\(\int \frac {-2-13 x+x^3-x^4+(2 x+x^2) \log (x)+(-5 x+x^2-x^3+x \log (x)) \log (e^{-x} (-60+12 x-12 x^2+12 \log (x)))}{(-5 x+x^2-x^3+x \log (x)) \log ^2(e^{-x} (-60+12 x-12 x^2+12 \log (x)))} \, dx\) [2609]

Optimal result

Integrand size = 103, antiderivative size = 24 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {2+x}{\log \left (12 e^{-x} \left (-5+x-x^2+\log (x)\right )\right )} \]

[Out]

(2+x)/ln(12*(ln(x)+x-x^2-5)/exp(x))

Rubi [F]

\[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {2+x}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \]

[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]

Maple [A] (verified)

Time = 1.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(-\frac {-40 x -80}{40 \ln \left (12 \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}\)	\(27\)
risch	\(\frac {4+2 x}{2 \ln \left (3\right )+4 \ln \left (2\right )+2 i \pi -2 \ln \left ({\mathrm e}^{x}\right )+2 \ln \left (-\ln \left (x \right )-x +x^{2}+5\right )-i \pi \,\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right )\right ) {\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right )\right ) \operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )+i \pi {\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )-2 i \pi {\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}^{2}-i \pi {\operatorname {csgn}\left (i \left (\ln \left (x \right )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )}^{3}}\)	\(203\)

[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]

-1/40*(-40*x-80)/ln(12*(ln(x)+x-x^2-5)/exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {x + 2}{\log \left (-12 \, {\left (x^{2} - x + 5\right )} e^{\left (-x\right )} + 12 \, e^{\left (-x\right )} \log \left (x\right )\right )} \]

[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))

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {x + 2}{\log {\left (\left (- 12 x^{2} + 12 x + 12 \log {\left (x \right )} - 60\right ) e^{- x} \right )}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=-\frac {x + 2}{x - \log \left (3\right ) - 2 \, \log \left (2\right ) - \log \left (-x^{2} + x + \log \left (x\right ) - 5\right )} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=-\frac {x + 2}{x - \log \left (12\right ) - \log \left (-x^{2} + x + \log \left (x\right ) - 5\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 8.85 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )} \, dx=\frac {x+2}{\ln \left ({\mathrm {e}}^{-x}\,\left (12\,x+12\,\ln \left (x\right )-12\,x^2-60\right )\right )} \]

[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))