3.62.34 \(\int \frac {\log (x) (5+\log (e^{-\frac {2-x}{e^9}} (-5+e^{\frac {2-x}{e^9}} x))) (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+(-25 x+5 e^{9+\frac {2-x}{e^9}} x) \log (x)+(-25 e^9+5 e^{9+\frac {2-x}{e^9}} x) \log (e^{-\frac {2-x}{e^9}} (-5+e^{\frac {2-x}{e^9}} x)))}{(-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2) \log (x)+(-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2) \log (x) \log (e^{-\frac {2-x}{e^9}} (-5+e^{\frac {2-x}{e^9}} x))} \, dx\)

Optimal. Leaf size=23 \[ 5 \log (x) \left (5+\log \left (-5 e^{-\frac {2-x}{e^9}}+x\right )\right ) \]

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Rubi [F]  time = 4.28, 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 {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(Log[x]*(5 + Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x)/E^9)])*(-125*E^9 + 25*E^(9 + (2 - x)/E^9)*x + (-25*x
+ 5*E^(9 + (2 - x)/E^9)*x)*Log[x] + (-25*E^9 + 5*E^(9 + (2 - x)/E^9)*x)*Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x
)/E^9)]))/((-25*E^9*x + 5*E^(9 + (2 - x)/E^9)*x^2)*Log[x] + (-5*E^9*x + E^(9 + (2 - x)/E^9)*x^2)*Log[x]*Log[(-
5 + E^((2 - x)/E^9)*x)/E^((2 - x)/E^9)]),x]

[Out]

(-5*x)/E^9 + 25*Log[x] + (5*x*Log[x])/E^9 - 5*E^(2/E^9)*Log[x]*Defer[Int][(5*E^(x/E^9) - E^(2/E^9)*x)^(-1), x]
- 5*E^(-9 + 2/E^9)*Log[x]*Defer[Int][x/(-5*E^(x/E^9) + E^(2/E^9)*x), x] + 5*Defer[Int][Log[-5*E^((-2 + x)/E^9
) + x]/x, x] + 5*E^(2/E^9)*Defer[Int][Defer[Int][(5*E^(x/E^9) - E^(2/E^9)*x)^(-1), x]/x, x] + 5*E^(-9 + 2/E^9)
*Defer[Int][Defer[Int][x/(-5*E^(x/E^9) + E^(2/E^9)*x), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5 \left (e^{9+\frac {2}{e^9}}-5 e^{\frac {x}{e^9}}\right ) x \log (x)+5 e^9 \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right ) \left (5+\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )\right )}{e^9 x \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right )} \, dx\\ &=\frac {\int \frac {-5 \left (e^{9+\frac {2}{e^9}}-5 e^{\frac {x}{e^9}}\right ) x \log (x)+5 e^9 \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right ) \left (5+\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )\right )}{x \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right )} \, dx}{e^9}\\ &=\frac {\int \left (\frac {5 e^{\frac {2}{e^9}} \left (e^9-x\right ) \log (x)}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x}+\frac {5 \left (5 e^9+x \log (x)+e^9 \log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )\right )}{x}\right ) \, dx}{e^9}\\ &=\frac {5 \int \frac {5 e^9+x \log (x)+e^9 \log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx}{e^9}+\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {\left (e^9-x\right ) \log (x)}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx\\ &=\frac {5 \int \left (\frac {5 e^9+x \log (x)}{x}+\frac {e^9 \log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x}\right ) \, dx}{e^9}-\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {-e^9 \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx-\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x} \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx\\ &=5 \int \frac {\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx+\frac {5 \int \frac {5 e^9+x \log (x)}{x} \, dx}{e^9}-\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \left (-\frac {e^9 \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx}{x}-\frac {\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x}\right ) \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx\\ &=5 \int \frac {\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx+\frac {5 \int \left (\frac {5 e^9}{x}+\log (x)\right ) \, dx}{e^9}+\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x} \, dx+\left (5 e^{\frac {2}{e^9}}\right ) \int \frac {\int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx}{x} \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx\\ &=25 \log (x)+5 \int \frac {\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx+\frac {5 \int \log (x) \, dx}{e^9}+\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x} \, dx+\left (5 e^{\frac {2}{e^9}}\right ) \int \frac {\int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx}{x} \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx\\ &=-\frac {5 x}{e^9}+25 \log (x)+\frac {5 x \log (x)}{e^9}+5 \int \frac {\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx+\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x} \, dx+\left (5 e^{\frac {2}{e^9}}\right ) \int \frac {\int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx}{x} \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 20, normalized size = 0.87 \begin {gather*} 5 \log (x) \left (5+\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[x]*(5 + Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x)/E^9)])*(-125*E^9 + 25*E^(9 + (2 - x)/E^9)*x + (
-25*x + 5*E^(9 + (2 - x)/E^9)*x)*Log[x] + (-25*E^9 + 5*E^(9 + (2 - x)/E^9)*x)*Log[(-5 + E^((2 - x)/E^9)*x)/E^(
(2 - x)/E^9)]))/((-25*E^9*x + 5*E^(9 + (2 - x)/E^9)*x^2)*Log[x] + (-5*E^9*x + E^(9 + (2 - x)/E^9)*x^2)*Log[x]*
Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x)/E^9)]),x]

[Out]

5*Log[x]*(5 + Log[-5*E^((-2 + x)/E^9) + x])

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fricas [A]  time = 0.86, size = 41, normalized size = 1.78 \begin {gather*} 5 \, \log \left ({\left (x e^{\left (-{\left (x - 9 \, e^{9} - 2\right )} e^{\left (-9\right )}\right )} - 5 \, e^{9}\right )} e^{\left ({\left (x - 9 \, e^{9} - 2\right )} e^{\left (-9\right )}\right )}\right ) \log \relax (x) + 25 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*log((x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x*exp(9)
*exp((2-x)/exp(9))-25*x)*log(x)+25*x*exp(9)*exp((2-x)/exp(9))-125*exp(9))*exp(log(log((x*exp((2-x)/exp(9))-5)/
exp((2-x)/exp(9)))+5)+log(log(x)))/((x^2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9))*log(x)*log((x*exp((2-x)/exp(9))-
5)/exp((2-x)/exp(9)))+(5*x^2*exp(9)*exp((2-x)/exp(9))-25*x*exp(9))*log(x)),x, algorithm="fricas")

[Out]

5*log((x*e^(-(x - 9*e^9 - 2)*e^(-9)) - 5*e^9)*e^((x - 9*e^9 - 2)*e^(-9)))*log(x) + 25*log(x)

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giac [A]  time = 0.40, size = 36, normalized size = 1.57 \begin {gather*} 5 \, {\left (e^{9} \log \left (x e^{\left (2 \, e^{\left (-9\right )}\right )} - 5 \, e^{\left (x e^{\left (-9\right )}\right )}\right ) \log \relax (x) + 5 \, e^{9} \log \relax (x) - 2 \, \log \relax (x)\right )} e^{\left (-9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*log((x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x*exp(9)
*exp((2-x)/exp(9))-25*x)*log(x)+25*x*exp(9)*exp((2-x)/exp(9))-125*exp(9))*exp(log(log((x*exp((2-x)/exp(9))-5)/
exp((2-x)/exp(9)))+5)+log(log(x)))/((x^2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9))*log(x)*log((x*exp((2-x)/exp(9))-
5)/exp((2-x)/exp(9)))+(5*x^2*exp(9)*exp((2-x)/exp(9))-25*x*exp(9))*log(x)),x, algorithm="giac")

[Out]

5*(e^9*log(x*e^(2*e^(-9)) - 5*e^(x*e^(-9)))*log(x) + 5*e^9*log(x) - 2*log(x))*e^(-9)

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maple [C]  time = 0.31, size = 212, normalized size = 9.22

method	result	size
risch	\(-5 \ln \relax (x ) \ln \left ({\mathrm e}^{-\left (x -2\right ) {\mathrm e}^{-9}}\right )+5 \ln \relax (x ) \ln \left (x \,{\mathrm e}^{-\left (x -2\right ) {\mathrm e}^{-9}}-5\right )-\frac {5 i \pi \ln \relax (x ) \mathrm {csgn}\left (i \left (x \,{\mathrm e}^{-\left (x -2\right ) {\mathrm e}^{-9}}-5\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x -2\right ) {\mathrm e}^{-9}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x -2\right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (x -2\right ) {\mathrm e}^{-9}}-5\right )\right )}{2}+\frac {5 i \pi \ln \relax (x ) \mathrm {csgn}\left (i \left (x \,{\mathrm e}^{-\left (x -2\right ) {\mathrm e}^{-9}}-5\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x -2\right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (x -2\right ) {\mathrm e}^{-9}}-5\right )\right )^{2}}{2}+\frac {5 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x -2\right ) {\mathrm e}^{-9}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x -2\right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (x -2\right ) {\mathrm e}^{-9}}-5\right )\right )^{2}}{2}-\frac {5 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{\left (x -2\right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (x -2\right ) {\mathrm e}^{-9}}-5\right )\right )^{3}}{2}+25 \ln \relax (x )\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*ln((x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x*exp(9)*exp((2
-x)/exp(9))-25*x)*ln(x)+25*x*exp(9)*exp((2-x)/exp(9))-125*exp(9))*exp(ln(ln((x*exp((2-x)/exp(9))-5)/exp((2-x)/
exp(9)))+5)+ln(ln(x)))/((x^2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9))*ln(x)*ln((x*exp((2-x)/exp(9))-5)/exp((2-x)/e
xp(9)))+(5*x^2*exp(9)*exp((2-x)/exp(9))-25*x*exp(9))*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-5*ln(x)*ln(exp(-(x-2)*exp(-9)))+5*ln(x)*ln(x*exp(-(x-2)*exp(-9))-5)-5/2*I*Pi*ln(x)*csgn(I*(x*exp(-(x-2)*exp(-
9))-5))*csgn(I*exp((x-2)*exp(-9)))*csgn(I*exp((x-2)*exp(-9))*(x*exp(-(x-2)*exp(-9))-5))+5/2*I*Pi*ln(x)*csgn(I*
(x*exp(-(x-2)*exp(-9))-5))*csgn(I*exp((x-2)*exp(-9))*(x*exp(-(x-2)*exp(-9))-5))^2+5/2*I*Pi*ln(x)*csgn(I*exp((x
-2)*exp(-9)))*csgn(I*exp((x-2)*exp(-9))*(x*exp(-(x-2)*exp(-9))-5))^2-5/2*I*Pi*ln(x)*csgn(I*exp((x-2)*exp(-9))*
(x*exp(-(x-2)*exp(-9))-5))^3+25*ln(x)

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maxima [A]  time = 0.49, size = 35, normalized size = 1.52 \begin {gather*} 5 \, {\left (e^{9} \log \left (x e^{\left (2 \, e^{\left (-9\right )}\right )} - 5 \, e^{\left (x e^{\left (-9\right )}\right )}\right ) \log \relax (x) + {\left (5 \, e^{9} - 2\right )} \log \relax (x)\right )} e^{\left (-9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*log((x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x*exp(9)
*exp((2-x)/exp(9))-25*x)*log(x)+25*x*exp(9)*exp((2-x)/exp(9))-125*exp(9))*exp(log(log((x*exp((2-x)/exp(9))-5)/
exp((2-x)/exp(9)))+5)+log(log(x)))/((x^2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9))*log(x)*log((x*exp((2-x)/exp(9))-
5)/exp((2-x)/exp(9)))+(5*x^2*exp(9)*exp((2-x)/exp(9))-25*x*exp(9))*log(x)),x, algorithm="maxima")

[Out]

5*(e^9*log(x*e^(2*e^(-9)) - 5*e^(x*e^(-9)))*log(x) + (5*e^9 - 2)*log(x))*e^(-9)

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mupad [B]  time = 5.01, size = 21, normalized size = 0.91 \begin {gather*} 5\,\ln \relax (x)\,\left (\ln \left (x-5\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{-9}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-9}}\right )+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(log(x)) + log(log(exp(exp(-9)*(x - 2))*(x*exp(-exp(-9)*(x - 2)) - 5)) + 5))*(125*exp(9) + log(x)*
(25*x - 5*x*exp(-exp(-9)*(x - 2))*exp(9)) + log(exp(exp(-9)*(x - 2))*(x*exp(-exp(-9)*(x - 2)) - 5))*(25*exp(9)
- 5*x*exp(-exp(-9)*(x - 2))*exp(9)) - 25*x*exp(-exp(-9)*(x - 2))*exp(9)))/(log(x)*(25*x*exp(9) - 5*x^2*exp(-e
xp(-9)*(x - 2))*exp(9)) + log(exp(exp(-9)*(x - 2))*(x*exp(-exp(-9)*(x - 2)) - 5))*log(x)*(5*x*exp(9) - x^2*exp
(-exp(-9)*(x - 2))*exp(9))),x)

[Out]

5*log(x)*(log(x - 5*exp(-2*exp(-9))*exp(x*exp(-9))) + 5)

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sympy [A]  time = 0.66, size = 31, normalized size = 1.35 \begin {gather*} 5 \log {\relax (x )} \log {\left (\left (x e^{\frac {2 - x}{e^{9}}} - 5\right ) e^{- \frac {2 - x}{e^{9}}} \right )} + 25 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*ln((x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x*exp(9)*
exp((2-x)/exp(9))-25*x)*ln(x)+25*x*exp(9)*exp((2-x)/exp(9))-125*exp(9))*exp(ln(ln((x*exp((2-x)/exp(9))-5)/exp(
(2-x)/exp(9)))+5)+ln(ln(x)))/((x**2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9))*ln(x)*ln((x*exp((2-x)/exp(9))-5)/exp(
(2-x)/exp(9)))+(5*x**2*exp(9)*exp((2-x)/exp(9))-25*x*exp(9))*ln(x)),x)

[Out]

5*log(x)*log((x*exp((2 - x)*exp(-9)) - 5)*exp(-(2 - x)*exp(-9))) + 25*log(x)

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