3.65.69 \(\int \frac {-e^{\frac {e^{e^3}}{3}} x^2+e^x (-4+6 x-x^2)}{e^x (4 x-x^2)+e^{\frac {e^{e^3}}{3}} (-4 x^2+x^3)} \, dx\)

Optimal. Leaf size=31 \[ -7-\log \left (\frac {-4+x}{e^{\frac {e^{e^3}}{3}}-\frac {e^x}{x}}\right ) \]

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Rubi [F]  time = 1.05, 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 {e^{e^3}}{3}} x^2+e^x \left (-4+6 x-x^2\right )}{e^x \left (4 x-x^2\right )+e^{\frac {e^{e^3}}{3}} \left (-4 x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 32, normalized size = 1.03 \begin {gather*} -\log (4-x)-\log (x)+\log \left (e^x-e^{\frac {e^{e^3}}{3}} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^(E^E^3/3)*x^2) + E^x*(-4 + 6*x - x^2))/(E^x*(4*x - x^2) + E^(E^E^3/3)*(-4*x^2 + x^3)),x]

[Out]

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

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fricas [A]  time = 0.67, size = 24, normalized size = 0.77 \begin {gather*} -\log \left (x^{2} - 4 \, x\right ) + \log \left (-x e^{\left (\frac {1}{3} \, e^{\left (e^{3}\right )}\right )} + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2*exp(1/3*exp(exp(3)))+(-x^2+6*x-4)*exp(x))/((x^3-4*x^2)*exp(1/3*exp(exp(3)))+(-x^2+4*x)*exp(x))
,x, algorithm="fricas")

[Out]

-log(x^2 - 4*x) + log(-x*e^(1/3*e^(e^3)) + e^x)

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giac [A]  time = 0.63, size = 24, normalized size = 0.77 \begin {gather*} \log \left (-x e^{\left (\frac {1}{3} \, e^{\left (e^{3}\right )}\right )} + e^{x}\right ) - \log \left (x - 4\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2*exp(1/3*exp(exp(3)))+(-x^2+6*x-4)*exp(x))/((x^3-4*x^2)*exp(1/3*exp(exp(3)))+(-x^2+4*x)*exp(x))
,x, algorithm="giac")

[Out]

log(-x*e^(1/3*e^(e^3)) + e^x) - log(x - 4) - log(x)

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maple [A]  time = 0.19, size = 25, normalized size = 0.81

method	result	size
risch	\(-\ln \left (x^{2}-4 x \right )+\ln \left ({\mathrm e}^{x}-x \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{3}}}{3}}\right )\)	\(25\)
norman	\(-\ln \relax (x )-\ln \left (x -4\right )+\ln \left (x \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{3}}}{3}}-{\mathrm e}^{x}\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2*exp(1/3*exp(exp(3)))+(-x^2+6*x-4)*exp(x))/((x^3-4*x^2)*exp(1/3*exp(exp(3)))+(-x^2+4*x)*exp(x)),x,met
hod=_RETURNVERBOSE)

[Out]

-ln(x^2-4*x)+ln(exp(x)-x*exp(1/3*exp(exp(3))))

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maxima [A]  time = 0.45, size = 24, normalized size = 0.77 \begin {gather*} \log \left (-x e^{\left (\frac {1}{3} \, e^{\left (e^{3}\right )}\right )} + e^{x}\right ) - \log \left (x - 4\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2*exp(1/3*exp(exp(3)))+(-x^2+6*x-4)*exp(x))/((x^3-4*x^2)*exp(1/3*exp(exp(3)))+(-x^2+4*x)*exp(x))
,x, algorithm="maxima")

[Out]

log(-x*e^(1/3*e^(e^3)) + e^x) - log(x - 4) - log(x)

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mupad [B]  time = 0.40, size = 24, normalized size = 0.77 \begin {gather*} \ln \left (x-{\mathrm {e}}^{-\frac {{\mathrm {e}}^{{\mathrm {e}}^3}}{3}}\,{\mathrm {e}}^x\right )-\ln \left (x-4\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(x^2 - 6*x + 4) + x^2*exp(exp(exp(3))/3))/(exp(x)*(4*x - x^2) - exp(exp(exp(3))/3)*(4*x^2 - x^3))
,x)

[Out]

log(x - exp(-exp(exp(3))/3)*exp(x)) - log(x - 4) - log(x)

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sympy [A]  time = 0.17, size = 22, normalized size = 0.71 \begin {gather*} - \log {\left (x^{2} - 4 x \right )} + \log {\left (- x e^{\frac {e^{e^{3}}}{3}} + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2*exp(1/3*exp(exp(3)))+(-x**2+6*x-4)*exp(x))/((x**3-4*x**2)*exp(1/3*exp(exp(3)))+(-x**2+4*x)*ex
p(x)),x)

[Out]

-log(x**2 - 4*x) + log(-x*exp(exp(exp(3))/3) + exp(x))

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