2)/x^3/exp(x)^2/exp(exp(x)),x, algorithm="fricas")

[Out]

(25*x^2*e^(e^x) + x*e^(3*x + 2*e^x) - 5*e^(2*x))*e^(-2*x - e^x)/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (50 \, x^{3} e^{\left (e^{x}\right )} - 5 \, x e^{\left (3 \, x\right )} - {\left (x^{2} e^{\left (3 \, x\right )} + {\left (x^{2} - x\right )} e^{\left (2 \, x\right )}\right )} e^{\left (x + 2 \, e^{x}\right )} - 10 \, e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x - e^{x}\right )}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(x)^3+(x^2-x)*exp(x)^2)*exp(exp(x))*exp(exp(x)+x)-50*x^3*exp(exp(x))+5*x*exp(x)^3+10*exp(x)
^2)/x^3/exp(x)^2/exp(exp(x)),x, algorithm="giac")

[Out]

integrate(-(50*x^3*e^(e^x) - 5*x*e^(3*x) - (x^2*e^(3*x) + (x^2 - x)*e^(2*x))*e^(x + 2*e^x) - 10*e^(2*x))*e^(-2
*x - e^x)/x^3, x)

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maple [A]  time = 0.04, size = 27, normalized size = 0.79

method	result	size
risch	\(25 \,{\mathrm e}^{-2 x}+\frac {{\mathrm e}^{{\mathrm e}^{x}+x}}{x}-\frac {5 \,{\mathrm e}^{-{\mathrm e}^{x}}}{x^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2*exp(x)^3+(x^2-x)*exp(x)^2)*exp(exp(x))*exp(exp(x)+x)-50*x^3*exp(exp(x))+5*x*exp(x)^3+10*exp(x)^2)/x^
3/exp(x)^2/exp(exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 26, normalized size = 0.76 \begin {gather*} \frac {x e^{\left (x + e^{x}\right )} - 5 \, e^{\left (-e^{x}\right )}}{x^{2}} + 25 \, e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(x)^3+(x^2-x)*exp(x)^2)*exp(exp(x))*exp(exp(x)+x)-50*x^3*exp(exp(x))+5*x*exp(x)^3+10*exp(x)
^2)/x^3/exp(x)^2/exp(exp(x)),x, algorithm="maxima")

[Out]

(x*e^(x + e^x) - 5*e^(-e^x))/x^2 + 25*e^(-2*x)

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mupad [B]  time = 7.20, size = 26, normalized size = 0.76 \begin {gather*} 25\,{\mathrm {e}}^{-2\,x}+\frac {{\mathrm {e}}^{x+{\mathrm {e}}^x}}{x}-\frac {5\,{\mathrm {e}}^{-{\mathrm {e}}^x}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*x)*exp(-exp(x))*(10*exp(2*x) + 5*x*exp(3*x) - 50*x^3*exp(exp(x)) + exp(x + exp(x))*exp(exp(x))*(x^
2*exp(3*x) - exp(2*x)*(x - x^2))))/x^3,x)

[Out]

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

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sympy [A]  time = 0.23, size = 29, normalized size = 0.85 \begin {gather*} 25 e^{- 2 x} + \frac {x^{2} e^{x} e^{e^{x}} - 5 x e^{- e^{x}}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2*exp(x)**3+(x**2-x)*exp(x)**2)*exp(exp(x))*exp(exp(x)+x)-50*x**3*exp(exp(x))+5*x*exp(x)**3+10*
exp(x)**2)/x**3/exp(x)**2/exp(exp(x)),x)

[Out]

25*exp(-2*x) + (x**2*exp(x)*exp(exp(x)) - 5*x*exp(-exp(x)))/x**3

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