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

Optimal. Leaf size=27 \[ e^{\frac {\left (-1+\frac {5 e^x (2-x)}{x}\right ) (1+x)}{\log (3)}}+x \]

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Rubi [F]  time = 2.96, 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 {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.60, size = 36, normalized size = 1.33 \begin {gather*} e^{-\frac {1}{\log (3)}-\frac {x}{\log (3)}-\frac {5 e^x (-2+x) (1+x)}{x \log (3)}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-Log[3]^(-1) - x/Log[3] - (5*E^x*(-2 + x)*(1 + x))/(x*Log[3])) + x

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fricas [A]  time = 0.69, size = 29, normalized size = 1.07 \begin {gather*} x + e^{\left (-\frac {x^{2} + 5 \, {\left (x^{2} - x - 2\right )} e^{x} + x}{x \log \relax (3)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^2*log(3) - (x^2 + 5*(x^3 - 2*x + 2)*e^x)*e^(-(x^2 + 5*(x^2 - x - 2)*e^x + x)/(x*log(3))))/(x^2*lo
g(3)), x)

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maple [A]  time = 0.13, size = 27, normalized size = 1.00

method	result	size
risch	\(x +{\mathrm e}^{-\frac {\left (x +1\right ) \left (5 \,{\mathrm e}^{x} x -10 \,{\mathrm e}^{x}+x \right )}{\ln \relax (3) x}}\)	\(27\)
norman	\(\frac {x^{2}+x \,{\mathrm e}^{\frac {\left (-5 x^{2}+5 x +10\right ) {\mathrm e}^{x}-x^{2}-x}{x \ln \relax (3)}}}{x}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x^3+10*x-10)*exp(x)-x^2)*exp(((-5*x^2+5*x+10)*exp(x)-x^2-x)/x/ln(3))+x^2*ln(3))/x^2/ln(3),x,method=_
RETURNVERBOSE)

[Out]

x+exp(-(x+1)*(5*exp(x)*x-10*exp(x)+x)/ln(3)/x)

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maxima [B]  time = 0.55, size = 56, normalized size = 2.07 \begin {gather*} \frac {x \log \relax (3) + e^{\left (-\frac {5 \, x e^{x}}{\log \relax (3)} - \frac {x}{\log \relax (3)} + \frac {5 \, e^{x}}{\log \relax (3)} + \frac {10 \, e^{x}}{x \log \relax (3)} - \frac {1}{\log \relax (3)}\right )} \log \relax (3)}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*log(3) + e^(-5*x*e^x/log(3) - x/log(3) + 5*e^x/log(3) + 10*e^x/(x*log(3)) - 1/log(3))*log(3))/log(3)

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mupad [B]  time = 3.78, size = 49, normalized size = 1.81 \begin {gather*} x+{\mathrm {e}}^{-\frac {x}{\ln \relax (3)}}\,{\mathrm {e}}^{\frac {10\,{\mathrm {e}}^x}{x\,\ln \relax (3)}}\,{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^x}{\ln \relax (3)}}\,{\mathrm {e}}^{-\frac {1}{\ln \relax (3)}}\,{\mathrm {e}}^{-\frac {5\,x\,{\mathrm {e}}^x}{\ln \relax (3)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(-x/log(3))*exp((10*exp(x))/(x*log(3)))*exp((5*exp(x))/log(3))*exp(-1/log(3))*exp(-(5*x*exp(x))/log(3))

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sympy [A]  time = 0.27, size = 26, normalized size = 0.96 \begin {gather*} x + e^{\frac {- x^{2} - x + \left (- 5 x^{2} + 5 x + 10\right ) e^{x}}{x \log {\relax (3 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp((-x**2 - x + (-5*x**2 + 5*x + 10)*exp(x))/(x*log(3)))

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