3.27.9 \(\int \frac {-2-13 x+x^3-x^4+(2 x+x^2) \log (x)+(-5 x+x^2-x^3+x \log (x)) \log (e^{-x} (-60+12 x-12 x^2+12 \log (x)))}{(-5 x+x^2-x^3+x \log (x)) \log ^2(e^{-x} (-60+12 x-12 x^2+12 \log (x)))} \, dx\)

Optimal. Leaf size=24 \[ \frac {2+x}{\log \left (12 e^{-x} \left (-5+x-x^2+\log (x)\right )\right )} \]

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Rubi [F]  time = 2.37, 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 {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 - 13*x + x^3 - x^4 + (2*x + x^2)*Log[x] + (-5*x + x^2 - x^3 + x*Log[x])*Log[(-60 + 12*x - 12*x^2 + 12*
Log[x])/E^x])/((-5*x + x^2 - x^3 + x*Log[x])*Log[(-60 + 12*x - 12*x^2 + 12*Log[x])/E^x]^2),x]

[Out]

13*Defer[Int][1/((5 - x + x^2 - Log[x])*Log[(-12*(5 - x + x^2 - Log[x]))/E^x]^2), x] + 2*Defer[Int][1/(x*(5 -
x + x^2 - Log[x])*Log[(-12*(5 - x + x^2 - Log[x]))/E^x]^2), x] - Defer[Int][x^2/((5 - x + x^2 - Log[x])*Log[(-
12*(5 - x + x^2 - Log[x]))/E^x]^2), x] + Defer[Int][x^3/((5 - x + x^2 - Log[x])*Log[(-12*(5 - x + x^2 - Log[x]
))/E^x]^2), x] - 2*Defer[Int][Log[x]/((5 - x + x^2 - Log[x])*Log[(-12*(5 - x + x^2 - Log[x]))/E^x]^2), x] - De
fer[Int][(x*Log[x])/((5 - x + x^2 - Log[x])*Log[(-12*(5 - x + x^2 - Log[x]))/E^x]^2), x] + Defer[Int][Log[(-12
*(5 - x + x^2 - Log[x]))/E^x]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2+13 x-x^3+x^4-\left (2 x+x^2\right ) \log (x)-\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (5 x-x^2+x^3-x \log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx\\ &=\int \left (\frac {13}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}+\frac {2}{x \left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}-\frac {x^2}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}+\frac {x^3}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}-\frac {(2+x) \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}+\frac {1}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}\right ) \, dx\\ &=2 \int \frac {1}{x \left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+13 \int \frac {1}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \frac {x^2}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {x^3}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \frac {(2+x) \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {1}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx\\ &=2 \int \frac {1}{x \left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+13 \int \frac {1}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \left (\frac {2 \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}+\frac {x \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}\right ) \, dx-\int \frac {x^2}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {x^3}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {1}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx\\ &=2 \int \frac {1}{x \left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-2 \int \frac {\log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+13 \int \frac {1}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \frac {x^2}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {x^3}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \frac {x \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {1}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 26, normalized size = 1.08 \begin {gather*} \frac {2+x}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 13*x + x^3 - x^4 + (2*x + x^2)*Log[x] + (-5*x + x^2 - x^3 + x*Log[x])*Log[(-60 + 12*x - 12*x^2
+ 12*Log[x])/E^x])/((-5*x + x^2 - x^3 + x*Log[x])*Log[(-60 + 12*x - 12*x^2 + 12*Log[x])/E^x]^2),x]

[Out]

(2 + x)/Log[(-12*(5 - x + x^2 - Log[x]))/E^x]

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fricas [A]  time = 0.67, size = 30, normalized size = 1.25 \begin {gather*} \frac {x + 2}{\log \left (-12 \, {\left (x^{2} - x + 5\right )} e^{\left (-x\right )} + 12 \, e^{\left (-x\right )} \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x^3+x^2-5*x)*log((12*log(x)-12*x^2+12*x-60)/exp(x))+(x^2+2*x)*log(x)-x^4+x^3-13*x-2)/(x*l
og(x)-x^3+x^2-5*x)/log((12*log(x)-12*x^2+12*x-60)/exp(x))^2,x, algorithm="fricas")

[Out]

(x + 2)/log(-12*(x^2 - x + 5)*e^(-x) + 12*e^(-x)*log(x))

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giac [A]  time = 0.42, size = 26, normalized size = 1.08 \begin {gather*} -\frac {x + 2}{x - \log \left (12\right ) - \log \left (-x^{2} + x + \log \relax (x) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x^3+x^2-5*x)*log((12*log(x)-12*x^2+12*x-60)/exp(x))+(x^2+2*x)*log(x)-x^4+x^3-13*x-2)/(x*l
og(x)-x^3+x^2-5*x)/log((12*log(x)-12*x^2+12*x-60)/exp(x))^2,x, algorithm="giac")

[Out]

-(x + 2)/(x - log(12) - log(-x^2 + x + log(x) - 5))

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maple [C]  time = 0.15, size = 201, normalized size = 8.38

method	result	size
risch	\(-\frac {2 i \left (2+x \right )}{-2 \pi \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )-\pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right )\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )^{2}-\pi \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )^{3}+2 \pi -4 i \ln \relax (2)-2 i \ln \relax (3)-2 i \ln \left (-\ln \relax (x )-x +x^{2}+5\right )+2 i \ln \left ({\mathrm e}^{x}\right )}\)	\(201\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)-x^3+x^2-5*x)*ln((12*ln(x)-12*x^2+12*x-60)/exp(x))+(x^2+2*x)*ln(x)-x^4+x^3-13*x-2)/(x*ln(x)-x^3+x
^2-5*x)/ln((12*ln(x)-12*x^2+12*x-60)/exp(x))^2,x,method=_RETURNVERBOSE)

[Out]

-2*I*(2+x)/(-2*Pi*csgn(I*(ln(x)+x-x^2-5)*exp(-x))^2-Pi*csgn(I*(ln(x)+x-x^2-5))*csgn(I*exp(-x))*csgn(I*(ln(x)+x
-x^2-5)*exp(-x))-Pi*csgn(I*(ln(x)+x-x^2-5))*csgn(I*(ln(x)+x-x^2-5)*exp(-x))^2+Pi*csgn(I*exp(-x))*csgn(I*(ln(x)
+x-x^2-5)*exp(-x))^2-Pi*csgn(I*(ln(x)+x-x^2-5)*exp(-x))^3+2*Pi-4*I*ln(2)-2*I*ln(3)-2*I*ln(-ln(x)-x+x^2+5)+2*I*
ln(exp(x)))

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maxima [A]  time = 0.75, size = 30, normalized size = 1.25 \begin {gather*} -\frac {x + 2}{x - \log \relax (3) - 2 \, \log \relax (2) - \log \left (-x^{2} + x + \log \relax (x) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x^3+x^2-5*x)*log((12*log(x)-12*x^2+12*x-60)/exp(x))+(x^2+2*x)*log(x)-x^4+x^3-13*x-2)/(x*l
og(x)-x^3+x^2-5*x)/log((12*log(x)-12*x^2+12*x-60)/exp(x))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.82, size = 26, normalized size = 1.08 \begin {gather*} \frac {x+2}{\ln \left ({\mathrm {e}}^{-x}\,\left (12\,x+12\,\ln \relax (x)-12\,x^2-60\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((13*x + log(exp(-x)*(12*x + 12*log(x) - 12*x^2 - 60))*(5*x - x*log(x) - x^2 + x^3) - log(x)*(2*x + x^2) -
x^3 + x^4 + 2)/(log(exp(-x)*(12*x + 12*log(x) - 12*x^2 - 60))^2*(5*x - x*log(x) - x^2 + x^3)),x)

[Out]

(x + 2)/log(exp(-x)*(12*x + 12*log(x) - 12*x^2 - 60))

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sympy [A]  time = 0.56, size = 22, normalized size = 0.92 \begin {gather*} \frac {x + 2}{\log {\left (\left (- 12 x^{2} + 12 x + 12 \log {\relax (x )} - 60\right ) e^{- x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)-x**3+x**2-5*x)*ln((12*ln(x)-12*x**2+12*x-60)/exp(x))+(x**2+2*x)*ln(x)-x**4+x**3-13*x-2)/(x
*ln(x)-x**3+x**2-5*x)/ln((12*ln(x)-12*x**2+12*x-60)/exp(x))**2,x)

[Out]

(x + 2)/log((-12*x**2 + 12*x + 12*log(x) - 60)*exp(-x))

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