3.56.15 \(\int \frac {e^{\frac {1}{10} (1-5 e^{\frac {-1+x^2}{x}} x)+\frac {-1+x^2}{x}} (1+x+x^2)}{40 x} \, dx\)

Optimal. Leaf size=28 \[ 1-\frac {1}{20} e^{\frac {1}{2} \left (\frac {1}{5}-e^{-\frac {1}{x}+x} x\right )} \]

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Rubi [F]  time = 0.72, 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^{\frac {1}{10} \left (1-5 e^{\frac {-1+x^2}{x}} x\right )+\frac {-1+x^2}{x}} \left (1+x+x^2\right )}{40 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((1 - 5*E^((-1 + x^2)/x)*x)/10 + (-1 + x^2)/x)*(1 + x + x^2))/(40*x),x]

[Out]

Defer[Int][E^((1 - 5*E^((-1 + x^2)/x)*x)/10 + (-1 + x^2)/x), x]/40 + Defer[Int][E^((1 - 5*E^((-1 + x^2)/x)*x)/
10 + (-1 + x^2)/x)/x, x]/40 + Defer[Int][E^((1 - 5*E^((-1 + x^2)/x)*x)/10 + (-1 + x^2)/x)*x, x]/40

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{40} \int \frac {e^{\frac {1}{10} \left (1-5 e^{\frac {-1+x^2}{x}} x\right )+\frac {-1+x^2}{x}} \left (1+x+x^2\right )}{x} \, dx\\ &=\frac {1}{40} \int \left (e^{\frac {1}{10} \left (1-5 e^{\frac {-1+x^2}{x}} x\right )+\frac {-1+x^2}{x}}+\frac {e^{\frac {1}{10} \left (1-5 e^{\frac {-1+x^2}{x}} x\right )+\frac {-1+x^2}{x}}}{x}+e^{\frac {1}{10} \left (1-5 e^{\frac {-1+x^2}{x}} x\right )+\frac {-1+x^2}{x}} x\right ) \, dx\\ &=\frac {1}{40} \int e^{\frac {1}{10} \left (1-5 e^{\frac {-1+x^2}{x}} x\right )+\frac {-1+x^2}{x}} \, dx+\frac {1}{40} \int \frac {e^{\frac {1}{10} \left (1-5 e^{\frac {-1+x^2}{x}} x\right )+\frac {-1+x^2}{x}}}{x} \, dx+\frac {1}{40} \int e^{\frac {1}{10} \left (1-5 e^{\frac {-1+x^2}{x}} x\right )+\frac {-1+x^2}{x}} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 24, normalized size = 0.86 \begin {gather*} -\frac {1}{20} e^{\frac {1}{10}-\frac {1}{2} e^{-\frac {1}{x}+x} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((1 - 5*E^((-1 + x^2)/x)*x)/10 + (-1 + x^2)/x)*(1 + x + x^2))/(40*x),x]

[Out]

-1/20*E^(1/10 - (E^(-x^(-1) + x)*x)/2)

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fricas [B]  time = 0.65, size = 44, normalized size = 1.57 \begin {gather*} -\frac {1}{20} \, e^{\left (-\frac {5 \, x^{2} e^{\left (\frac {x^{2} - 1}{x}\right )} - 10 \, x^{2} - x + 10}{10 \, x} - \frac {x^{2} - 1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/40*(x^2+x+1)*exp((x^2-1)/x)*exp(-1/2*x*exp((x^2-1)/x)+1/10)/x,x, algorithm="fricas")

[Out]

-1/20*e^(-1/10*(5*x^2*e^((x^2 - 1)/x) - 10*x^2 - x + 10)/x - (x^2 - 1)/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} + x + 1\right )} e^{\left (-\frac {1}{2} \, x e^{\left (\frac {x^{2} - 1}{x}\right )} + \frac {x^{2} - 1}{x} + \frac {1}{10}\right )}}{40 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/40*(x^2+x+1)*exp((x^2-1)/x)*exp(-1/2*x*exp((x^2-1)/x)+1/10)/x,x, algorithm="giac")

[Out]

integrate(1/40*(x^2 + x + 1)*e^(-1/2*x*e^((x^2 - 1)/x) + (x^2 - 1)/x + 1/10)/x, x)

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maple [A]  time = 0.22, size = 19, normalized size = 0.68

method	result	size
norman	\(-\frac {{\mathrm e}^{-\frac {x \,{\mathrm e}^{\frac {x^{2}-1}{x}}}{2}+\frac {1}{10}}}{20}\)	\(19\)
risch	\(-\frac {{\mathrm e}^{-\frac {x \,{\mathrm e}^{\frac {\left (x -1\right ) \left (x +1\right )}{x}}}{2}+\frac {1}{10}}}{20}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/40*(x^2+x+1)*exp((x^2-1)/x)*exp(-1/2*x*exp((x^2-1)/x)+1/10)/x,x,method=_RETURNVERBOSE)

[Out]

-1/20*exp(-1/2*x*exp((x^2-1)/x)+1/10)

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maxima [A]  time = 0.57, size = 16, normalized size = 0.57 \begin {gather*} -\frac {1}{20} \, e^{\left (-\frac {1}{2} \, x e^{\left (x - \frac {1}{x}\right )} + \frac {1}{10}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/40*(x^2+x+1)*exp((x^2-1)/x)*exp(-1/2*x*exp((x^2-1)/x)+1/10)/x,x, algorithm="maxima")

[Out]

-1/20*e^(-1/2*x*e^(x - 1/x) + 1/10)

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mupad [B]  time = 3.59, size = 16, normalized size = 0.57 \begin {gather*} -\frac {{\mathrm {e}}^{1/10}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{-\frac {1}{x}}\,{\mathrm {e}}^x}{2}}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/10 - (x*exp((x^2 - 1)/x))/2)*exp((x^2 - 1)/x)*(x + x^2 + 1))/(40*x),x)

[Out]

-(exp(1/10)*exp(-(x*exp(-1/x)*exp(x))/2))/20

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sympy [A]  time = 0.37, size = 19, normalized size = 0.68 \begin {gather*} - \frac {e^{- \frac {x e^{\frac {x^{2} - 1}{x}}}{2} + \frac {1}{10}}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/40*(x**2+x+1)*exp((x**2-1)/x)*exp(-1/2*x*exp((x**2-1)/x)+1/10)/x,x)

[Out]

-exp(-x*exp((x**2 - 1)/x)/2 + 1/10)/20

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