3.6.42 \(\int \frac {1-4 x^3-x^4+e^{e^5} (2 x+x^2)+(-2 x-x^2) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx\)

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

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Rubi [F]  time = 0.78, 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 {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 - 4*x^3 - x^4 + E^E^5*(2*x + x^2) + (-2*x - x^2)*Log[x])/(-x - E^E^5*x^2 + x^4 + x^2*Log[x]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 30, normalized size = 1.07 \begin {gather*} -x-\log (x)-\log \left (1+e^{e^5} x-x^3-x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 4*x^3 - x^4 + E^E^5*(2*x + x^2) + (-2*x - x^2)*Log[x])/(-x - E^E^5*x^2 + x^4 + x^2*Log[x]),x]

[Out]

-x - Log[x] - Log[1 + E^E^5*x - x^3 - x*Log[x]]

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fricas [A]  time = 0.67, size = 30, normalized size = 1.07 \begin {gather*} -x - 2 \, \log \relax (x) - \log \left (\frac {x^{3} - x e^{\left (e^{5}\right )} + x \log \relax (x) - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-2*x)*log(x)+(x^2+2*x)*exp(exp(5))-x^4-4*x^3+1)/(x^2*log(x)-x^2*exp(exp(5))+x^4-x),x, algorith
m="fricas")

[Out]

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

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giac [A]  time = 0.46, size = 26, normalized size = 0.93 \begin {gather*} -x - \log \left (x^{3} - x e^{\left (e^{5}\right )} + x \log \relax (x) - 1\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-2*x)*log(x)+(x^2+2*x)*exp(exp(5))-x^4-4*x^3+1)/(x^2*log(x)-x^2*exp(exp(5))+x^4-x),x, algorith
m="giac")

[Out]

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

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maple [A]  time = 0.08, size = 29, normalized size = 1.04

method	result	size
norman	\(-\ln \relax (x )-x -\ln \left (-x^{3}+x \,{\mathrm e}^{{\mathrm e}^{5}}-x \ln \relax (x )+1\right )\)	\(29\)
risch	\(-x -2 \ln \relax (x )-\ln \left (\ln \relax (x )-\frac {-x^{3}+x \,{\mathrm e}^{{\mathrm e}^{5}}+1}{x}\right )\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2-2*x)*ln(x)+(x^2+2*x)*exp(exp(5))-x^4-4*x^3+1)/(x^2*ln(x)-x^2*exp(exp(5))+x^4-x),x,method=_RETURNVER
BOSE)

[Out]

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

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maxima [A]  time = 0.54, size = 30, normalized size = 1.07 \begin {gather*} -x - 2 \, \log \relax (x) - \log \left (\frac {x^{3} - x e^{\left (e^{5}\right )} + x \log \relax (x) - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-2*x)*log(x)+(x^2+2*x)*exp(exp(5))-x^4-4*x^3+1)/(x^2*log(x)-x^2*exp(exp(5))+x^4-x),x, algorith
m="maxima")

[Out]

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

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mupad [B]  time = 0.59, size = 34, normalized size = 1.21 \begin {gather*} -\ln \left (x\,\ln \relax (x)-x\,{\mathrm {e}}^{{\mathrm {e}}^5}+x^3-1\right )-\frac {x^2\,\ln \relax (x)+x^3}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(2*x + x^2) - exp(exp(5))*(2*x + x^2) + 4*x^3 + x^4 - 1)/(x + x^2*exp(exp(5)) - x^2*log(x) - x^4),
x)

[Out]

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

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sympy [A]  time = 0.20, size = 26, normalized size = 0.93 \begin {gather*} - x - 2 \log {\relax (x )} - \log {\left (\log {\relax (x )} + \frac {x^{3} - x e^{e^{5}} - 1}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2-2*x)*ln(x)+(x**2+2*x)*exp(exp(5))-x**4-4*x**3+1)/(x**2*ln(x)-x**2*exp(exp(5))+x**4-x),x)

[Out]

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

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