3.6.29 \(\int \frac {e^{3-x-\frac {e^{3-x} (1+5 e^{-3+x} x)}{x}} (-1-x-e^{-3+x} x+6 e^{-3+x+\frac {e^{3-x} (1+5 e^{-3+x} x)}{x}} x)}{5 x} \, dx\)

Optimal. Leaf size=33 \[ -1-e^3+x+\frac {1}{5} \left (x-e^{-5-\frac {e^{3-x}}{x}} x\right ) \]

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Rubi [F]  time = 1.46, 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^{3-x-\frac {e^{3-x} \left (1+5 e^{-3+x} x\right )}{x}} \left (-1-x-e^{-3+x} x+6 e^{-3+x+\frac {e^{3-x} \left (1+5 e^{-3+x} x\right )}{x}} x\right )}{5 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(3 - x - (E^(3 - x)*(1 + 5*E^(-3 + x)*x))/x)*(-1 - x - E^(-3 + x)*x + 6*E^(-3 + x + (E^(3 - x)*(1 + 5*E
^(-3 + x)*x))/x)*x))/(5*x),x]

[Out]

(6*x)/5 - Defer[Int][E^(-5 - E^(3 - x)/x), x]/5 - Defer[Int][E^(-2 - E^(3 - x)/x - x), x]/5 - Defer[Int][E^(-2
- E^(3 - x)/x - x)/x, x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{3-x-\frac {e^{3-x} \left (1+5 e^{-3+x} x\right )}{x}} \left (-1-x-e^{-3+x} x+6 e^{-3+x+\frac {e^{3-x} \left (1+5 e^{-3+x} x\right )}{x}} x\right )}{x} \, dx\\ &=\frac {1}{5} \int \left (6-e^{-5-\frac {e^{3-x}}{x}}-\frac {e^{-2-\frac {e^{3-x}}{x}-x} (1+x)}{x}\right ) \, dx\\ &=\frac {6 x}{5}-\frac {1}{5} \int e^{-5-\frac {e^{3-x}}{x}} \, dx-\frac {1}{5} \int \frac {e^{-2-\frac {e^{3-x}}{x}-x} (1+x)}{x} \, dx\\ &=\frac {6 x}{5}-\frac {1}{5} \int e^{-5-\frac {e^{3-x}}{x}} \, dx-\frac {1}{5} \int \left (e^{-2-\frac {e^{3-x}}{x}-x}+\frac {e^{-2-\frac {e^{3-x}}{x}-x}}{x}\right ) \, dx\\ &=\frac {6 x}{5}-\frac {1}{5} \int e^{-5-\frac {e^{3-x}}{x}} \, dx-\frac {1}{5} \int e^{-2-\frac {e^{3-x}}{x}-x} \, dx-\frac {1}{5} \int \frac {e^{-2-\frac {e^{3-x}}{x}-x}}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.98, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{5} \left (6 x-e^{-5-\frac {e^{3-x}}{x}} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 - x - (E^(3 - x)*(1 + 5*E^(-3 + x)*x))/x)*(-1 - x - E^(-3 + x)*x + 6*E^(-3 + x + (E^(3 - x)*(1
+ 5*E^(-3 + x)*x))/x)*x))/(5*x),x]

[Out]

(6*x - E^(-5 - E^(3 - x)/x)*x)/5

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fricas [B]  time = 0.77, size = 63, normalized size = 1.91 \begin {gather*} -\frac {1}{5} \, {\left (x e^{\left (x - 3\right )} - 6 \, x e^{\left (\frac {{\left ({\left (x^{2} + 2 \, x\right )} e^{\left (x - 3\right )} + 1\right )} e^{\left (-x + 3\right )}}{x}\right )}\right )} e^{\left (-\frac {{\left ({\left (x^{2} + 2 \, x\right )} e^{\left (x - 3\right )} + 1\right )} e^{\left (-x + 3\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(6*x*exp(x-3)*exp((5*x*exp(x-3)+1)/x/exp(x-3))-x*exp(x-3)-x-1)/x/exp(x-3)/exp((5*x*exp(x-3)+1)/x
/exp(x-3)),x, algorithm="fricas")

[Out]

-1/5*(x*e^(x - 3) - 6*x*e^(((x^2 + 2*x)*e^(x - 3) + 1)*e^(-x + 3)/x))*e^(-((x^2 + 2*x)*e^(x - 3) + 1)*e^(-x +
3)/x)

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giac [A]  time = 0.44, size = 23, normalized size = 0.70 \begin {gather*} -\frac {1}{5} \, x e^{\left (-\frac {5 \, x + e^{\left (-x + 3\right )}}{x}\right )} + \frac {6}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(6*x*exp(x-3)*exp((5*x*exp(x-3)+1)/x/exp(x-3))-x*exp(x-3)-x-1)/x/exp(x-3)/exp((5*x*exp(x-3)+1)/x
/exp(x-3)),x, algorithm="giac")

[Out]

-1/5*x*e^(-(5*x + e^(-x + 3))/x) + 6/5*x

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maple [A]  time = 0.08, size = 29, normalized size = 0.88

method	result	size
risch	\(\frac {6 x}{5}-\frac {x \,{\mathrm e}^{-\frac {\left (5 x \,{\mathrm e}^{x -3}+1\right ) {\mathrm e}^{3-x}}{x}}}{5}\)	\(29\)
norman	\(\left (-\frac {x \,{\mathrm e}^{x -3}}{5}+\frac {6 x \,{\mathrm e}^{x -3} {\mathrm e}^{\frac {\left (5 x \,{\mathrm e}^{x -3}+1\right ) {\mathrm e}^{3-x}}{x}}}{5}\right ) {\mathrm e}^{3-x} {\mathrm e}^{-\frac {\left (5 x \,{\mathrm e}^{x -3}+1\right ) {\mathrm e}^{3-x}}{x}}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(6*x*exp(x-3)*exp((5*x*exp(x-3)+1)/x/exp(x-3))-x*exp(x-3)-x-1)/x/exp(x-3)/exp((5*x*exp(x-3)+1)/x/exp(x
-3)),x,method=_RETURNVERBOSE)

[Out]

6/5*x-1/5*x*exp(-(5*x*exp(x-3)+1)/x*exp(3-x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {6}{5} \, x - \frac {1}{5} \, \int \frac {{\left (x e^{3} + x e^{x} + e^{3}\right )} e^{\left (-x - \frac {e^{\left (-x + 3\right )}}{x} - 5\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(6*x*exp(x-3)*exp((5*x*exp(x-3)+1)/x/exp(x-3))-x*exp(x-3)-x-1)/x/exp(x-3)/exp((5*x*exp(x-3)+1)/x
/exp(x-3)),x, algorithm="maxima")

[Out]

6/5*x - 1/5*integrate((x*e^3 + x*e^x + e^3)*e^(-x - e^(-x + 3)/x - 5)/x, x)

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mupad [B]  time = 0.58, size = 21, normalized size = 0.64 \begin {gather*} \frac {6\,x}{5}-\frac {x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x}}\,{\mathrm {e}}^{-5}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3 - x)*exp(-(exp(3 - x)*(5*x*exp(x - 3) + 1))/x)*(x/5 + (x*exp(x - 3))/5 - (6*x*exp(x - 3)*exp((exp(
3 - x)*(5*x*exp(x - 3) + 1))/x))/5 + 1/5))/x,x)

[Out]

(6*x)/5 - (x*exp(-(exp(-x)*exp(3))/x)*exp(-5))/5

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sympy [A]  time = 11.21, size = 26, normalized size = 0.79 \begin {gather*} \frac {6 x}{5} - \frac {x e^{- \frac {\left (5 x e^{x - 3} + 1\right ) e^{3 - x}}{x}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(6*x*exp(x-3)*exp((5*x*exp(x-3)+1)/x/exp(x-3))-x*exp(x-3)-x-1)/x/exp(x-3)/exp((5*x*exp(x-3)+1)/x
/exp(x-3)),x)

[Out]

6*x/5 - x*exp(-(5*x*exp(x - 3) + 1)*exp(3 - x)/x)/5

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