3.82.66 \(\int \frac {e^{5+x} (4-4 \log (x-\log (x))) (5-6 x+x^3+(4 x^2-3 x^3-x^4+(-4 x+3 x^2+x^3) \log (x)) \log (x-\log (x))+(-4 x^2+3 x^3+x^4+(4 x-3 x^2-x^3) \log (x)) \log ^2(x-\log (x)))}{\log (x-\log (x)) ((x^2-x \log (x)) \log (x-\log (x))+(-x^2+x \log (x)) \log ^2(x-\log (x)))} \, dx\)

Optimal. Leaf size=29 \[ e^{5+x} \left (5-x-x^2\right ) \left (-4+\frac {4}{\log (x-\log (x))}\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 6.17, 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^{5+x} (4-4 \log (x-\log (x))) \left (5-6 x+x^3+\left (4 x^2-3 x^3-x^4+\left (-4 x+3 x^2+x^3\right ) \log (x)\right ) \log (x-\log (x))+\left (-4 x^2+3 x^3+x^4+\left (4 x-3 x^2-x^3\right ) \log (x)\right ) \log ^2(x-\log (x))\right )}{\log (x-\log (x)) \left (\left (x^2-x \log (x)\right ) \log (x-\log (x))+\left (-x^2+x \log (x)\right ) \log ^2(x-\log (x))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(5 + x)*(4 - 4*Log[x - Log[x]])*(5 - 6*x + x^3 + (4*x^2 - 3*x^3 - x^4 + (-4*x + 3*x^2 + x^3)*Log[x])*Lo
g[x - Log[x]] + (-4*x^2 + 3*x^3 + x^4 + (4*x - 3*x^2 - x^3)*Log[x])*Log[x - Log[x]]^2))/(Log[x - Log[x]]*((x^2
- x*Log[x])*Log[x - Log[x]] + (-x^2 + x*Log[x])*Log[x - Log[x]]^2)),x]

[Out]

-20*E^(5 + x) + 4*E^(5 + x)*x + 4*E^(5 + x)*x^2 - 24*Defer[Int][E^(5 + x)/((x - Log[x])*Log[x - Log[x]]^2), x]
+ 20*Defer[Int][E^(5 + x)/(x*(x - Log[x])*Log[x - Log[x]]^2), x] + 4*Defer[Int][(E^(5 + x)*x^2)/((x - Log[x])
*Log[x - Log[x]]^2), x] + 16*Defer[Int][E^(5 + x)/Log[x - Log[x]], x] - 12*Defer[Int][(E^(5 + x)*x)/Log[x - Lo
g[x]], x] - 4*Defer[Int][(E^(5 + x)*x^2)/Log[x - Log[x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^{5+x} (1-x) \left (5-x-x^2+x (4+x) (x-\log (x)) \log (x-\log (x))-x (4+x) (x-\log (x)) \log ^2(x-\log (x))\right )}{x (x-\log (x)) \log ^2(x-\log (x))} \, dx\\ &=4 \int \frac {e^{5+x} (1-x) \left (5-x-x^2+x (4+x) (x-\log (x)) \log (x-\log (x))-x (4+x) (x-\log (x)) \log ^2(x-\log (x))\right )}{x (x-\log (x)) \log ^2(x-\log (x))} \, dx\\ &=4 \int \left (e^{5+x} (-1+x) (4+x)+\frac {e^{5+x} \left (5-6 x+x^3\right )}{x (x-\log (x)) \log ^2(x-\log (x))}-\frac {e^{5+x} (-1+x) (4+x)}{\log (x-\log (x))}\right ) \, dx\\ &=4 \int e^{5+x} (-1+x) (4+x) \, dx+4 \int \frac {e^{5+x} \left (5-6 x+x^3\right )}{x (x-\log (x)) \log ^2(x-\log (x))} \, dx-4 \int \frac {e^{5+x} (-1+x) (4+x)}{\log (x-\log (x))} \, dx\\ &=4 \int \left (-4 e^{5+x}+3 e^{5+x} x+e^{5+x} x^2\right ) \, dx+4 \int \left (-\frac {6 e^{5+x}}{(x-\log (x)) \log ^2(x-\log (x))}+\frac {5 e^{5+x}}{x (x-\log (x)) \log ^2(x-\log (x))}+\frac {e^{5+x} x^2}{(x-\log (x)) \log ^2(x-\log (x))}\right ) \, dx-4 \int \left (-\frac {4 e^{5+x}}{\log (x-\log (x))}+\frac {3 e^{5+x} x}{\log (x-\log (x))}+\frac {e^{5+x} x^2}{\log (x-\log (x))}\right ) \, dx\\ &=4 \int e^{5+x} x^2 \, dx+4 \int \frac {e^{5+x} x^2}{(x-\log (x)) \log ^2(x-\log (x))} \, dx-4 \int \frac {e^{5+x} x^2}{\log (x-\log (x))} \, dx+12 \int e^{5+x} x \, dx-12 \int \frac {e^{5+x} x}{\log (x-\log (x))} \, dx-16 \int e^{5+x} \, dx+16 \int \frac {e^{5+x}}{\log (x-\log (x))} \, dx+20 \int \frac {e^{5+x}}{x (x-\log (x)) \log ^2(x-\log (x))} \, dx-24 \int \frac {e^{5+x}}{(x-\log (x)) \log ^2(x-\log (x))} \, dx\\ &=-16 e^{5+x}+12 e^{5+x} x+4 e^{5+x} x^2+4 \int \frac {e^{5+x} x^2}{(x-\log (x)) \log ^2(x-\log (x))} \, dx-4 \int \frac {e^{5+x} x^2}{\log (x-\log (x))} \, dx-8 \int e^{5+x} x \, dx-12 \int e^{5+x} \, dx-12 \int \frac {e^{5+x} x}{\log (x-\log (x))} \, dx+16 \int \frac {e^{5+x}}{\log (x-\log (x))} \, dx+20 \int \frac {e^{5+x}}{x (x-\log (x)) \log ^2(x-\log (x))} \, dx-24 \int \frac {e^{5+x}}{(x-\log (x)) \log ^2(x-\log (x))} \, dx\\ &=-28 e^{5+x}+4 e^{5+x} x+4 e^{5+x} x^2+4 \int \frac {e^{5+x} x^2}{(x-\log (x)) \log ^2(x-\log (x))} \, dx-4 \int \frac {e^{5+x} x^2}{\log (x-\log (x))} \, dx+8 \int e^{5+x} \, dx-12 \int \frac {e^{5+x} x}{\log (x-\log (x))} \, dx+16 \int \frac {e^{5+x}}{\log (x-\log (x))} \, dx+20 \int \frac {e^{5+x}}{x (x-\log (x)) \log ^2(x-\log (x))} \, dx-24 \int \frac {e^{5+x}}{(x-\log (x)) \log ^2(x-\log (x))} \, dx\\ &=-20 e^{5+x}+4 e^{5+x} x+4 e^{5+x} x^2+4 \int \frac {e^{5+x} x^2}{(x-\log (x)) \log ^2(x-\log (x))} \, dx-4 \int \frac {e^{5+x} x^2}{\log (x-\log (x))} \, dx-12 \int \frac {e^{5+x} x}{\log (x-\log (x))} \, dx+16 \int \frac {e^{5+x}}{\log (x-\log (x))} \, dx+20 \int \frac {e^{5+x}}{x (x-\log (x)) \log ^2(x-\log (x))} \, dx-24 \int \frac {e^{5+x}}{(x-\log (x)) \log ^2(x-\log (x))} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.70, size = 31, normalized size = 1.07 \begin {gather*} \frac {4 e^{5+x} \left (-5+x+x^2\right ) (-1+\log (x-\log (x)))}{\log (x-\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + x)*(4 - 4*Log[x - Log[x]])*(5 - 6*x + x^3 + (4*x^2 - 3*x^3 - x^4 + (-4*x + 3*x^2 + x^3)*Log[
x])*Log[x - Log[x]] + (-4*x^2 + 3*x^3 + x^4 + (4*x - 3*x^2 - x^3)*Log[x])*Log[x - Log[x]]^2))/(Log[x - Log[x]]
*((x^2 - x*Log[x])*Log[x - Log[x]] + (-x^2 + x*Log[x])*Log[x - Log[x]]^2)),x]

[Out]

(4*E^(5 + x)*(-5 + x + x^2)*(-1 + Log[x - Log[x]]))/Log[x - Log[x]]

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 33, normalized size = 1.14 \begin {gather*} -{\left (x^{2} + x - 5\right )} e^{\left (x + \log \left (-\frac {4 \, {\left (\log \left (x - \log \relax (x)\right ) - 1\right )}}{\log \left (x - \log \relax (x)\right )}\right ) + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-3*x^2+4*x)*log(x)+x^4+3*x^3-4*x^2)*log(x-log(x))^2+((x^3+3*x^2-4*x)*log(x)-x^4-3*x^3+4*x^2)*
log(x-log(x))+x^3-6*x+5)*exp(log((-4*log(x-log(x))+4)/log(x-log(x)))+5+x)/((x*log(x)-x^2)*log(x-log(x))^2+(-x*
log(x)+x^2)*log(x-log(x))),x, algorithm="fricas")

[Out]

-(x^2 + x - 5)*e^(x + log(-4*(log(x - log(x)) - 1)/log(x - log(x))) + 5)

________________________________________________________________________________________

giac [B]  time = 0.25, size = 75, normalized size = 2.59 \begin {gather*} \frac {4 \, {\left (x^{2} e^{\left (x + 5\right )} \log \left (x - \log \relax (x)\right ) - x^{2} e^{\left (x + 5\right )} + x e^{\left (x + 5\right )} \log \left (x - \log \relax (x)\right ) - x e^{\left (x + 5\right )} - 5 \, e^{\left (x + 5\right )} \log \left (x - \log \relax (x)\right ) + 5 \, e^{\left (x + 5\right )}\right )}}{\log \left (x - \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-3*x^2+4*x)*log(x)+x^4+3*x^3-4*x^2)*log(x-log(x))^2+((x^3+3*x^2-4*x)*log(x)-x^4-3*x^3+4*x^2)*
log(x-log(x))+x^3-6*x+5)*exp(log((-4*log(x-log(x))+4)/log(x-log(x)))+5+x)/((x*log(x)-x^2)*log(x-log(x))^2+(-x*
log(x)+x^2)*log(x-log(x))),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.36, size = 245, normalized size = 8.45

method	result	size
risch	\(\frac {4 \left (\ln \left (x -\ln \relax (x )\right ) x^{2}-x^{2}+x \ln \left (x -\ln \relax (x )\right )-x -5 \ln \left (x -\ln \relax (x )\right )+5\right ) {\mathrm e}^{5+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (\ln \left (x -\ln \relax (x )\right )-1\right )}{\ln \left (x -\ln \relax (x )\right )}\right )^{3}}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (\ln \left (x -\ln \relax (x )\right )-1\right )}{\ln \left (x -\ln \relax (x )\right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \left (x -\ln \relax (x )\right )}\right )}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (\ln \left (x -\ln \relax (x )\right )-1\right )}{\ln \left (x -\ln \relax (x )\right )}\right )^{2} \mathrm {csgn}\left (i \left (\ln \left (x -\ln \relax (x )\right )-1\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (\ln \left (x -\ln \relax (x )\right )-1\right )}{\ln \left (x -\ln \relax (x )\right )}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (x -\ln \relax (x )\right )}\right ) \mathrm {csgn}\left (i \left (\ln \left (x -\ln \relax (x )\right )-1\right )\right )}{2}-i \pi \mathrm {csgn}\left (\frac {i \left (\ln \left (x -\ln \relax (x )\right )-1\right )}{\ln \left (x -\ln \relax (x )\right )}\right )^{2}+x}}{\ln \left (x -\ln \relax (x )\right )}\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^3-3*x^2+4*x)*ln(x)+x^4+3*x^3-4*x^2)*ln(x-ln(x))^2+((x^3+3*x^2-4*x)*ln(x)-x^4-3*x^3+4*x^2)*ln(x-ln(x)
)+x^3-6*x+5)*exp(ln((-4*ln(x-ln(x))+4)/ln(x-ln(x)))+5+x)/((x*ln(x)-x^2)*ln(x-ln(x))^2+(-x*ln(x)+x^2)*ln(x-ln(x
))),x,method=_RETURNVERBOSE)

[Out]

4*(ln(x-ln(x))*x^2-x^2+x*ln(x-ln(x))-x-5*ln(x-ln(x))+5)/ln(x-ln(x))*exp(5+1/2*I*Pi*csgn(I/ln(x-ln(x))*(ln(x-ln
(x))-1))^3+1/2*I*Pi*csgn(I/ln(x-ln(x))*(ln(x-ln(x))-1))^2*csgn(I/ln(x-ln(x)))+1/2*I*Pi*csgn(I/ln(x-ln(x))*(ln(
x-ln(x))-1))^2*csgn(I*(ln(x-ln(x))-1))-1/2*I*Pi*csgn(I/ln(x-ln(x))*(ln(x-ln(x))-1))*csgn(I/ln(x-ln(x)))*csgn(I
*(ln(x-ln(x))-1))-I*Pi*csgn(I/ln(x-ln(x))*(ln(x-ln(x))-1))^2+x)

________________________________________________________________________________________

maxima [B]  time = 0.43, size = 56, normalized size = 1.93 \begin {gather*} \frac {4 \, {\left ({\left (x^{2} e^{5} + x e^{5} - 5 \, e^{5}\right )} e^{x} \log \left (x - \log \relax (x)\right ) - {\left (x^{2} e^{5} + x e^{5} - 5 \, e^{5}\right )} e^{x}\right )}}{\log \left (x - \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-3*x^2+4*x)*log(x)+x^4+3*x^3-4*x^2)*log(x-log(x))^2+((x^3+3*x^2-4*x)*log(x)-x^4-3*x^3+4*x^2)*
log(x-log(x))+x^3-6*x+5)*exp(log((-4*log(x-log(x))+4)/log(x-log(x)))+5+x)/((x*log(x)-x^2)*log(x-log(x))^2+(-x*
log(x)+x^2)*log(x-log(x))),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{x+\ln \left (-\frac {4\,\ln \left (x-\ln \relax (x)\right )-4}{\ln \left (x-\ln \relax (x)\right )}\right )+5}\,\left (\ln \left (x-\ln \relax (x)\right )\,\left (\ln \relax (x)\,\left (x^3+3\,x^2-4\,x\right )+4\,x^2-3\,x^3-x^4\right )-6\,x-{\ln \left (x-\ln \relax (x)\right )}^2\,\left (\ln \relax (x)\,\left (x^3+3\,x^2-4\,x\right )+4\,x^2-3\,x^3-x^4\right )+x^3+5\right )}{\ln \left (x-\ln \relax (x)\right )\,\left (x\,\ln \relax (x)-x^2\right )-{\ln \left (x-\ln \relax (x)\right )}^2\,\left (x\,\ln \relax (x)-x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + log(-(4*log(x - log(x)) - 4)/log(x - log(x))) + 5)*(log(x - log(x))*(log(x)*(3*x^2 - 4*x + x^3)
+ 4*x^2 - 3*x^3 - x^4) - 6*x - log(x - log(x))^2*(log(x)*(3*x^2 - 4*x + x^3) + 4*x^2 - 3*x^3 - x^4) + x^3 + 5)
)/(log(x - log(x))*(x*log(x) - x^2) - log(x - log(x))^2*(x*log(x) - x^2)),x)

[Out]

int(-(exp(x + log(-(4*log(x - log(x)) - 4)/log(x - log(x))) + 5)*(log(x - log(x))*(log(x)*(3*x^2 - 4*x + x^3)
+ 4*x^2 - 3*x^3 - x^4) - 6*x - log(x - log(x))^2*(log(x)*(3*x^2 - 4*x + x^3) + 4*x^2 - 3*x^3 - x^4) + x^3 + 5)
)/(log(x - log(x))*(x*log(x) - x^2) - log(x - log(x))^2*(x*log(x) - x^2)), x)

________________________________________________________________________________________

sympy [B]  time = 0.57, size = 51, normalized size = 1.76 \begin {gather*} \frac {\left (4 x^{2} \log {\left (x - \log {\relax (x )} \right )} - 4 x^{2} + 4 x \log {\left (x - \log {\relax (x )} \right )} - 4 x - 20 \log {\left (x - \log {\relax (x )} \right )} + 20\right ) e^{x + 5}}{\log {\left (x - \log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**3-3*x**2+4*x)*ln(x)+x**4+3*x**3-4*x**2)*ln(x-ln(x))**2+((x**3+3*x**2-4*x)*ln(x)-x**4-3*x**3+4
*x**2)*ln(x-ln(x))+x**3-6*x+5)*exp(ln((-4*ln(x-ln(x))+4)/ln(x-ln(x)))+5+x)/((x*ln(x)-x**2)*ln(x-ln(x))**2+(-x*
ln(x)+x**2)*ln(x-ln(x))),x)

[Out]

(4*x**2*log(x - log(x)) - 4*x**2 + 4*x*log(x - log(x)) - 4*x - 20*log(x - log(x)) + 20)*exp(x + 5)/log(x - log
(x))

________________________________________________________________________________________