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

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

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Rubi [F]  time = 8.55, 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 {(-3+x) \log (3-x)+\frac {e^{e^{e^5}+\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}} \left (-e^3 x-x^2+\left (-3 x+x^2\right ) \log (3-x)\right )}{\log (3-x)}}{\left (-3 x+x^2\right ) \log (3-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.54, size = 26, normalized size = 0.96 \begin {gather*} e^{\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 67, normalized size = 2.48 \begin {gather*} {\left (e^{\left (e^{\left (e^{5}\right )}\right )} \log \relax (x) + e^{\left (\frac {{\left (x + e^{3}\right )} e^{\left (e^{\left (e^{5}\right )}\right )} + e^{\left (e^{5}\right )} \log \left (-x + 3\right ) - \log \left (-x + 3\right ) \log \left (\log \left (-x + 3\right )\right )}{\log \left (-x + 3\right )}\right )} \log \left (-x + 3\right )\right )} e^{\left (-e^{\left (e^{5}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 22, normalized size = 0.81

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 4.77, size = 35, normalized size = 1.30 \begin {gather*} \ln \relax (x)+{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^5}}\,{\mathrm {e}}^3}{\ln \left (3-x\right )}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^5}}}{\ln \left (3-x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.45, size = 20, normalized size = 0.74 \begin {gather*} e^{\frac {\left (x + e^{3}\right ) e^{e^{e^{5}}}}{\log {\left (3 - x \right )}}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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