3.20.80 \(\int \frac {-6-3 x-x^3 \log (4 e^e)}{3 x-e^x x^3+x^3 \log (4 e^e)} \, dx\)

Optimal. Leaf size=22 \[ -x+\log \left (-e^x+\frac {3}{x^2}+\log \left (4 e^e\right )\right ) \]

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Rubi [F]  time = 0.80, 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 {-6-3 x-x^3 \log \left (4 e^e\right )}{3 x-e^x x^3+x^3 \log \left (4 e^e\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-6 - 3*x - x^3*Log[4*E^E])/(3*x - E^x*x^3 + x^3*Log[4*E^E]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 30, normalized size = 1.36 \begin {gather*} -x-2 \log (x)+\log \left (3+e x^2-e^x x^2+x^2 \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 - 3*x - x^3*Log[4*E^E])/(3*x - E^x*x^3 + x^3*Log[4*E^E]),x]

[Out]

-x - 2*Log[x] + Log[3 + E*x^2 - E^x*x^2 + x^2*Log[4]]

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fricas [A]  time = 0.70, size = 32, normalized size = 1.45 \begin {gather*} -x + \log \left (-\frac {x^{2} e - x^{2} e^{x} + 2 \, x^{2} \log \relax (2) + 3}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*log(4*exp(exp(1)))-3*x-6)/(x^3*log(4*exp(exp(1)))-exp(x)*x^3+3*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.31, size = 31, normalized size = 1.41 \begin {gather*} -x + \log \left (x^{2} e - x^{2} e^{x} + 2 \, x^{2} \log \relax (2) + 3\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*log(4*exp(exp(1)))-3*x-6)/(x^3*log(4*exp(exp(1)))-exp(x)*x^3+3*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 29, normalized size = 1.32

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^3*ln(4*exp(exp(1)))-3*x-6)/(x^3*ln(4*exp(exp(1)))-exp(x)*x^3+3*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.66, size = 30, normalized size = 1.36 \begin {gather*} -x + \log \left (-\frac {x^{2} {\left (e + 2 \, \log \relax (2)\right )} - x^{2} e^{x} + 3}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*log(4*exp(exp(1)))-3*x-6)/(x^3*log(4*exp(exp(1)))-exp(x)*x^3+3*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.27, size = 31, normalized size = 1.41 \begin {gather*} \ln \left (\frac {x^2\,\mathrm {e}-x^2\,{\mathrm {e}}^x+2\,x^2\,\ln \relax (2)+3}{x^2}\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + x^3*log(4*exp(exp(1))) + 6)/(3*x - x^3*exp(x) + x^3*log(4*exp(exp(1)))),x)

[Out]

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

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sympy [A]  time = 0.21, size = 27, normalized size = 1.23 \begin {gather*} - x + \log {\left (e^{x} + \frac {- e x^{2} - 2 x^{2} \log {\relax (2 )} - 3}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**3*ln(4*exp(exp(1)))-3*x-6)/(x**3*ln(4*exp(exp(1)))-exp(x)*x**3+3*x),x)

[Out]

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

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