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

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

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Rubi [F]  time = 1.57, 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 {2-2 x+e^x \left (-2 x^2-x^3\right )}{-2 x^2-e^x x^3+2 x \log \left (\frac {x}{3 e^5}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x + 2*Log[x] - 2*(11 + Log[9])*Defer[Int][1/(x*(2*x + E^x*x^2 + 10*(1 + Log[3]/5) - 2*Log[x])), x] + 2*Defer[I
nt][Log[x/3]/(2*x + E^x*x^2 + 10*(1 + Log[3]/5) - 2*Log[x]), x] + 4*Defer[Int][Log[x]/(x*(2*x + E^x*x^2 + 10*(
1 + Log[3]/5) - 2*Log[x])), x] + 12*Defer[Int][(-2*x - E^x*x^2 - 10*(1 + Log[3]/5) + 2*Log[x])^(-1), x] + 2*De
fer[Int][x/(-2*x - E^x*x^2 - 10*(1 + Log[3]/5) + 2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2+x}{x}+\frac {2 \left (-6 x-x^2-11 \left (1+\frac {2 \log (3)}{11}\right )+x \log \left (\frac {x}{3}\right )+2 \log (x)\right )}{x \left (2 x+e^x x^2+10 \left (1+\frac {\log (3)}{5}\right )-2 \log (x)\right )}\right ) \, dx\\ &=2 \int \frac {-6 x-x^2-11 \left (1+\frac {2 \log (3)}{11}\right )+x \log \left (\frac {x}{3}\right )+2 \log (x)}{x \left (2 x+e^x x^2+10 \left (1+\frac {\log (3)}{5}\right )-2 \log (x)\right )} \, dx+\int \frac {2+x}{x} \, dx\\ &=2 \int \left (\frac {-11-\log (9)}{x \left (2 x+e^x x^2+10 \left (1+\frac {\log (3)}{5}\right )-2 \log (x)\right )}+\frac {\log \left (\frac {x}{3}\right )}{2 x+e^x x^2+10 \left (1+\frac {\log (3)}{5}\right )-2 \log (x)}+\frac {2 \log (x)}{x \left (2 x+e^x x^2+10 \left (1+\frac {\log (3)}{5}\right )-2 \log (x)\right )}+\frac {6}{-2 x-e^x x^2-10 \left (1+\frac {\log (3)}{5}\right )+2 \log (x)}+\frac {x}{-2 x-e^x x^2-10 \left (1+\frac {\log (3)}{5}\right )+2 \log (x)}\right ) \, dx+\int \left (1+\frac {2}{x}\right ) \, dx\\ &=x+2 \log (x)+2 \int \frac {\log \left (\frac {x}{3}\right )}{2 x+e^x x^2+10 \left (1+\frac {\log (3)}{5}\right )-2 \log (x)} \, dx+2 \int \frac {x}{-2 x-e^x x^2-10 \left (1+\frac {\log (3)}{5}\right )+2 \log (x)} \, dx+4 \int \frac {\log (x)}{x \left (2 x+e^x x^2+10 \left (1+\frac {\log (3)}{5}\right )-2 \log (x)\right )} \, dx+12 \int \frac {1}{-2 x-e^x x^2-10 \left (1+\frac {\log (3)}{5}\right )+2 \log (x)} \, dx-(2 (11+\log (9))) \int \frac {1}{x \left (2 x+e^x x^2+10 \left (1+\frac {\log (3)}{5}\right )-2 \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 19, normalized size = 0.70 \begin {gather*} \log \left (10+2 x+e^x x^2+\log (9)-2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[10 + 2*x + E^x*x^2 + Log[9] - 2*Log[x]]

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fricas [A]  time = 0.57, size = 20, normalized size = 0.74 \begin {gather*} \log \left (-x^{2} e^{x} - 2 \, x + 2 \, \log \left (\frac {1}{3} \, x e^{\left (-5\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-2*x^2)*exp(x)-2*x+2)/(2*x*log(1/3*x/exp(5))-exp(x)*x^3-2*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 19, normalized size = 0.70 \begin {gather*} \log \left (-x^{2} e^{x} - 2 \, x + 2 \, \log \left (\frac {1}{3} \, x\right ) - 10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-2*x^2)*exp(x)-2*x+2)/(2*x*log(1/3*x/exp(5))-exp(x)*x^3-2*x^2),x, algorithm="giac")

[Out]

log(-x^2*e^x - 2*x + 2*log(1/3*x) - 10)

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maple [A]  time = 0.04, size = 19, normalized size = 0.70

method	result	size
risch	\(\ln \left (-\frac {{\mathrm e}^{x} x^{2}}{2}-x +\ln \left (\frac {x \,{\mathrm e}^{-5}}{3}\right )\right )\)	\(19\)
norman	\(\ln \left ({\mathrm e}^{x} x^{2}-2 \ln \left (\frac {x \,{\mathrm e}^{-5}}{3}\right )+2 x \right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3-2*x^2)*exp(x)-2*x+2)/(2*x*ln(1/3*x/exp(5))-exp(x)*x^3-2*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(-1/2*exp(x)*x^2-x+ln(1/3*x*exp(-5)))

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maxima [A]  time = 0.48, size = 29, normalized size = 1.07 \begin {gather*} 2 \, \log \relax (x) + \log \left (\frac {x^{2} e^{x} + 2 \, x + 2 \, \log \relax (3) - 2 \, \log \relax (x) + 10}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-2*x^2)*exp(x)-2*x+2)/(2*x*log(1/3*x/exp(5))-exp(x)*x^3-2*x^2),x, algorithm="maxima")

[Out]

2*log(x) + log((x^2*e^x + 2*x + 2*log(3) - 2*log(x) + 10)/x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {2\,x+{\mathrm {e}}^x\,\left (x^3+2\,x^2\right )-2}{x^3\,{\mathrm {e}}^x-2\,x\,\ln \left (\frac {x\,{\mathrm {e}}^{-5}}{3}\right )+2\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.36, size = 26, normalized size = 0.96 \begin {gather*} 2 \log {\relax (x )} + \log {\left (e^{x} + \frac {2 x - 2 \log {\left (\frac {x}{3 e^{5}} \right )}}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3-2*x**2)*exp(x)-2*x+2)/(2*x*ln(1/3*x/exp(5))-exp(x)*x**3-2*x**2),x)

[Out]

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

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